Mathematics and Such

Topology: Product Spaces (II)

Posted on February 16, 2013 by limsup

The Box Topology

Following an earlier article on products of two topological spaces, we’ll now talk about a product of possibly infinitely many topological spaces. Suppose $\{X_i : i\in I\}$ is a collection of topological spaces indexed by I, and we wish to define a topology on the set $X := \prod_{i\in I} X_i.$ What would be a good choice?

If we follow our instincts, the most natural definition would be:

Definition. The box topology on $X = \prod_i X_i$ is defined by the basis:

$\Sigma = \{\prod_{i\in I} U_i : \text{ each } U_i\subseteq X_i \text{ is open }\}.$

Since $(\prod_i U_i) \cap (\prod_i V_i) = \prod_i (U_i\cap V_i),$ our collection Σ is indeed a basis, so the definition makes sense. The only problem is that its behaviour has some anomalies.

Problem 1 : As Terence Tao pointed out in his blog post, if we consider the space $X = \mathbf{R}^\mathbf{N}$ of all sequences of real numbers, then there are sequences which one would expect to converge but fail to under the box topology. For example:

$x_1 = (e^{-n})_{n\ge 1}, \ x_2 = (e^{-2n})_{n\ge 1},\ x_3 = (e^{-3n})_{n\ge 1}, \ldots$

Reasonably, one would expect the x_i‘s to converge to (0, 0, … ), but if we pick the open subset

$U=\prod_{m=1}^\infty U_m,$ where $\ U_m = (-e^{-m^2}, +e^{-m^2}),$

then no term of the sequence actually lies in U. The problem is that X has too many open subsets.

Problem 2 : Let I be any index set and X be a topological space. The diagonal map $\Delta : X\to X^I$ takes $x\in X$ to the constant tuple $(x_i : x_i=x)_{i\in I}.$ It turns out Δ is not continuous in general if we endow $X^I$ with the box topology! Indeed, the open subset $\prod_i U_i$ for open $U_i\subseteq X_i$ gives

$\Delta^{-1}(\prod_{i\in I} U_i) = \cap_{i \in I} U_i,$

which is not open in X in general if I is infinite. The fact that such a natural map is non-continuous makes things rather awkward.

The Product Topology

Let’s go back to the drawing board and decide what kind of subsets of $X=\prod_i X_i$ can be open. Firstly, we most definitely want the projection maps

$\pi_i : X \to X_i, \ (x_i)_{i\in I} \mapsto x_i,$

to be continuous. This means for any open subset $U_i\subseteq X_i,$ the “open slice” given by $\pi_i^{-1}(U_i) = (\prod_{j\ne i} X_j) \times U_i$ is an open subset of X. Let’s define the coarsest topology for which this is true.

Definition. The product topology on X is given by the subbasis of “open slices”:

$\{ (\prod_{j\ne i} X_j)\times U_i : U_i\subseteq X_i \text{ open }, i\in I\}.$

[ Open slices of X × Y × Z ]

Taking intersections of finitely many elements of this subbasis, we get the basis:

$\Sigma = \{ \prod_{i\in I} U_i : U_i\subseteq X_i \text{ open, } X_i\ne U_i \text{ for only finitely many } i\}.$

Note that for a finite product $X_1 \times X_2 \times\ldots\times X_n,$ the box topology and the product topology are identical.

Taking the complement of an “open slice” gives the “closed slice”

$C = (\prod_{j\ne i} X_j) \times C_i,\$ for a closed subset $\ C_i\subseteq X_i.$

Now if $C_i\subseteq X_i$ is a collection of closed subsets, then the product $\prod_i C_i$ is an intersection of the “closed slices”, so we get:

Proposition 1. The product $\prod_i C_i$ of closed subsets is closed in the product topology.

Since the box topology is an even finer topology, $\prod_i C_i$ is also closed in the box topology.

Next, we’ll proceed to give some nice properties of the product topology. Hopefully, these will convince you that we’ve made the “right” choice.

Properties of Product Topology

We’ll give the most important property first.

Universal Property of the Product. Let $X=\prod_i X_i$ with projection maps $\pi_i : X\to X_i.$ Now, for any topological space Z and function $f:Z\to X,$

f is continuous at $z\in Z$ if and only if $\pi_i\circ f:Z\to X_i$ is continuous at z, for each i.

Proof.

The forward direction is obvious since each $\pi_i$ is continuous.

For the converse, we need to show that for any open subset V of X containing f(x), $f^{-1}(V)$ contains an open subset U which contains z. If we fix a subbasis of X, we may assume V belongs to this subbasis. Hence let $V=\pi_i^{-1}(U_i)$ for some open subset $U_i\subseteq X_i$ and index i. This gives:

$f^{-1}(V) = f^{-1}\pi_i^{-1}(U_i) = (\pi_i\circ f)^{-1}(U_i),$

which indeed contains an open subset containing z since $\pi_i\circ f$ is continuous. ♦

In particular, we have:

Universal Property of the Product (II) : if Z is a topological space and f:Z → X is a map, then f is continuous if and only if π_if is continuous for each index i.

In examining the proof of the universal property, the reader may suspect that this property uniquely defines the product topology, and he’d be right!

In what follows, we’ll list some results which can be proven by invoking the universal property.

Applications of the Universal Property

We’ll use the above universal property of the product topology to prove a variety of results.

Proposition 2. The diagonal map $\Delta : X\to X^I$ is a homeomorphism onto its image.

Proof.

Continuity: by the universal property, it suffices to show that composing with each projection map gives a continuous map $\pi_i \circ \Delta : X\to X$ for each i. But this map is simply the identity, which is clearly continuous.

To show that the inverse map Δ(X) → X is continuous, note that it is simply the restriction of a projection map $\pi_i : \prod_i X_i \to X_i$ to Δ(X). ♦

Proposition 3. Let $f_i : X_i \to Y_i$ be a collection of continuous maps indexed by i. Then the map $f:\prod_i X_i \to \prod_i Y_i$ which takes:

$(x_i)_{i\in I} \mapsto (f(x_i))_{i\in I},$

is continuous.

Proof

Let $X = \prod_i X_i, Y = \prod_i Y_i$ and $\pi_i : X\to X_i, p_i : Y\to Y_i$ be the projection maps.

To prove f is continuous, the universal property tells us it suffices to show $p_i\circ f$ is continuous for each i. But then $p_i\circ f = f_i\circ \pi_i$ so it is indeed continuous. ♦

Proposition 4. Let J be a directed set and $(x_i)_j$ be a net in $\prod_{i\in I} X_i$ indexed by $j\in J.$ Then

the net converges converges to $(a_i)\in \prod_i X_i$ if and only if for each fixed i, the net $x_{ij} \to a_i$ converges in X_i.

[ Feel free to substitute “net” with “sequence” if you’re not too comfortable with it. ]

Proof.

As we noted earlier, $(x_i)_j \to (a_i)$ if and only if the map

$f:J^* \to \prod_i X_i,\$ which takes $j \mapsto (x_i)_j,\ \infty \mapsto (a_i),$

is continuous at ∞. By the universal property, this is true if and only if for each i, the composition $\pi_i\circ f : J^* \to X_i$ is continuous. But this map takes $j\mapsto x_{ij},\ \infty\mapsto a_i,$ which is continuous at ∞ if and only if $(x_{ij})\to a_i\in X_i$ for each i. ♦

Application: consider the earlier sequence $(x_i)_j = e^{-ij}$ for positive integers i, j. If we fix i, then as j → ∞, $(x_i)_j \to 0.$ Hence, $(x_i)_j \to (0, 0, \ldots).$

Proposition 5. Suppose we have a collection of collections of topological spaces $\{X_{ij}\},$ where the index set J(i) for j may depend on i. Then:

$\prod_{i\in I} \left(\prod_{j\in J(i)} X_{ij}\right) = \prod_{i\in I, j\in J(i)} X_{ij},$

as topological spaces.

Proof.

Note that set-theoretically, there’s a natural bijection between the two sides. Let f : LHS → RHS and g : RHS → LHS be this bijection and its inverse. We need to show that they’re both continuous.

There are three collections of projection maps, all of which are continuous.

$\pi_{ij} : (\prod_{i,j} X_{ij}) \to X_{ij},\quad \pi_j : (\prod_j X_{ij})\to X_{ij},\quad \pi_i : \prod_i (\prod_j X_{ij})\to (\prod_j X_{ij}),$

for various $i\in I, j\in J(i).$

To show that f is continuous, we need to show $\pi_{ij}\circ f$ is continuous for any i, j. But this map is simply $\pi_j\circ \pi_i,$ so it is indeed continuous.
To show that g is continuous, we need to show $\pi_i\circ g$ is continuous for each i. For that, we need to show that $\pi_j\circ\pi_i \circ g$ is continuous for each $i\in I, j\in J(i).$ But this is simply $\pi_{ij}$ so it is indeed continuous. ♦

Exercise (on Universal Properties)

Prove the following universal property for subspace. Suppose $Y\subseteq X$ gets the subspace topology from X. Let $i:Y\hookrightarrow X$ be the inclusion map. Then:

for any topological space Z, a function f : Z → Y is continuous if and only if $i\circ f : Z\to X$ is continuous.

Suppose $Y_i\subseteq X_i$ is a collection of subsets of topological spaces. There are two ways to form a topology on $\prod_i Y_i:$

take the subspace topology on each $Y_i\subseteq X_i$ and take the product topology $\prod_i Y_i$ or
take the product topology on $\prod_i X_i$ then take the subspace topology $\prod_i Y_i\subseteq \prod_i X_i.$

Prove that both constructions give the same topology, using the universal properties of the product and the subspace topologies.

Interior and Closure

Next, we attempt to generalise the prior two articles on interior/closure of products.

Proposition 6. If $Y_i\subseteq X_i$ is a collection of subsets of topological spaces, then

$\text{int}(\prod_i Y_i) \ne \prod_i \text{int}(Y_i)$ in general;

$\text{cl}(\prod_i Y_i) = \prod_i \text{cl}(Y_i).$

Proof.

For the first property, suppose each Y_i is open in X_i. Then the RHS is $\prod_i \text{int}(Y_i) = \prod_i Y_i$ which is not open in general unless Y_i= X_i for all but finitely many i. Hence it is not equal to the LHS.

For the second property, the RHS is a closed subset containing $\prod_i Y_i$ so it must also contain $\text{cl}(\prod_i Y_i).$ For the reverse inclusion, if $(x_i)\not\in \text{LHS}$ then it does not lie in $\prod_i Y_i$ and is not a point of accumulation for it. Thus, we can pick a basic open subset:

$(x_i)\in \prod_i U_i\subseteq \prod_i X_i,\$ where U_i=X_i except for finitely many i,

such that $(\prod_i U_i) \cap (\prod_i Y_i) = \emptyset.$ Thus for some i, $U_i\cap Y_i = \emptyset$ and x_i is not a point of accumulation of Y_i. Furthermore, since $x_i\in U_i,$ we have $x_i\not\in Y_i$ so

$x_i \not \in Y_i \cup Y_i^{acc} = \text{cl}(Y_i).$ ♦

Exercise

Prove that in the box topology, we do have:

$\text{int}(\prod_i Y_i) = \prod_i \text{int}(Y_i)\$ and $\ \text{cl}(\prod_i Y_i) =\prod_i \text{cl}(Y_i).$

Thus there’s at least one aspect of the box topology which triumphs the product topology.

