## Motivation

The separation axioms attempt to answer the following.

Question. Given a topological space X, how far is it from being metrisable?

We had a hint earlier: all metric spaces are Hausdorff, i.e. distinct points can be separated by two disjoint open subsets. But that’s only part of the story. Metric spaces, in fact, satisfy more.

Theorem 1. If (X, d) is a metric space and $C, D\subset X$ are disjoint non-empty closed subsets, then we can find open subsets U, V of X such that:

$U\supseteq C,\ V\supseteq D,\ U\cap V=\emptyset.$

To prove this, let’s have a little lemma. [ Note: this will be important later; it pays to take note of it now. ]

Lemma. Let C be a closed subset of a metric space (X, d) and $x\in X-C.$ Then the distance from x to C:

$d(x, C) := \inf\{ d(x,y) : y\in C\}$

is positive.

Proof of Lemma.

Since $x\in X-C$ and XC is open, there is an ε>0 such that $N(x,\epsilon) \subseteq X-C.$ It follows that any y in C satisfies d(xy) ≥ ε so $d(x,C)\ge \epsilon.$ ♦

Now we’re ready to prove the theorem.

Proof of Theorem.

For each $x\in C$, we have d(xD) > 0. Take the union of all $N(x, \frac 1 2 d(x,D))$ over $x\in C$ and call this set U; this is an open set containing C. Likewise, take the union of all $N(y, \frac 1 2 d(y,C))$ over all $y\in D$ and call this set V, which is an open set containing D.

We claim U and V are disjoint. If not, z lies in both $N(x, \frac 1 2 d(x,D))$ and $N(y, \frac 1 2 d(y,C))$ for some $x\in C, y\in D.$ This gives:

$d(x,z) < \frac 1 2 d(x,D),\ d(y,z) < \frac 1 2 d(y,C) \implies d(x,y)< \frac 1 2(d(x,D) + d(y,C)).$

But by definition d(xy) ≥ d(xD) and d(yx) ≥ d(yC), which is a contradiction. ♦

## Separation Axioms

Let X be any topological space now. The following labels are given to X if it satisfies the corresponding condition:

Definition.

• T1 : for any distinct points $x,y\in X,$ we can find an open subset U of X containing x but not y.
• T2 : for any distinct points $x,y\in X,$ we can find open subsets U, V of X, such that $x\in U, y\in V, U\cap V=\emptyset.$
• T3 : T1 and regular (X is regular if, for any point $x\in X$ and closed subset C of X not containing x, we can find open subsets U, V of X, such that $x\in U, C\subseteq V, U\cap V=\emptyset$).
• T4 : T1 and normal (X is normal if, for any disjoint closed subsets $C, D\subseteq X,$ we can find open subsets U, V of X such that $C\subseteq U, D\subseteq V, U\cap V=\emptyset$).

Note that T2 is just the Hausdorff condition. Theorem 1 thus says: a metric space satisfies all the separation axioms.

Let’s examine these axioms one at a time.

Proposition 2 (T1). The space X is T1 if and only if all singleton sets {x} are closed.

Proof.

(→) If X is T1, let C = {x}. For any y outside x, the T1 axiom tells us there’s an open subset V containing y but not x, i.e. $y\in V\subseteq X-C.$ Hence XC is open.

(←) If xy are distinct points in X, then X-{y} is open and it contains x but not y. ♦

Note.

This proposition explains why, for T3 and T4, we have to specify the condition of T1 in addition to regularity / normality. This ensures that each {x} is closed and so T4 → T3 → T2 → T1, in increasing order of generality.

Furthermore, none of the implications is reversible. E.g. the topology (X, T) with X = {1, 2} and T = {{}, {1}, {1, 2}} satisfies T1 but not T2. For other examples, the reader may search the “spacebook” which is based on the famous book “Counterexamples in Topology“.

Proposition 3 (T3). A locally compact Hausdorff space is T3.

Proof.

Note that a space X is regular if and only if for any open subset U of X and $x\in U,$ there’s an open subset V of X satisfying $x\in V\subseteq \text{cl}(V)\subseteq U.$

Now suppose X is locally compact and Hausdorff. Let U be an open subset of X containing x. By theorem 1 herex is contained in an open subset V of X such that cl(V) is compact and contained in U. ♦

The converse is not true: there’re T3 spaces which are not locally compact. Also, there’re locally compact spaces which are not T4.

Proposition 4 (T4). A compact Hausdorff space is T4.

Proof

Suppose X is compact; then it’s locally compact so it’s T3. Let CD be disjoint closed subsets of X. For each $y\in D,$ since X is T3, we can pick open subsets:

$U_y\supseteq C,\ V_y\ni y,\$ such that $U_y\cap V_y = \emptyset.$

Now the collection of {Vy} covers D. Since D is a closed subset of compact X, it’s compact as well, so there’s a finite subcover $\{V_{y_k}\}\subseteq \{V_y\}.$ Let:

$U := \cap_{k=1}^n U_{y_k},\ V := \cup_{k=1}^n V_{y_k}.$

Since Uy and Vy are disjoint for each y, so are U and V. We get: $U = \cap_k U_{y_k} \supseteq C$ and $V = \cup_k V_{y_k} \supseteq D.$ ♦

Clearly, not every T4 space is compact since there’re metric spaces which are not compact.

