Topology: Product Spaces (II)

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).