Product of Metric Spaces

Finally, we ask: if each $(X_i, d_i)$ is a metric space, is the topological product necessarily metrisable also? For finite products, we knew this to be true. It turns out for countably infinite products $X = \prod_{i=1}^\infty X_i$ this is still true.

The first step is to replace each d_i with a bounded metric. Recall that for any metric space (X, d), replacing d with $d'(x, y) = \frac{d(x,y)}{1 + d(x,y)}$ gives a metric d’ which is topologically equivalent and bounded by 1. Multiplying by a suitable constant, we ensure that each d_i is bounded by 2^–i for i = 1, 2, 3, … and define a metric on $X = \prod_{i=1}^\infty X_i$ by:

$d( (x_i), (y_i)) := \sum_{i=1}^\infty d_i(x_i, y_i).$

It’s not hard to check that d is a metric bounded by 1.

The only question is whether the induced topology is identical to the product topology.

First, let’s consider each open slice: $U=(\prod_{j\ne i} X_j) \times U_i$ where $U_i\subseteq X_i$ is open. If $a=(a_i)$ is in this slice, then $a_i \in U_i$ so there’s an open ball $N(a_i, \epsilon) \subseteq U_i.$ Now it’s clear that the open ball in X : $N(a, \epsilon) \subseteq U.$ So U is indeed open in the metric topology.

Conversely, consider an open ball N(a, ε) in the metric topology, where $a=(a_i)$ as before. Pick N such that 2^–N < ε/2. Then the set:

$N(a_1, \frac\epsilon {2N}) \times N(a_2, \frac\epsilon {2N}) \times \ldots\times N(a_N, \frac\epsilon {2N}) \times X_{N+1} \times X_{N+2} \times\ldots$

is open in the product topology, and the distance between a and any point in this set is bounded by $N\cdot \frac\epsilon{2N} + 2^{-N-1} + 2^{-N-2} + \ldots = \frac\epsilon 2 + 2^{-N} < \epsilon.$ Thus this set is contained in the open ball N(a, ε).

Posted in Notes | Tagged advanced, box topology, closures, interiors, metrisable topology, points of accumulation, product topology, subspaces, topology, universal properties | Leave a comment

Topology: Interior

Posted on February 13, 2013 by limsup

Let Y be a subset of a topological space X. In the previous article, we defined the closure of Y as the smallest closed subset of X containing Y. Dually, we shall now define the interior of Y to be the largest open subset contained in it. The construction is similar.

Let Σ be the collection of all open $U\subseteq X$ contained in Y.
Since $\emptyset \in\Sigma,$ we see that Σ is not empty.
Hence, it makes sense to define the interior of Y as the union of all U in Σ.

Thus, $\text{int}(Y) = \cup_{U\in\Sigma} U.$ As before, the interior satisfies the following.

$\text{int}(Y)\in\Sigma$ : since a union of open subsets is open, so int(Y) must be open;
if $U\in\Sigma,$ then since int(Y) is a union of many sets including U, we must have $U\subseteq \text{int}(Y).$

This justifies describing int(Y) as the largest open subset contained in Y.

As an immediate property:

Basic Fact. If $Y\subseteq Z$ are subsets of X, then $\text{int}(Y)\subseteq \text{int}(Z).$

Now int(Y) is an open subset contained in Y and hence Z. Thus, int(Y) must be contained in int(Z).

Examples

If Y is open in X, then int(Y) = Y.
Take the half-open interval $Y=[0, 1)\subseteq \mathbf{R}.$ The interior is (0, 1).
The the set of rationals Y = Q in X = R. The interior is empty.
Suppose Y is the set of even positive integers in N*, together with ∞. What is the interior of Y? [ Answer: empty set. ]

Suppose X is a metric space. Now the open ball N(a, ε) is always open, but the interior of the closed ball N(a, ε)* is not the open ball N(a, ε) in general. For example, suppose X is the subspace [-1, 1] of R. Then the closed ball Y := N(0, 1)* is already open, so int(Y) = Y, whereas the open ball N(0, 1) is a proper subset of Y.

Properties of Interior

One can relate the interior of Y with the closure of the complement of Y.

Proposition 1. For any subset $Y\subseteq X$ , we have:

$\text{int}(Y) = X - \text{cl}(X-Y).$

Proof.

Now int(Y) is the union of all open U contained in Y. Thus its complement is the intersection of all closed C which contains Y. This gives $X-\text{int}(Y) = \text{cl}(X-Y),$ which is exactly what we desired to prove. ♦

Proposition 2. If $Y, Z\subseteq X$ are subsets, then

$\text{int}(Y\cap Z) = \text{int}(Y)\cap\text{int}(Z).$

Proof.

We’ll turn this into a relation of closure by invoking proposition 1. Taking the complement turns the LHS into:

$\begin{aligned}X-\text{int}(Y\cap Z) &=\text{cl}(X - (Y\cap Z)) = \text{cl}((X-Y)\cup(X-Z))= \text{cl}(X-Y)\cup\text{cl}(X-Z)\\&= (X-\text{int}(Y))\cup (X-\text{int}(Z)) = X - (\text{int}(Y)\cap\text{int}(Z)),\end{aligned}$

where the third equality followed from proposition 2 of the previous article. ♦

By induction, this can be generalised to finite intersections:

$\text{int}(Y_1) \cap \text{int}(Y_2) \cap\ldots\cap \text{int}(Y_n) = \text{int}(Y_1\cap Y_2 \cap \ldots \cap Y_n)$

for arbitrary subsets $Y_1, Y_2, \ldots, Y_n\subseteq X.$

This does not hold for infinite intersections. For example, if $Y_n = [-\frac 1 n, +\frac 1 n],$ then the intersection $\cap_n \text{int}(Y_n) = \cap_n (-\frac 1 n, +\frac 1 n) = \{0\}.$ On the other hand, $\text{int}(\cap_n Y_n) = \text{int}(\{0\}) = \emptyset.$

It also doesn’t work for finite union. E.g. let Y = Q and Z = R–Q, in the real line X = R. Then int(Y) and int(Z) are both empty so $\text{int}(Y)\cup\text{int}(Z) = \emptyset.$ But $\text{int}(Y\cup Z) = \text{int}(\mathbf{R}) = \mathbf{R}.$

However, we do have:

$\text{int}(\cup_i Y_i) \supseteq \cup_i \text{int}(Y_i)\$ and $\text{int}(\cap Y_i) \subseteq \cap_i \text{int}(Y_i)$

for any collection of subsets $Y_i\subseteq X.$

For the first inclusion: since for each i, we have $Y_i\subseteq \cup_i Y_i,$ we get $\text{int}(Y_i)\subseteq \text{int}(\cup_i Y_i).$ Since this works for each i, we’re done.
For the second inclusion: for each i, we have $\cap_i Y_i \subseteq Y_i$ so this gives $\text{int}(\cap_i Y_i) \subseteq \text{int}(Y_i).$ Since this holds for each i, we’re done.

Further Properties

Unfortunately, the interior of Y in a smaller subspace cannot be deduced from its interior in a bigger space.

Proposition 3. Let $Y\subseteq Z\subseteq X$ be subsets of a topological space X. Then:

$\text{int}_Z(Y) \supseteq Z\cap\text{int}_X(Y).$

In general equality doesn’t hold.

Proof.

Since the RHS is an open subset of Z and is contained in Y, it must be contained in the LHS.

To see that equality doesn’t hold in general, let X = R, Z = Q and Y = Q ∩ (0, 1). Then since (0, 1) is open in R, Y is open in Z. Hence, $\text{int}_Z(Y) = Y.$ On the other hand, $\text{int}_X(Y) = \emptyset,$ so the RHS is empty. ♦

Thankfully, product still works:

Proposition 4. If $Y_1\subseteq X_1$ and $Y_2\subseteq X_2$ are subsets of topological spaces, then

$\text{int}(Y_1 \times Y_2) = \text{int}(Y_1)\times \text{int}(Y_2),$

where the LHS interior is taken in $X_1\times X_2.$

Proof.

Since $\text{int}(Y_1)\times \text{int}(Y_2)$ is an open subset of $X_1\times X_2$ and is contained in $Y_1\times Y_2,$ we see that the RHS is contained in the LHS.

Conversely, if $(x,y)\in \text{int}(Y_1\times Y_2)$ then there’s a basic open subset U×V such that $(x,y)\in U\times V\subseteq \text{int}(Y_1 \times Y_2).$ Then $U\subseteq Y_1$ and $V\subseteq Y_2$ so $x\in \text{int}(Y_1)$ and $y\in \text{int}(Y_2).$ So the LHS is contained in the RHS. ♦

Finally, since $\text{int}(Y) \subseteq Y\subseteq \text{cl}(Y)$ we define the boundary of Y to be the difference

bd(Y) := cl(Y) – int(Y).

Note that since bd(Y) is the intersection of cl(Y) and the complement of int(Y), both closed, bd(Y) is closed too.

Note: some people also call cl(Y)-int(Y) the frontier of Y, since the term “boundary” has a different meaning in algebraic topology. We opt not to use that term, since in practice there’s little cause for confusion.

Comparing Interior and Closure

The following table summarises the contents of this and the previous article. Some of these relations are equivalent by virtue of the fact that X-int(Y) = cl(X–Y).

Closure of Y	Interior of Y
If $Y\subseteq Z,$ then $\text{cl}(Y)\subseteq \text{cl}(Z).$	If $Y\subseteq Z,$ then $\text{int}(Y) \subseteq \text{int}(Z).$
Finite union: $\text{cl}(Y\cup Z) = \text{cl}(Y) \cup \text{cl}(Z).$	Finite intersection: $\text{int}(Y \cap Z) = \text{int}(Y) \cap \text{int}(Z).$
Arbitrary union: $\text{cl}(\cup_i Y_i) \supseteq \cup_i \text{cl}(Y_i).$	Arbitrary union: $\text{int}(\cup_i Y_i) \supseteq \cup_i \text{int}(Y_i).$
Arbitrary intersection: $\text{cl}(\cap_i Y_i) \subseteq \cap_i \text{cl}(Y_i).$	Arbitrary intersection: $\text{int}(\cap_i Y_i) \subseteq \cap_i\text{int}(Y_i).$
If $Y\subseteq Z\subseteq X$ , then $\text{cl}_Z(Y) = Z\cap \text{cl}_X(Y).$	If $Y\subseteq Z\subseteq X,$ then $\text{int}_Z(Y) \supseteq Z\cap \text{int}_X(Y).$
For subsets $Y_1\subseteq X_1, Y_2\subseteq X_2,$ we have $\text{cl}(Y_1\times Y_2) = \text{cl}(Y_1)\times \text{cl}(Y_2).$	For subsets $Y_1\subseteq X_1, Y_2\subseteq X_2,$ we have $\text{int}(Y_1\times Y_2) = \text{int}(Y_1)\times \text{int}(Y_2).$

Exercises: deduce the corresponding entries for boundary bd(Y).

Answers:

If Y is contained in Z, it’s not always true that bd(Y) is contained in bd(Z). Indeed, take Y = (0, 1) ∩ Q and Z = (0, 1). Then the boundary of Y is cl(Y) – int(Y) = [0, 1] – (empty set) = [0, 1], but the boundary of Z is {0, 1}. Surprised?
For finite union, bd(union of Y, Z) is contained in the union of bd(Y) and bd(Z).
For finite intersection, bd(Y ∩ Z) is contained in bd(Y) ∩ bd(Z).
For arbitrary union, things get hairy. E.g. if Y_n = [0, (n-1)/n] for n=2, 3, …, then union of bd(Y_n) is {0, 1/2, 2/3, 3/4, … }, while bd(union of Y_n) = {0, 1}. Neither is contained in the other.
Same for arbitrary intersection.
bd_Z(Y) is contained in Z ∩ bd_X(Y).
bd(Y₁ × Y₂) is the union of bd(Y₁) × cl(Y₂) and cl(Y₁) × bd(Y₂).

Posted in Notes | Tagged advanced, boundaries, closures, interiors, product topology, subspaces, topology | 2 Comments

Topology: Closure

Posted on February 10, 2013 by limsup

Suppose Y is a subset of a topological space X. We define cl(Y) to be the “smallest” closed subset containing Y. Its formal definition is as follows.

Let Σ be the collection of all closed subsets $C\subseteq X$ containing Y.
Note that $X\in \Sigma$ , so Σ is not empty.
Thus, we can take define $\text{cl}(Y) := \cap_{C\in\Sigma} C.$

This is called the closure of Y in X. Note that the notation cl(Y) does not indicate the ambient space X. If there’s any possibility of confusion, we denote it by cl_X(Y) instead. The closure satisfies the following properties:

$\text{cl}(Y) \in \Sigma$ : since an intersection of closed subsets is still closed;
if $C\in \Sigma,$ then since cl(Y) is the intersection of many sets, including C, we must have $\text{cl}(Y)\subseteq C.$

This justifies our calling cl(Y) the smallest closed subset containing Y.

As an immediate property, we have:

Basic Fact. If $Y\subseteq Z$ are subsets of X, then $\text{cl}(Y)\subseteq \text{cl}(Z).$

Indeed, cl(Z) is a closed subset of X containing Z, and hence Y. Thus, it must contain cl(Y).

Examples

If Y is closed in X, then cl(Y) = Y.
Take the half-open interval $Y = [0, 1)\subseteq \mathbf{R}.$ The closure is [0, 1].
Take the set of rationals Y = Q in X = R. The closure is the whole real line.
Suppose Y = {3, 5, 7} in N*. What is the closure of Y (see here for the definition of N*)? [ Answer: {1, 2, 3, 4, 5, 6, 7}. ]

Warning: in a metric space X, the closed ball $N(a, \epsilon)^* := \{x\in X: d(x, a) \le \epsilon\}$ is a closed subset since the complement (which contains all x such that d(x, a) > ε) is open in X.

However, the closure of the open ball $N(a, \epsilon)$ is not the closed ball $N(a, \epsilon)^*$ in general. For a rather pathological example, consider $X = [-1, +1] \cup \{2\}$ and take the open ball Y = N(0, 2). Then Y = [-1, +1] is already closed in X so cl(Y) = Y, while the closed ball N(0, 2)* = X ≠ cl(Y).

Properties of Closure

The closure of Y can be characterised as follows.

Proposition 1. We have $\text{cl}(Y) = Y\cup Y^{acc}$ where Y^acc is the collection of all points of accumulation of Y.

Proof.

Let’s first prove that $C:=Y\cup Y^{acc}$ is closed in X.

If $x\in X-C,$ then in particular x is not a point of accumulation of Y, so there exists an open subset $U\subseteq X$ containing x such that U ∩ Y contains at most x. But x is not in Y, so $U\cap Y=\emptyset,$ i.e. $U\subseteq X-Y.$
Next we wish to show $U\subseteq X-Y^{acc}.$ Suppose on the contrary $y\in U\cap Y^{acc},$ so in particular y is a point of accumulation of Y. Since U is an open subset of X containing y, we have U ∩ Y containing a point z ≠ y, which contradicts what we just proved. Thus $U\subseteq X - (Y\cup Y^{acc}) = X-C,$ so X–C is a union of open subsets and is thus open.

To complete the proof, since C is a closed subset containing Y, it also contains cl(Y). On the other hand, since cl(Y) is closed, $\text{cl}(Y) \supseteq \text{cl}(Y)^{acc} \supseteq Y^{acc}.$ This gives $\text{cl}(Y) \supseteq Y\cup Y^{acc} = C,$ which completes the proof. ♦

Proposition 2. If $Y, Z\subseteq X$ are subsets, then

$\text{cl}(Y\cup Z) = \text{cl}(Y) \cup \text{cl}(Z).$

Proof.

Since the RHS is closed and contains $Y\cup Z$ , it must contain the LHS also.

Conversely, since $Y\subseteq Y\cup Z,$ we have $\text{cl}(Y) \subseteq \text{cl}(Y\cup Z).$ Similarly, $\text{cl}(Z) \subseteq\text{cl}(Y\cup Z)$ so the LHS contains the RHS. ♦

Unions and Intersections of Closures

By induction, one can prove that:

$\text{cl}(Y_1 \cup Y_2 \cup \ldots \cup Y_n) = \text{cl}(Y_1) \cup \text{cl}(Y_2) \cup \ldots \cup \text{cl}(Y_n)$

for any subsets $Y_1, Y_2, \ldots, Y_n \subseteq X.$ However, this is as far as we go, as we cannot extend this to the union of infinitely many sets. E.g. if $Y_n = [\frac 1 n, 1]$ in X=R, then since each $Y_n\subseteq X$ is closed, we have $\cup_n\text{cl}(Y_n) = \cup_n Y_n=(0, 1]$ which isn’t closed, so it cannot possibly be $\text{cl}(\cup_n Y_n).$

Also, the result is not true for intersection. E.g. let Y = Q and Z = R–Q be subsets of the real line R. Then Y ∩ Z is empty, and so is cl(Y ∩ Z). On the other hand, cl(Y) ∩ cl(Z) = R ∩ R = R. So we have cl(Y) ∩ cl(Z) ≠ cl(Y ∩ Z).

That being said, we do at least have a one-sided inclusions:

$\text{cl}(\cap_i Y_i) \subseteq \cap_i \text{cl}(Y_i), \quad \text{cl}(\cup_i Y_i) \supseteq \cup_i \text{cl}(Y_i)$

for any collection of subsets $Y_i\subseteq X.$

For the first inclusion: since $\cap_i Y_i \subseteq Y_i$ for each i, we have $\text{cl}(\cap_i Y_i)\subseteq \text{cl}(Y_i).$ Since this holds for every i, we get the desired inclusion.
For the second inclusion: since $Y_i\subseteq \cup_i Y_i$ for each i, we have $\text{cl}(Y_i) \subseteq \text{cl}(\cup_i Y_i).$ Since this holds for each i, we’re done. ♦

Further Properties

Next, the closure of Y in a smaller subspace can be deduced from the closure in a bigger space.

Proposition 3. If $Y\subseteq Z \subseteq X$ are subsets of a topological space X, then the closure of Y in Z is the intersection of Z with the closure of Y in X.

$\text{cl}_Z(Y) = Z\cap \text{cl}_X(Y).$

Proof

Since the RHS is a closed subset of Z containing Y, it must contain the LHS.

Conversely, since LHS is closed in Z it must be of the form C ∩ Z for some closed subset C of X. Then C is a closed subset of X containing Y and thus must contain cl_X(Y). So C ∩ Z must contain the RHS. ♦

Finally, closure also respects the product.

Proposition 4. If $Y_1 \subseteq X_1$ and $Y_2\subseteq X_2$ are subsets of topological spaces, then:

$\text{cl}(Y_1\times Y_2) = \text{cl}(Y_1) \times \text{cl}(Y_2)$

where the LHS closure is taken in $X_1\times X_2.$

Proof.

Since the RHS is closed in $X_1 \times X_2,$ it must contain the LHS. For the reverse inclusion, we prove by contradiction by picking an element (x, y) outside the LHS, cl(Y₁ × Y₂).

In particular, $(x, y)\not\in (Y_1\times Y_2)^{acc},$ so there is an open subset containing (x, y) which doesn’t intersect Y₁ × Y₂.
This open subset must contain a basic open set of the form U × V which contains (x, y).
Note that we still have $(U\times V)\cap (Y_1\times Y_2) = \emptyset$ and so $U\cap Y_1 = \emptyset$ or $V\cap Y_2 = \emptyset.$
Assume the former; then $x\not\in Y_1^{acc}.$ Since $x\in U,$ we also have $x\not\in Y_1,$ thus giving $x\not\in Y_1\cup Y_1^{acc} = \text{cl}(Y_1).$

So (x, y) is outside the RHS also, and we’re done. ♦

Posted in Notes | Tagged advanced, closed balls, closed subsets, closures, open balls, points of accumulation, product topology, topology | Leave a comment

Thoughts on a Problem III

Posted on February 8, 2013 by limsup

I saw an interesting problem recently and can’t resist writing it up. The thought process for this problem was exceedingly unusual as you’ll see later.

First, here’s the source:

But here’s the full problem (rephrased a little) if you’d rather not watch a video.

Problem. From a cylindrical cake with chocolate icing on top, cut successive slices of a fixed angle x. Each slice is then inverted and inserted back into the cake. Find all angles x, such that after finitely many iterations, all the icing is back on top of the cake.

The whole point of the video is to present us with problems whose solutions violate our intuition. This problem is a prime example. At first glance, one might conjecture that the answer is any rational multiple of the full rotation. The correct answer is a little surprising, to say the least.

[ Warning: spoiler ahead with detailed solution. Actually, since this article is about my thoughts on a problem, what else did you expect? 😛 ]

The following solution is to be performed by trained professionals only. Readers please do not attempt to try this on an actual cake.

Solution

The correct answer is: any angle. Yes, even irrational multiples of the full rotation! This completely bowled me over since for these “irrational angles”, you’ll never cut the same position twice. Fortunately, the professor in the video stopped short of explaining why all angles work, which gives us viewers an opportunity to think this through.

It took me a full 10 minutes to convince myself that “irrational angles” are possible. I realised that after one full round, in passing through the original baseline, the process of flipping the slice also introduces a rather unexpected turn of events.

What I had thought:

What actually transpires:

Note that even the initial cutting line had been flipped within the sector.

Upon this realisation, one sees that cutting lines may repeat even though the positions at which you cut don’t, since the cutting lines themselves aren’t stationary. Needless to say, the next step is to see what happens with the second round of cuts. As we continue round, we get: x (brown), x (brown) …

Now it should be noted that a = 2π-kx for some integer k such that kx ≤ 2π < (k+1)x. And b = x–a = (k+1)x – 2π. Hence, upon the second round, a sector of the cake is inverted which includes “b” above:

The pattern actually got more complex! As the cutting goes through more and more rounds, we’ll certainly expect more interspersing of the colours. How do we prove that it eventually returns to the original pattern?

Turns out there’s a nice way. First we let a and b be as above: a = 2π – kx, where k is the largest positive integer such that kx ≤ 2π < (k+1)x. Then let b = x – a = (k+1)x – 2π. This gives kb + (k+1)a = k(a+b) + a = kx+a = 2π. Draw a pie chart with k+1 slices of angle a and k slices of angle b. Also highlight the sector where one of edges is the common edge between two slices of angle a (see diagram below); this is the slice we’ll invert.

Observe that after inverting this slice, the two angles a and b are swapped and the next sector is highlighted. This new sector also has one of its edges lying on the common edge between two slices of angle a. Thus, if we consider the total set S of all possible ways to colour each slice with either brown and white ( $|S| = 2^{2k+1}$ ), then inverting a slice corresponds to a bijective function on S. By elementary theory of permutations, we must eventually get back the original configuration, i.e. all chocolate icing up.

This pretty much solves our problem, although there’s a nagging feeling that the proof may not work if angle x > π. For this case, we imagine the corresponding problem with angle x’ = 2π – x < π. Instead of inverting a slice of angle x, one imagines inverting a slice of angle x’ then flipping the whole cake over. Hence if, for angle x’, we get back the cake with chocolate icing up in N moves, then for the case of angle x, performing N moves will give us the cake with chocolate icing either all up or all down. In the former case, that’s good; in the latter case, another N moves will flip the cake back up.

Conclusion (with apologies to Valve):

Posted in Thought | Tagged basic, combinatorics, intuition, thoughts, video | Leave a comment

Topology: Cauchy Sequences and Uniform Continuity

Posted on February 7, 2013 by limsup

[ Updated on 8 Mar 13 to include Cauchy-continuity and added answers to exercises. ]

We wish to generalise the concept of Cauchy sequences to metric spaces. Recall that on an intuitive level, a Cauchy sequence is one where the elements get “closer and closer”.

Definition. Let (X, d) be a metric space. A sequence $(x_n)$ in X is said to be Cauchy if

for any ε>0, there exists N such that whenever m, n > N, we have $d(x_m, x_n) < \epsilon.$

Let’s list some basic results.

Proposition 1. A convergent sequence is Cauchy.

Proof.

Suppose $(x_n) \to a$ in the metric space (X, d). For any ε>0, there exists N such that whenever n > N, we have $d(x_n, a)<\epsilon/2.$ Thus, whenever m, n > N, we have:

$d(x_m, x_n) \le d(x_m, a) + d(a, x_n) < \frac\epsilon 2 + \frac\epsilon 2 = \epsilon.$ ♦

Proposition 2. Let Y be a metric subspace of (X, d). Then a sequence $(x_n)$ in Y is Cauchy if and only if it’s Cauchy in X.

There’s nothing to prove here since Y inherits the distance function from X. However, this serves to highlight the fact that while a sequence from Y which converges in X may not be convergent in Y, the same problem doesn’t hold for Cauchy sequences, i.e. being Cauchy is a property of the sequence itself, regardless of the ambient space.

Proposition 3. If $(x_n), (y_n)$ are sequences of metric spaces $(X, d_X), (Y, d_Y)$ respectively, then $(x_n, y_n)$ is a Cauchy sequence of X × Y if and only if each of $(x_n), (y_n)$ is Cauchy in the respective metric space.

For the metric of X × Y, we can pick any one of the following:

$((x, y), (x', y')) \mapsto \sqrt{d_X(x, x')^2 + d_Y(y, y')^2}$ ;

$((x, y), (x', y')) \mapsto d_X(x, x') + d_Y(y, y')$ ;

$((x, y), (x', y')) \mapsto\max(d_X(x, x'), d_Y(y, y'))$ .

Proof

First suppose $(x_n)$ and $(y_n)$ are Cauchy. For any ε>0,

there exists M such that when m, n > M, we have $d_X(x_m, x_n) < \frac\epsilon 2;$
there exists N such that when m, n > N, we have $d_Y(y_m, y_n) < \frac\epsilon 2.$

Thus, when m, n > max(M, N), we have – for any metric d on X × Y in the above list – $d((x_m, y_m), (x_n, y_n)) < \epsilon.$

For the converse, we use the fact that

$d((x,y), (x',y')) \ge d_X(x, x')$ and $d((x,y), (x', y')) \ge d_Y(y, y')$

for any one of three choices of d. ♦

One might ask if it’s necessary to consider all three metrics on X × Y since they all give rise to the same topology anyway. And this is where we drop the bombshell.

The concept of Cauchy sequences actually relies heavily on the metric and not just the underlying topology. In other words, it’s possible for two metrics on the same space to be topologically equivalent, but a sequence is Cauchy in one and not the other.

For example, consider the homeomorphism f : R⁺ → R⁺ of the space of positive reals given by f(x) = 1/x. The sequence $x_n = 1/n$ is Cauchy, but the resulting sequence $f(x_n) = n$ is not. Put in another way, we can define two metrics on R⁺ via $(r, r') \mapsto |r-r'|$ and $(r, r')\mapsto |r^{-1} - r'^{-1}|.$ Then the sequence $x_n = 1/n$ is Cauchy under the first metric but not the second.

The same example also tells us:

Definition. A function $f:(X, d_X) \to (Y, d_Y)$ of metric spaces is said to be Cauchy-continuous if whenever $(x_n)$ is a Cauchy sequence in X, $(f(x_n))$ is a Cauchy sequence in Y.

Warning. Not all continuous functions are Cauchy-continuous.

To rectify that, we need a stronger form of continuity.

Uniform Continuity

The answer to our problem is the following definition. We had already seen it earlier in the case of R.

Definition. A map $f:(X, d_X)\to (Y, d_Y)$ of metric spaces is said to be uniformly continuous if

for any ε>0, there exists δ>0 such that whenever $x, x'\in X$ satisfies $d_X(x, x')<\delta,$ we have $d_Y(f(x), f(x')) < \epsilon.$

Clearly, a uniformly continuous function is also continuous (at every point of X). But here’s an example where the converse is not true.

Take the function f : R⁺ → R⁺ given by f(x) = 1/x as before. This is clearly continuous. To show that it’s not uniformly continuous, negating the definition means we need to find an ε>0 such that for any δ>0, there exist x and x’ such that |x – x’| < δ but |f(x) – f(x’)| ≥ ε.

This is not too hard: set ε=1. Now for any δ>0, pick a positive integer n > 1/δ and let $x = \frac 1 n, x' = \frac 1 {n+1}.$ We now have:

$d(x, x') = |x - x'| = \frac 1 {n(n+1)} < \frac 1 n <\delta$ but $d(f(x), f(x')) = 1.$

Next, the main result we’d like to prove is:

Theorem 4. If $f:(X, d_X) \to (Y, d_Y)$ is a uniformly continuous map of metric spaces, then it is Cauchy-continuous.

Proof.

Let ε>0. Then:

by uniform continuity of f, there exists δ>0 such that whenever $x, x'\in X$ satisfies $d_X(x, x')<\delta$ , we have $d_Y(f(x), f(x')) < \epsilon;$
there exist N such that when m, n > N, we get $d_X(x_m, x_n) < \delta.$

Thus, when m, n > N, we get $d_Y(f(x_m), f(x_n)) < \epsilon.$ So $(f(x_n))$ is indeed a Cauchy sequence in Y. ♦

The following are some easy properties of uniformly continuous functions.

Proposition 5.

If Y is a metric subspace of (X, d), then the inclusion map $Y\hookrightarrow X$ is uniformly continuous.

If $f:X\to Y$ and $g:Y\to Z$ are uniformly continuous functions of metric spaces, then so is $g\circ f:X\to Z.$

The projection maps $p_1: X\times Y\to X$ and $p_2 : X\times Y\to Y$ are uniformly continuous, where the metric on X × Y is one of the three in proposition 3.

Proof.

The first two statements are obvious. The last follows from the inequality we saw earlier:

$d((x, y), (x', y')) \ge d_X(x, x')$ and $d((x,y), (x', y'))\ge d_Y(y, y'),$

for any one of the three d. ♦

Finally, we end this article with some exercises on Cauchy and convergent sequences.

Exercises

Prove that if $(x_n)$ is a Cauchy sequence in a metric space X, then every subsequence is also Cauchy.
Prove that if $(x_n)$ is a convergent sequence in a topological space X, then every subsequence is also convergent.
Suppose $(x_n)$ is a Cauchy sequence in a metric space X. Prove that if a subsequence of $(x_n)$ is convergent, so is the entire sequence.
Prove that a Cauchy-continuous map $f:(X, d_X)\to (Y, d_Y)$ of metric spaces is continuous.
Find a Cauchy-continuous f which is not uniformly continuous.

Answers (Highlight to Read)

Suppose (x_n) is Cauchy, and the subsequence indexed by n[1] < n[2] < n[3] … converges to x. For each ε>0, pick N such that (i) when m, n > N, d(x_m, x_n) < ε/2 and (ii) when k>N, d(x_n[k], x) < ε/2. Thus, when n>N, we have d(x_n, x) ≤ d(x_n, x_n[k]) + d(x_n[k], x) < ε for some k>N. This proves that (x_n) → x.
By theorem 6 in the previous article, it suffices to show that for any convergent (x_n) → x in X, we have (f(x_n)) → f(x). Now construct a new sequence (y_n) by interspersing (x_n) and x : y_2n = x_n and y_2n-1 = x, for n = 1, 2, … . Then (y_n) still converges to x, so it’s Cauchy; since f is Cauchy-continuous, (f(y_n)) is Cauchy too. But (f(y_n)) has a subsequence (f(x), f(x), … ), so by Q3, (f(y_n)) → f(x).
Take f : R → R, given by f(x) = x².
- Not uniformly continuous since we can set ε=2; if x=n, y=n + (1/n), then |x–y| = 1/n, but |f(x)-f(y)| > 2.
- Cauchy-continuous since any Cauchy sequence (x_n) in R must be convergent so (f(x_n)) is also convergent, and hence Cauchy.

Summary

We have three related notions of continuity (for maps $f:(X, d_X)\to (Y, d_Y)$ of metric spaces).

$\left\{\begin{matrix}\text{Uniform}\\ \text{continuity}\end{matrix}\right\}\implies \left\{\begin{matrix} \text{Cauchy-}\\ \text{continuity}\end{matrix}\right\}\implies \left\{\text{Continuity}\right\}.$

All implications are non-reversible since there’re counter-examples. The first two are only defined for maps between metric spaces while the last is a topological property.

Posted in Notes | Tagged advanced, cauchy sequences, cauchy-continuity, metric spaces, product topology, topology, uniform continuity | Leave a comment

Topology: Nets and Points of Accumulation

Posted on February 5, 2013 by limsup

Recall that a sequence $(x_n)$ in a topological space X converges to a in X if the function f : N* → X which takes $n\mapsto x_n, \infty\mapsto a$ is continuous at $\infty\in\mathbf{N}^*$ . Unrolling the definition, it means that for any open subset U of X containing a, the set $f^{-1}(U)\subseteq \mathbf{N}^*$ contains (N, ∞] for some N. In other words,

for any open subset $U\subseteq X$ containing a, there exists N such that when n > N, $x_n\in U$ .

We also saw that if X is Hausdorff, then the limit a is unique, but we won’t make that assumption now.

The key question here is:

Suppose $Y\subseteq X$ is a subspace and $(x_n)$ is a sequence in Y converging to $a\in X$ . Does it mean $a \in Y$ ?

Even in the case where X = R, this is not true for some subspaces, e.g. Y = (0, 1), where the sequence $x_n = 1/n$ converges to 0 which lies outside Y. In an earlier article, we saw that this is related to the requirement that Y is closed in X. Let’s attempt to reproduce that result for topological spaces.

Definition. Let a be a point in topological space X. We say it is a point of accumulation for Y if for every open subset $U\subseteq X$ containing a, there exists $x\in U\cap Y$ , x≠a.

The Case of Metric Spaces

Let’s examine this definition for a metric space X. In this case, a is a point of accumulation for Y iff for each ε>0, the intersection N(a, ε) ∩ Y contains some point other than a. Indeed, (LHS) → (RHS) follows from the fact that N(a, ε) is an open subset containing a, while (RHS) → (LHS) follows from the fact that every open subset containing a must contain an open ball N(a, ε).

Example

Suppose X = R and Y is the set of rational numbers 0 < r < 1. Then any real number in [0, 1] is a point of accumulation for Y since for any ε>0, the open ball N(r, ε) must contain a rational number. [ The full rigourous argument needs a bit more care, but we won’t harp on this any more. ]

This result is reminiscent of an earlier one.

Theorem 1. In a metric space X and subspace Y, the following are equivalent.

If $(x_n)$ is a sequence in Y converging to $a\in X$ , then $a\in Y$ .

Y contains all its points of accumulation.

Y is closed in X.

Proof.

(1) → (2) : let a be a point of accumulation of Y. For each n = 1, 2, …, N(a, 1/n) ∩ Y contains a point x_n ≠ a. Then (x_n) is a sequence in Y converging to a; thus a is in Y.

(2) → (3) : let $a\in X-Y$ , so it is not a point of accumulation of Y. By definition, there exists an open subset U containing a such that U ∩ Y has no point other than possibly a. But a is not in Y anyway, so U ∩ Y is empty, i.e. $U\subseteq X-Y$ and X–Y is open in X.

(3) → (1) : suppose (x_n) is a sequence in Y converging to a. If $a\in X-Y$ , which is open in X, then there exists N such that when n>N, x_n lies in X–Y, which is absurd. ♦

The Case of Topological Spaces

We will see later that in the case of topological spaces, (2) and (3) are still equivalent, but (1) fails. The problem lies in the definition of sequences, which is essentially a function f : N* – {∞} → X, and the space N* is given unnecessary prominence. To overcome this one needs a more general form of sequences for topological spaces.

Definition. A directed set is a partially ordered set I such that for any $i, j\in I$ , there exists $k\in I$ such that $i\le k, j\le k$ . In words, this means that every finite subset has an upper bound.

A net in a topological space X (indexed by I) is a function f : I → X, written as $(x_i)_{i\in I}$ .

We say that a net $(x_i)$ has limit $a\in X$ if for any open subset $U\subseteq X$ containing a, there exists an index i such that for all j ≥ i, $x_j \in U.$

Now the following is true.

Theorem 2. Let Y be a subspace of a topological space X. The following are equivalent.

If $(x_i)$ is a net in Y which converges to $a\in X$ , then $a\in Y.$

Y contains all its points of accumulation.

Y is closed in X.

Proof

(1) → (2) : suppose a is a point of accumulation of Y. Let the index set I correspond to the collection of open subsets U containing a, ordered by reverse inclusion: $i \le j$ iff $U_i \supseteq U_j.$ Since a is a point of accumulation of Y, for each $i\in I$ we can pick $x_i \in U_i\cap Y, x_i \ne a.$ Then $(x_i)_{i\in I} \to a$ so $a\in Y.$

(2) → (3) : identical to earlier.

(3) → (1) : let $(x_i)$ be a net in Y converging to $a\in X-Y.$ Since X–Y is open in X, there exists index i such whenever j≥i, $x_j \in X-Y,$ which is a contradiction. ♦

Interlude: Alternate Look at Nets

At this point, the reader may wonder why we need I to be a directed set, i.e. wouldn’t an ordinary partially ordered set (poset) suffice? Indeed, the above theorem seems to work even when I is just a poset.

To answer this question, we provide an alternate definition for nets and limits. Given a directed set I, let $I^* := I \cup \{\infty\}$ , where ∞ is just a dummy symbol. A basis for open subsets of I* is given by:

$P_i := \{j\in I: j\ge i\} \cup \{\infty\}$ , for various $i\in I.$

To check that this is a basis, we claim that $P_i \cap P_j = \cup_{k\ge i, k\ge j} P_k.$

Indeed, clearly RHS is contained in LHS since $k\ge i \implies P_k \subseteq P_i.$
Conversely, if $k\in P_i\cap P_j$ is an index, then k ≥ i and k ≥ j so $k\in P_k$ is found in the RHS. The only case left is $\infty \in P_i \cap P_j.$ Since I is a directed set, there exists k≥i and k≥j. Thus, the RHS is a non-empty union and ∞ is in the RHS. This proves that LHS is contained in RHS.

Thus, the fact that I is a directed set is necessary for P_i‘s to form a basis. Otherwise, they’d only form a subbasis in which case it’s possible for {∞} to be open in I*.

Next, the condition that a net $(x_i) \to a$ is exactly the same as that $\lim_{i\to\infty} x_i = a$ where the limit was defined in the previous article. Since {∞} is not open in I*, we conclude:

Proposition 3. If X is Hausdorff, a converging net in X has a unique limit.

Finally, we have the following.

Proposition 4. If g : X → Y is a continuous map of topological spaces, then every convergent net $(x_i) \to a$ gives $(g(x_i)) \to g(a).$

The proof follows from the proposition here.

Proposition 5. If $(x_i, y_i)$ is a net in X × Y, then $(x_i, y_i) \to (a, b) \in X\times Y$ if and only if $(x_i)\to a$ and $(y_i)\to b.$

Proof.

We have $(x_i, y_i) \to (a, b)$ if and only if the function I* → X × Y which takes $i\mapsto (x_i, y_i)$ and $\infty \mapsto (a, b)$ is continuous. Now we use the following property, whose proof is left as an exercise.

If X, Y, Z are topological spaces, then f : Z → X × Y is continuous if and only if composing it with the projection maps X × Y → X and X × Y → Y, we get continuous maps Z → X and Z → Y.

So the above function I* → X × Y is continuous if and only if composing with the projection maps give continuous functions I* → X and I* → Y. And this is true if and only if $(x_i) \to a$ and $(y_i)\to b.$ ♦

One Final Result

Another sign that nets are a good way to generalise sequences:

Theorem 6.

If f : X → Y is a map of metric spaces, then f is continuous if and only if for any convergent sequence $(x_n) \to a$ , we also have $(f(x_n)) \to f(a).$

If f : X → Y is a map of topological spaces, then f is continuous if and only if for any convergent net $(x_i) \to a$ , we also have $(f(x_i)) \to f(a).$

Proof.

The forward direction in both cases has been proven in proposition 4 above and here. For the converse:

First statement: if f is not continuous at a, then there exists ε>0 such that for any δ>0, we can find x in X such that d_X(x, a) < δ but d_Y(f(x), f(a)) ≥ ε. Letting δ = 1/n for n=1, 2, 3, … we get a sequence $x_n\in X$ such that $d_X(x_n, a) < \frac 1 n$ but $d_Y(f(x_n), f(a)) \ge \epsilon.$ Then $(x_n) \to a$ but $(f(x_n)) \not\to a.$

Second statement: if f is not continuous at a, there exists an open subset $V\subseteq Y,$ $f(a)\in V,$ such that $f^{-1}(V)$ doesn’t contain an open subset of U containing a. As before, let the index set I correspond to the collection of open subsets of X containing a, ordered by reverse inclusion: $i\le j$ iff $U_i\supseteq U_j.$ Since no U_i is contained in f^-1(V), we can pick $x_i \in U_i - f^{-1}(V)$ for each i. Then $(x_i)$ is a net approaching a but $(f(x_i))$ doesn’t approach f(a) since they lie outside V. ♦

Posted in Notes | Tagged advanced, closed subsets, continuity, convergence, limits, metric spaces, nets, points of accumulation, sequences, topology | Leave a comment

Topology: Limits and Convergence

Posted on February 2, 2013 by limsup

Following what we did for real analysis, we have the following definition of limits.

Definition of Limits. Let X, Y be topological spaces and $a\in X$ . If f : X-{a} → Y is a function, then we write $\lim_{x\to a} f(x) = b \in Y$ if the function:

$g:X\to Y,\ g(x) = \begin{cases} f(x), &\text{if } x\ne a, \\ b, &\text{if }x=a.\end{cases}$

is continuous at x=a. In words, we say that f(x) approaches b as x approaches a.

Definition of Convergence. If X is a topological space, a sequence of elements $x_n \in X$ is said to converge to a if the function f : N* – {∞} → X, f(n) = x_n, approaches a as n approaches ∞. A sequence which has a limit is called a convergent sequence.

[ Refer here for the definition of N*. ]

We shall show that this convergence is consistent with our earlier definition of convergence.

Proposition. Let (X, d) be a metric space. Then a sequence $x_n\in X$ converges to a (in the above definition) if and only if:

for any ε>0, there exists N such that whenever n>N, $d(x_n, a) < \epsilon$ .

Proof.

Suppose $x_n\in\mathbf{R}$ converges to a. Given any ε>0, N(a, ε) is an open ball containing a. By definition, if g : N* → X is defined by g(n)=x_n, for n=1, 2, 3, … and g(∞)=a, then g is continuous at ∞. Thus $U:=g^{-1}(N(a,\epsilon))$ contains an open subset of N* containing ∞. So U contains some U_N = {N, N+1, N+2, … } and $n>N\implies n\in U\implies x_n=g(n)\in N(a,\epsilon).$
Conversely, assume the condition in the proposition holds; let’s prove that g is continuous at ∞. Now any open subset $V\subseteq X$ containing a must contain an open ball $N(a,\epsilon) \subseteq V$ for some ε>0. Then there exists N such that whenever n>N, $d(x_n, a)<\epsilon \implies x_n \in N(a,\epsilon).$ Thus $g^{-1}(V)$ contains the set U_N₊₁ = {N+1, N+2, N+3, …, ∞} which is an open subset containing ∞. ♦

In addition to N*, let’s consider the extended real line $\overline{\mathbf{R}} = \mathbf{R}\cup \{-\infty, \infty\}.$ Again we should think of the infinities as dummy symbols. Its topology is defined via the basis:

$(a,b)$ for real a < b;
$(a,\infty] = (a, \infty) \cup \{\infty\}$ for real a;
$[-\infty, b) = (-\infty, b)\cup \{-\infty\}$ for real b.

Then all prior limits can be expressed via the extended real line. For example:

Proposition.

If $x_n$ is a sequence of real numbers, then $\lim_{n\to\infty} x_n = \infty$ in the classical sense iff the function $f:\mathbf{N}^*-\{\infty\}\to \overline{\mathbf{R}}$ which takes $n\mapsto x_n$ tends to ∞ as n tends to ∞.

If $f: \mathbf{R}\to\mathbf{R}$ is a function, then $\lim_{x\to\infty} g(x) = L \in \mathbf{R}$ in the classical sense iff the function $f:\overline{\mathbf{R}} - \{\infty\} \to \mathbf{R}$ tends to L as x→∞.

Proof.

We’ll prove (LHS) → (RHS) and leave the converse to the reader.

Let’s prove the first statement. Suppose LHS holds; we need to show $g:\mathbf{N}^* \to \overline{\mathbf{R}}$ which takes $n\mapsto x_n, \infty\mapsto \infty$ is continuous at ∞. Indeed, any open subset of $\overline{\mathbf{R}}$ containing ∞ must contain some (L, ∞]. By classical definition, there exists N such that whenever n>N, we have x_n>L. Then $g^{-1}((L,\infty])$ contains $U_{N+1}\cup\{\infty\}$ , an open set containing ∞.

For the second statement, again assume LHS holds. Let $g:\overline{\mathbf{R}} \to \mathbf{R}$ be the function which takes real x to f(x) and ∞ to L; our job is to prove continuity of g at ∞. Let V be an open subset of R containing L, which contains (L-ε, L+ε) for some ε>0. By classical definition, there exists M such that whenever x>M, f(x) lies in (L-ε, L+ε) and hence V. Thus, $(M,\infty] \subseteq f^{-1}(V)$ is an open subset of $\overline{\mathbf{R}}$ containing ∞, and g is continuous at ∞. ♦

The point we’re trying to say is this.

Summary. The various multi-case definitions of limits and convergence can all be subsumed under a single definition by considering various topological spaces.

Basic Properties of Limits

The following is basic.

Proposition. Let f : X-{a} → Y be a function where $\lim_{x\to a} f(x) = b$ . If g : Y → Z is continuous at b, then gf : X-{a} → Z satisfies $\lim_{x\to a} gf(x) = g(b)$ .

Proof.

Define $h:X\to Y$ and $i:X\to Z$ via:

$h(x)=\begin{cases} f(x), &\text{if } x\ne a,\\ b, &\text{if }x=a,\end{cases}$ and $i(x) = \begin{cases} gf(x), &\text{if } x\ne a,\\ g(b), &\text{if }x=a.\end{cases}$

From the condition, we know that h is continuous at a. Since g is continuous at b=h(a), i = gh is also continuous at a. ♦

Corollary. If $x_n \to a$ in the topological space X, and f : X → Y is continuous, then $f(x_n)\to f(a)$ in Y.

Proof

Since $x_n\to a$ , the function $g:\mathbf{N}^* \to X$ which takes $n\mapsto x_n$ has a limit $\lim_{n\to\infty} g(n) = a$ . By the previous proposition, we get $\lim_{n\to\infty} f(g(n)) = f(a)$ and so $f(x_n)\to f(a).$ ♦

Next we shall approach the basic question: is the limit $\lim_{x\to a} f(x)$ unique? Phrased in such generality, the answer is no, as we saw counter-examples even for the case where X is a subset of R.

So let’s first consider sequences.

Theorem. Let X be a topological space. Assume:

for any two distinct points $x, y\in X$ , there exist open subsets U and V, $x\in U$ , $y\in V$ , $U\cap V=\emptyset$ .

Then every sequence $x_n \in X$ has at most one limit in X. A topological space satisfying the above property is said to be Hausdorff.

Thus, the Hausdorff property is a sufficient condition for unique convergence, but it’s known that the condition is not necessary.

Proof.

Suppose a, b are distinct limits of $(x_n)$ . Pick open subsets U, V such that $a\in U$ , $b\in V$ , $U\cap V=\emptyset$ . Then there exists M, N such that (i) whenever n>M, we have $x_n \in U$ , (ii) whenever n>N, we have $x_n\in V$ . The two statements clearly contradict. ♦

One obvious source of Hausdorff topological spaces is via metric spaces, i.e. metric spaces are Hausdorff. Indeed, if x, y are distinct points in a metric space X, then letting ε=d(x, y)/2 > 0, we have $N(x,\epsilon) \cap N(y,\epsilon) = \emptyset$ , for if z satisfies d(x, z) < ε and d(y, z) < ε, then d(x, y) ≤ d(x, z) + d(z, y) < 2ε = d(x, y), which is absurd. In particular, any convergent sequence in a metric space has a unique limit.

For uniqueness of general limits, obviously we have to care about the domain space X.

Theorem. Let f : X-{a} → Y be a map of topological spaces. Assume the following.

Y is Hausdorff.

The singleton set {a} is not open in X.

Then there’s at most one limit for $\lim_{x\to a} f(x)$ .

Proof.

Suppose $b,c\in Y$ are distinct limits. Hence the functions $g_1, g_2:X\to Y$ ,

$g_1(x) = \begin{cases} f(x), &\text{if } x\ne a,\\ b, &\text{if }x=a\end{cases}$ and $g_2(x)=\begin{cases} f(x), &\text{if }x\ne a,\\ c &\text{if }x=a\end{cases}$

are continuous. Pick open subsets U and V of Y, such that $b\in U$ , $c\in V$ and $U\cap V=\emptyset$ . Now $g_1^{-1}(U) = f^{-1}(U)\cup\{a\}$ must contain an open subset which contains a. Same for $g_2^{-1}(V) = f^{-1}(V) \cup\{a\}$ . But since $f^{-1}(U)\cap f^{-1}(V) = f^{-1}(U\cap V)=\emptyset$ , the intersection of these two open subsets is {a}, which contradicts the second condition. ♦

Note: it’s easy to find a counter-example when {a} is open in X. In an earlier article, we saw the following:

Here $X=(-\infty, -1)\cup \{0\}\cup (1, \infty)$ and f : X-{0} → R is defined by f(x)=x. But because {0} is an open subset of X, f(0) can take any value without violating its continuity.

Posted in Notes | Tagged advanced, continuity, convergence, extended reals, Hausdorff, limits, sequences, topology | Leave a comment

Topology: Continuous Maps

Posted on January 31, 2013 by limsup

Continuity in Metric Spaces

Following the case of real analysis, let’s define continuous functions via the usual ε-δ definition.

Definition. Let (X, d) and (Y, d’) be two metric spaces. A function f : X → Y is said to be continuous at $a\in X$ if:

for each ε>0, there exists δ>0 such that whenever $x\in X$ satisfies d(x, a) < δ, we have d’(f(x), f(a)) < ε.

Equivalently, we can replace the condition “whenever … ” by the set-theoretic statement $f(N_d(a, \delta))\subseteq N_{d'}(f(a), \epsilon)$ which is often more convenient.

If f is continuous at every a, we just say that f is continuous.

Theorem. The function f : X → Y is continuous at a iff for any open subset $V\subseteq Y$ containing f(a), there’s an open subset $U\subseteq f^{-1}(V)$ containing a.

In particular, f is continuous if and only if for any open subset $V\subseteq Y$ , the pull-back $f^{-1}(V)\subseteq X$ is open.

Proof.

Suppose f is continuous at a. If V is an open subset of Y containing f(a), then $N_{d'}(f(a),\epsilon)\subseteq V$ for some ε>0. By continuity at a, there exists δ>0 such that $f(N_d(a,\delta))\subseteq N_{d'}(f(a),\epsilon)$ . Hence $N_d(a,\delta) \subseteq f^{-1}(V)$ is an open subset containing a.

For the converse, suppose the stated condition holds. Given ε>0, $V := N_{d'}(f(a),\epsilon)$ is an open subset of Y containing f(a), so there is an open $U\subseteq f^{-1}(V)$ containing a. Since U is open, there’s an open ball $N(a,\delta)\subseteq U$ . It thus follows that $f(N(a,\delta))\subseteq V$ .

For the second statement, suppose f is continuous and $V\subseteq Y$ is open. By what we just proved, each element a of $U:= f^{-1}(V)$ is contained in an open subset $U_a \subseteq X$ which is contained in U. Thus U is a union of open subsets and is hence open.

Conversely, suppose whenever $V\subseteq Y$ is open, so is $U:=f^{-1}(V)\subseteq X.$ Let a be any point in X. From what we just proved, f is continuous at a. ♦

The theorem tells us that continuity is fundamentally a statement on the underlying topologies. In other words, if f : (X, d) → (Y, d’) is a continuous map of metric spaces, then we can replace d or d’ by any topologically equivalent metric and it wouldn’t make any difference.

Next, we prove that the metric itself is continuous (considering this map is so fundamental, it’d be pretty weird if it weren’t).

Proposition. In a metric space (X, d), the map $d:X\times X\to\mathbf{R}$ is continuous.

Proof

First, recall that there’s no fixed way of defining a metric on X × X, so we pick one of the many topologically equivalent ones, say $d_1((x,y), (x',y')) = d(x,x') + d(y,y').$

Let $(x,y)\in X\times X$ . To show continuity at that point, suppose ε>0. Letting δ = ε/2, we see that whenever $d_1((x, y), (x',y')) < \delta$ , we have: d(x, x’) < δ and d(y, y’) < δ and thus the triangular inequality gives:

d(x, y) – d(x’, y’) ≤ d(y’, y) + d(x, x’) < 2δ = ε;
d(x’, y’) – d(x, y) ≤ d(y, y’) + d(x’, x) < 2δ = ε.

Hence, |d(x’, y’) – d(x, y)| ≤ ε. This shows d is continuous at (x, y). ♦

Note

In addition, we had already proved that the standard arithmetic operations on real numbers are continuous. Thus, addition and product $\mathbf{R}\times \mathbf{R}\to \mathbf{R}$ are continuous, as is reciprocal $\mathbf{R}^*\to\mathbf{R}^*.$

Continuity in Topology

Since the notion of continuity can be described by open sets, let’s generalise it to topological spaces.

Definition. Let X and Y be topological spaces. A function f : X → Y on the underlying sets is said to be continuous at $a\in X$ if:

for any open subset $V\subseteq Y$ containing f(a), there’s an open subset $U\subseteq f^{-1}(V)$ containing a.

In particular, f is continuous (at every point of X) if for any open $V\subseteq Y$ , the pullback $f^{-1}(V)$ is open in X.

The following properties of continuity are obvious for any topological spaces X, Y and Z.

The identity map id : X→ X is continuous.
If f : X → Y is continuous at x and g : Y → Z is continuous at f(x), then gf : X → Z is continuous at x.
If T and T’ are both topologies on X, then id : (X, T) → (X, T’) is continuous if and only if T is finer than T’.

From the third property, one sees that if f : X → Y is bijective and continuous, the inverse $f^{-1}$ may not be continuous. Specifically, if T is strictly finer than T’, then the identity map (X, T) → (X, T’) is continuous but (X, T’) → (X, T) is not. Recall that if f : X → Y is bijective, continuous and has a continuous inverse, then we say f is a homeomorphism, in which case we get a bijection between the collection of open subsets of X and that of Y.

To check that a map is continuous, we don’t have to look at every open subset of Y.

Proposition. Let f : X→ Y be a map of topological spaces and $S'\subseteq \mathbf{P}(Y)$ be a subbasis of Y. Then f is continuous iff for any $V\in S'$ , $f^{-1}(V) \subseteq X$ is open.

Proof.

The forward direction is obvious. The converse follows from the fact that the pullback preserves arbitrary intersection and union:

$f^{-1}(W_1 \cap W_2 \cap\ldots \cap W_k) = f^{-1}(W_1) \cap f^{-1}(W_2) \cap\ldots \cap f^{-1}(W_k),$

$f^{-1}(\cup V_i) = \cup_i f^{-1}(V_i).$ ♦

For example, to check that f : X → R is continuous, it only suffices to prove that $f^{-1}((a, \infty))$ and $f^{-1}((-\infty, b))$ are open in X for any real a, b.

Next, we wish to prove that the standard maps are continuous.

Proposition.

If $Y\subseteq X$ is a subspace of a topological space, then the inclusion map $i:Y\hookrightarrow X$ is continuous.

If X and Y are topological spaces, then the projection maps $p_1: X\times Y\to X$ and $p_2 : X\times Y\to Y$ are continuous.

If $X_i$ is a collection of topological spaces and $X=\coprod_i X_i$ is the disjoint union, then every inclusion map $\iota_i : X_i \to X$ is continuous.

Proof

For the first statement, each open $U\subseteq X$ pulls back to $i^{-1}(U) = U\cap Y$ which is open in Y by definition of subspace. For the second statement, each open $U\subseteq X$ pulls back to $p_1^{-1}(U) = U\times Y$ which is open in X × Y by definition of product space. For the last statement, each open subset $U =\coprod_i U_i$ of X pulls back to $\iota_i^{-1}(U) = U_i$ which is open in X_i. ♦

In retrospect, one could even define topological spaces specifically to satisfy the above properties. And one defines the topology “just enough” such that the properties are satisfied. This will be expounded in a separate article.

Finally, we wish to show that continuity of a function depends “locally” on small neighbourhoods.

Theorem. If $f:X \to Y$ is a function of topological spaces and $X=\cup_i U_i$ is a union of open subsets $U_i \subseteq X$ , then

f is continuous if and only if $f|_{U_i} : U_i \to Y$ is continuous for every i.

Proof.

The forward direction is easy: since the inclusion map $\iota_i:U_i \hookrightarrow X$ is continuous, so is the composition $f\circ\iota_i = f|_{U_i}$ . Conversely, suppose each $f|_{U_i}$ is continuous. Let V be an open subset of Y. Then

$f^{-1}(V) = \cup_i (f^{-1}(V) \cap U_i) = \cup_i (f|_{U_i})^{-1}(V).$

Now each $(f|_{U_i})^{-1}(V)$ is open in $U_i$ which is in turn open in X. Thus, $(f|_{U_i})^{-1}(V)$ is open in X. So $f^{-1}(V)$ , being a union of open subsets of X, is open in X. ♦

Exercises

Suppose f : X → Y is a continuous function of topological spaces and $X_1\subseteq X$ and $Y_1\subseteq Y$ are subspaces such that $f(X_1)\subseteq Y_1$ . Prove that the restriction $g := f|_{X_1} : X_1\to Y_1$ is also continuous.
Suppose f : X₁ → Y₁ and g: X₂ → Y₂ are continuous maps. Then the concatenation $h:X_1\times X_2 \to Y_1\times Y_2,$ which takes $(x_1, x_2)\mapsto (f(x_1), g(x_2))$ , is also continuous.
Let Y be a topological space written as a union of subspaces $Y = \cup_i X_i.$ Prove that we get a continuous map $\coprod_i X_i \to Y$ which takes an element $x_i \in X_i$ to the corresponding image in Y.

Answers (Highlight to read)

Let V₁ be an open subset of Y₁. So we have V₁ = V ∩ Y₁ for some open subset V of Y. Then g^-1(V₁) = f^-1(V) ∩ X₁; by continuity of f, f^-1(V) is open in X, so f^-1(V) ∩ X₁ is open in X₁.
Let U₁ and U₂ be open subsets of Y₁ and Y₂ respectively; thus U₁ × U₂ is a basic open subset of Y₁ × Y₂. Then h^-1(U₁ × U₂) = f^-1(U₁) × g^-1(U₂) which is open in X₁ × X₂ since f and g are continuous. Thus the result follows.
Let V be an open subset of Y. The inverse image in disjoint union of X_i is the disjoint union of U_i, where each U_i = X_i ∩ V is open in X_i by definition of subspace.

Posted in Notes | Tagged advanced, continuity, disjoint union topology, homeomorphism, metric spaces, product topology, subspaces, topology | Leave a comment

Topology: Disjoint Unions

Posted on January 29, 2013 by limsup

Disjoint Unions

Let X and Y be topological spaces and $Z := X \coprod Y$ be a set-theoretic disjoint union. We wish to define a topology on Z in a most natural way.

Definition. The topology on $Z := X \coprod Y$ is defined to be:

$T = \{U \coprod V : U \subseteq X, V \subseteq Y \text{ open }\}.$

It’s almost trivial to check that this satisfies the axioms of a topology. E.g. to check T is closed under union of arbitrarily many elements, we have

$\cup_i (U_i \coprod V_i) = (\cup_i U_i)\coprod (\cup_i V_i) \in T.$

Pictorially, the definition for T is quite straight-forward. One imagines placing the two spaces side-by-side and separate them by an insurmountable gulf.

More generally, one can take the disjoint union of an arbitrary collection of topological spaces $X = \coprod_i X_i$ , where the open subsets of U are precisely those of the form $U = \coprod_i U_i$ for open subsets $U_i\subseteq X_i$ .

The bases and subbases of X_i give rise to that of X in a natural manner:

Proposition. Let $X_i$ be a collection of topological spaces and $X := \coprod_i X_i$ .

If $B_{X_i}$ is a basis of $X_i$ for each i, then $B:=\coprod_i B_{X_i}$ is a basis for X.

If $S_{X_i}$ is a basis of $X_i$ for each i, then $S:=\coprod_i S_{X_i}$ is a basis for X.

Proof (of first property)

Let T be the topology for X obtained via disjoint union. Clearly, each $B_{X_i}\subseteq T$ since an open subset of X_i is also open in X. Thus $B\subseteq T.$ Conversely, every open subset U of X is a (disjoint) union of open subsets of X_i, each of which is a union of elements of $B_{X_i}$ . Hence U is a union of elements from B and B is a basis for T. ♦

The proof for the second property is similar and we won’t repeat ourselves. Since the disjoint union can be thought of as “addition” between topological spaces, it shouldn’t surprise us to find the distributive property for product.

Theorem (Distributive Property). Let $X, Y_i$ be a collection of topological spaces. Then:

$X \times (\coprod_i Y_i) \cong \coprod_i (X\times Y_i).$

Proof.

We clearly have a set-theoretical bijection between both sides. It remains to see that the collection of open subsets on both sides correspond.

Now, a basis of the LHS is given by $U\times (\coprod_i V_i)$ for some open subsets $U\subseteq X$ , $V_i\subseteq Y_i$ . But this is just $\coprod_i (U\times V_i)$ and since each $U\times V_i\subseteq X\times Y_i$ is open, so is their (disjoint) union. Thus, the LHS topology is contained in the RHS.

Conversely, a basis of the each term $X\times Y_i$ of the RHS is given by $\{U\times V_i\}$ , where $U\subseteq X$ , $V_i\subseteq Y_i$ are open. By the above proposition, the set $\cup_i \{U\times V_i : U\subseteq X, V_i\subseteq Y_i\}$ is a basis of RHS. Since each element of this basis is open in LHS, the RHS topology is contained in the LHS. ♦

Note.

The fact that multiplication is distributive over an infinite sum should not be taken for granted. Indeed, the corresponding case for algebra requires careful consideration of convergence on both sides.

Corollary. Let X be a topological space. If I is an index set, endowed with the discrete topology, then X × I is isomorphic to a disjoint union of $\{X_i\}_{i\in I}$ , where each $X_i \cong X$ .

Proof.

Consider the singleton set {i} with the obvious topology. Now I is the disjoint union of {i}, over $i\in I$ . Hence:

$X \times I \cong X\times \left(\coprod_{i\in I} \{i\}\right)\cong \coprod_{i\in I} \left(X \times \{i\}\right),$

where each component in the rightmost expression is isomorphic to X. ♦

Metrisability of Disjoint Unions

Next, we’re interested in the following question.

If X and Y are metrisable topological spaces, is $Z = X\coprod Y$ metrisable too?

On an intuitive level, the space Z comprises of components X and Y such that points $x\in X, y\in Y$ in different components are extremely far from each other. But one can’t set the distance between them to be infinite, since the metric can only take real values. [ One might understandably think of extending the metric function to include infinite distances, or even “nonstandard reals”, but that opens up a whole new can of worms. ]

However, it turns out that if X is a metrisable topological space, then we can pick a metric such that d(x, y) < 1 for any x, y in X.

Proposition. If (X, d) is a metric space, then the function $d' : X\times X\to \mathbf{R}$ , $d'(x, y) = \frac{d(x,y)}{1+d(x,y)} < 1$ , is also a metric.

Proof.

It suffices to show that if r, s, t are non-negative real numbers satisfying $r+s\ge t$ , then $\frac{r}{1+r}+\frac{s}{1+s} \ge \frac{t}{1+t}$ . Since $x\mapsto \frac{x}{1+x}$ is a strictly increasing function, we may assume t = r+s. Then clearing denominators gives:

$r(1+s)(1+r+s)+s(1+r)(1+r+s) - t(1+r)(1+s) = r^2 s + s^2 r + 2rs \ge 0.$ ♦

Hence, if each X_i is a metrisable topological space, then we can pick a metric d_i which is bounded by 1. [ We say that a metric d on X is bounded by B if d(x, y) ≤ B for any x, y in X. ] Now we just define a metric on the disjoint union $X := \coprod_i X_i$ via:

for $x_i\in X_i, x_j \in X_j$ , we have $d(x_i, x_j) = \begin{cases}d_i(x_i, x_j), &\text{ if } i=j,\\ 2,&\text{ if } i\ne j.\end{cases}$

Let’s summarise what we’ve learnt in the last few paragraphs.

If X is a metrisable metric space, then we may pick a metric which is bounded by 1. In other words, the condition that a metric is bounded is not a topological invariant.
If each X_i is a metrisable topological space, then so is their disjoint union.

Connected Spaces

On the other hand, let’s start with some generic topological space X. We’d like to partition X as a disjoint union of topological subspaces. Note that if $X:= \coprod_i X_i$ , then each $X_i\subseteq X$ is an open subset. The following theorem shows that this suffices to characterise X as a disjoint union.

Theorem. Suppose $X = \cup_i U_i$ , where each $U_i\subseteq X$ is open and $U_i\cap U_j=\emptyset$ for any i≠j. Then $X \cong \coprod_i U_i$ .

Example

Consider the subset $Y=(-\infty, -1]\cup\{0\}\cup [1,\infty)$ of R. Then $Y = U \cup V \cup W$ where $U=(-\infty, -1]$ , $V = \{0\}$ and $W = [1,\infty)$ . Each of these sets is open in Y since

$U=(-\infty, -1/2)\cap Y$ , $V=(-1/2,+1/2)\cap Y$ and $W=(1/2,\infty)\cap Y.$

Since they form a disjoint union of Y, we have $Y \cong U\coprod V\coprod W$ .

Proof.

From the inclusion maps $U_i\hookrightarrow X$ , we get a set-theoretic map $f:\coprod_i U_i \to X$ . The map is surjective since X is the union of U_i‘s. It is injective since any two distinct U_i‘s don’t intersect.

To show that f is a homeomorphism, each open subset of $\coprod_i U_i$ is of the form $\coprod_i V_i$ for open $V_i\subseteq U_i$ . Since $U_i\subseteq X$ is also open, we see that $V_i\subseteq X$ is open. Thus $f(\coprod V_i)=\cup_i V_i$ is open in X.

Conversely, if V is an open subset of X, then $f^{-1}(V) = \coprod_i V_i$ , where $V_i = V\cap U_i$ is open in U_i. Thus $f^{-1}(V) = \coprod_i V_i$ is open in $\coprod_i U_i.$ ♦

Note that in the theorem, each U_i is also closed since it’s complement is $\cup_{j\ne i} U_j$ which is open. Thus the U_i are clopen subsets of X: if X is a disjoint union of more than one components, than each component is a clopen subset.

Conversely, suppose $U\subseteq X$ is clopen such that $U\ne X,\emptyset$ . Then $X = U \cup (X-U)$ satisfies the conditions of the above theorem, so $X \cong U \coprod (X-U)$ . Hence, we have the following.

Corollary. If X has no non-trivial clopen subsets, then we cannot write $X \cong Y_1 \coprod Y_2$ in a non-trivial manner. When that happens, we say that X is connected.

In a nutshell, connected spaces are those which cannot be broken up any further into disjoint unions. We’ll be looking at these in greater detail later.

Posted in Notes | Tagged advanced, bases, connected spaces, disjoint union topology, metrisable topology, product topology, sub-bases, topology | Leave a comment

Topology: Product Spaces (I)

Posted on January 27, 2013 by limsup

In this article, we consider the product of two topological spaces. To motivate our definition, we first begin with metric spaces (X, d_X) and (Y, d_Y). Letting Z = X × Y be the set-theoretic product, we wish to define a metric on Z from d_X and d_Y. Here’re some possibilities, all of which seem rather reasonable. For any $(x,y), (x',y')\in X\times Y$ , let:

$d((x,y), (x',y')) = \sqrt{ d(x,x')^2 + d(y,y')^2}$ ;
$d_1((x,y), (x', y')) = d_X(x,x') + d_Y(y,y')$ ;
$d_\infty((x,y), (x',y')) = \max(d_X(x,x'), d_Y(y,y'))$ .

It’s not hard to prove that all the above are metrics, but which one to choose? It turns out all these are topologically equivalent anyway. E.g. to show equivalence of d and d₁, note that for any $p, p'\in X\times Y$ ,

$d(p, p') \le d_1(p, p')$ and $d_1(p, p') \le \sqrt 2 \cdot d(p, p').$

Since this is not the main point, we’ll leave the details of the proof and the remaining cases to the reader. Instead, the point we wish to make is the following:

Theorem. The above metrics all define the topology T with basis given by

$B= \{U\times V: U \text{ open in } X, V \text{ open in } Y\}.$

Proof.

Since we can pick any metric and use the corresponding open balls as a basis, let’s pick:

$\begin{aligned}N_{d_\infty}( (x,y),\epsilon) &= \{(x',y')\in X\times Y: d_X(x,x')<\epsilon \text{ and } d_Y(y,y')<\epsilon\}\\ &= N(x,\epsilon) \times N(y,\epsilon).\end{aligned}$

Let B’ denote this collection of open balls. Clearly, B’ is contained in B. Conversely, each element of B is obviously a union of elements of B’. Thus, they generate the same topology. ♦

So let’s generalise the definition of product space to arbitrary topological spaces.

Definition. If X and Y are topological spaces, then the corresponding topology on X × Y is defined by the basis

$B= \{U\times V: U \text{ open in } X, V \text{ open in } Y\}.$

Note that in any topological space, our B is truly a basis since (i) the union of all elements of B is X × Y (in fact, X × Y lies in B), and (ii) the intersection of (U × V) and (U’ × V’) is (U ∩ U’) × (V ∩ V’) which is in B. So the definition is valid.

Note. If $C\subseteq X, D\subseteq Y$ are closed subsets, then C × D is closed in X × Y, because its complement is $((X-C)\times Y) \cup (X\times (Y-D))$ , which is a union of two open subsets.

Relationship with Bases and Subbases

If we’re given bases or subbases of X and Y, then these can be used to define a corresponding basis or subbasis of X × Y.

Theorem. Let X and Y be topological spaces.

If B_X and B_Y are given bases of X and Y respectively, then $B=\{U\times V: U\in B_X, V\in B_Y\}$ is a basis of X × Y.

If S_X and S_Y are given subbases of X and Y respectively, then $S=\{U\times Y:U\in S_X\} \cup \{X\times V: V\in S_Y\}$ is a subbasis of X × Y.

Proof.

For the first statement, we already saw that $B' = \{U\times V: U \text{ open in } X, V\text{ open in } Y\}$ is a basis of X × Y. But since B_X and B_Y are bases of X and Y, we can write $U = \cup_i U_i$ , $V =\cup_j V_j$ for some $U_i\in B_X, V_j\in B_Y$ . This gives $U\times V = \cup_{i,j} (U_i\times V_j)$ so every element of B’ is expressible as a union of elements of B.

For the second, suppose finite intersections of S_X (resp. S_Y) give the basis B_X (resp. B_Y). Hence, if $U\in B_X, V\in B_Y$ , we can write $U = U_1\cap\ldots\cap U_m$ and $V=V_1\cap \ldots\cap V_n$ for some $U_i \in S_X$ , $V_j\in S_Y$ . This gives $U\times V = (U\times Y)\cap (X\times V) = \left(\cap_i (U_i\times Y)\right) \cap \left(\cap_j (X\times V_j)\right).$ Thus every element of the collection $B =\{U\times V: U\in B_X, V\in B_Y\}$ is a finite intersection of elements of S; by the first statement, S is a subbasis for X × Y. ♦

Consistency of Successive Products

Next, let’s consider some basic properties on commutativity and associativity of products. Recall that we mentioned two topological spaces X and Y are homeomorphic if there’s a bijective map f : X → Y, such that a subset U of X is open if and only if f(U) is open in Y.

Proposition. Let X, Y, Z be topological spaces. Then the following spaces are homeomorphic.

$X\times Y\cong Y\times X$ , via $(x,y)\mapsto (y,x)$ ;

$X\times (Y\times Z)\cong (X\times Y)\times Z$ , via $(x, (y,z)) \mapsto ((x,y),z)$ .

Proof.

The first property is obvious since the map takes U × V to V × U, for open subsets $U\subseteq X$ , $V\subseteq Y$ . For the second property, since B = {U × V : U open in X, V open in Y} is a basis for X × Y, the above theorem tells us that:

{(U × V) × W : U open in X, V open in Y, W open in Z}

is a basis for (X × Y) × Z. Likewise, the collection of U × (V × W) is a basis for X × (Y × Z). Since the above map takes (U × V) × W to U × (V × W), this gives a homeomorphism. ♦

From the proof, it follows that for the topology on X × Y × Z, one can take a basis comprising of U × V × W, for open subsets $U\subseteq X, V\subseteq Y, W\subseteq Z.$ Also, given a finite number of topological spaces $X_1, \ldots, X_k$ , one can unreservedly take their product $X_1 \times \ldots \times X_k$ since product of topological spaces is commutative and associative.

Consistency with Subspace

Finally, we have the following.

Proposition. Let X, Y be topological spaces and $X' \subseteq X$ , $Y'\subseteq Y$ be subsets. One can define a topology on X’ × Y’ in two distinct ways.

Take subspace topologies on X’ and Y’, then form the product topology X’ × Y’.

Take the product topology on X × Y, then the subspace topology on X’ × Y’.

These two are identical.

Proof.

For the first case, the subspace topology on X’ is given by {U ∩ X’ : U open in X} and that on Y’ is given by {V ∩ Y’ : V open in Y}. Hence, the product topology on X’ × Y’ is given by the basis {(U ∩ X’) × (V ∩ Y’)}, for open subsets U of X and V of Y.

For the second case, the product topology on X × Y has basis U × V for open subsets U of X and V of Y. From our previous article, a basis on X’ × Y’ can thus be given by the collection of:

(U × V) ∩ (X’ × Y’) = (U ∩ X’) × (V ∩ Y’)

for open subsets U of X and V of Y. ♦

Examples

The product of two (or finitely many) discrete topological spaces is still discrete. We’ll see later that this is not true for an infinite product of discrete spaces.
The product of Rⁿ and R^m, with topology given by the usual Euclidean metric, is R^n+m with the same topology. In particular, each Rⁿ has the product topology of n copies of R.
If X and Y are metrisable, then the initial paragraphs of this article show that X × Y is metrisable.
Suppose X is any topological space and Y = {1, 2} with the discrete topology. Then the picture of X × Y is that of two identical copies of X. We’ll expound on disjoint unions in the next article.
Let X = S¹, the set of points (x, y) in R² satisfying x² + y² = 1. Then the space X × X is called a torus.

Posted in Notes | Tagged advanced, metric spaces, product topology, subspaces, topology, torus | Leave a comment

The Box Topology

The Product Topology

Properties of Product Topology

Applications of the Universal Property

Interior and Closure

Product of Metric Spaces

Properties of Interior

Further Properties

Comparing Interior and Closure

Properties of Closure

Unions and Intersections of Closures

Further Properties

Solution

Uniform Continuity

Summary

The Case of Metric Spaces

The Case of Topological Spaces

One Final Result

Basic Properties of Limits

Continuity in Metric Spaces

Continuity in Topology

Disjoint Unions

Metrisability of Disjoint Unions

Connected Spaces

Relationship with Bases and Subbases

Consistency of Successive Products

Consistency with Subspace

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