## Combined Properties

The separation axioms satisfy the following properties. We skip the case of T2 since it had been done earlier.

Theorem 3. If Y is a subspace of X, and X satisfies T1 (resp. T2, T3), then so does Y.

Proof.

(T1). If $y\in Y,$ then {y} is closed in X since X satisfies T1. So {y} ∩ Y = {y} is closed in Y.

(T3). Let D be a closed subset of Y and $x\in Y-D.$ Denote $C = \text{cl}_X(D)$ which is closed in X. Since

$D = \text{cl}_Y(D) = \text{cl}_X(D) \cap Y = C\cap Y$

we see that C doesn’t contain x. Since X is regular, we can enclose xC in disjoint open subsets U, V of X. Then $D = C\cap Y\subseteq V\cap Y$ and $x\in U\cap Y$ where U ∩ Y and V ∩ Y are disjoint open subsets of Y. ♦

Exercise.

Prove that if Y is a closed subset of X and X satisfies T4, then so does Y. [ A subspace of a T4 space is not necessarily T4, but counter-examples are hard to find. ]

Theorem 4. If $\{X_i\}$ is a collection of topological spaces which satisfy T1 (resp. T2, T3), then so is their product $X:= \prod_i X_i.$

Proof.

(T1). Let $(x_i) \in X.$ If each Xi is T1, then the singleton set $\{x_i\}\subset X_i$ is closed. Since a product of closed subsets is closed, it follows that $\{(x_i)\} \subset X$ is also closed. Hence, X is T1.

(T3) Let $\mathbf{x} := (x_i) \in X$ and $C\subseteq X$ be a closed subset of X not containing x. Then XC is an open subset of X containing x so it contains a basic open subset:

$\mathbf{x} \in \prod_i U_i \subseteq X_i,$ for some open $U_i\subseteq X_i,$

where equality holds for $i\ne i_1,\ldots, i_n.$ Then $x_i\in U_i$ for each i. For each $i=i_1, \ldots, i_n,$ since Xi is regular, there’s an open subset Vi of Xi such that $x_i\in V_i\subseteq \text{cl}(U_i)\subseteq U_i.$ For the remaining $i\ne i_1, \ldots, i_n$ we set Vi := Xi.

Now $\prod_i V_i$ is open and:

$\mathbf{x}=(x_i) \in \prod_i V_i \subseteq \text{cl}(\prod_i V_i) = \prod_i \text{cl}(V_i) \subseteq \prod_i U_i.$ ♦

Note

Unfortunately, a product of T4 spaces is not necessarily T4.

## Urysohn’s Lemma

Urysohn’s Lemma. Suppose X is normal. Then for any disjoint non-empty closed subsets C, D of X, there is a continuous map $f:X\to [0,1]$ such that f(C)={0} and f(D)={1}.

[ In particular, this holds for T4 spaces. ]

Note.

If X is connected, then f must be surjective since f(X) is a connected subset of [0, 1] containing 0 and 1. So we can label points of X continuously from 0 to 1 without gaps!

Proof.

[ All closed/open subsets are assumed with reference to X. ]

Note that normality is equivalent to: for any closed C and open U containing it, there’s an open V and closed D such that $C\subseteq V\subseteq D\subseteq U.$

Armed with this, first define $C_0 := C$ and $V_1 := X-D \supseteq C_0.$ For the first step, let $V_{1/2}$ be open and $C_{1/2}$ be closed such that $C_0 \subseteq V_{1/2} \subseteq C_{1/2}\subseteq V_1.$

Now proceed inductively, suppose we have a sequence of sets (each “C” is closed and “V” is open):

$C_0 \subseteq V_{1/2^k} \subseteq C_{1/2^k} \subseteq \ldots \subseteq V_{r/2^k} \subseteq C_{r/2^k}\subseteq V_{(r+1)/2^k} \subseteq \ldots \subseteq V_1.$

For each 0 ≤ r/2< 1, given the inclusion $C_{r/2^k} \subseteq V_{(r+1)/2^k},$ find open $V_{(2r+1)/2^{k+1}}$ and closed $C_{(2r+1)/2^{k+1}}$ such that:

$C_{r/2^k} \subseteq V_{(2r+1)/2^{k+1}}\subseteq C_{(2r+1)/2^{k+1}} \subseteq V_{(r+1)/2^k}.$

This creates a corresponding chain for r/2k+1. In this way, we create a series of $V_r$ and $C_r$ for each dyadic rational number (i.e. of the form r/2k) between 0 and 1.

Now define the function:

$f:X \to [0,1], \quad f(x) = \inf\{t : x\in V_t \text{ or } t=1\}.$

Clearly f(C) = {0} and f(D) = {1}. It remains to show f is continuous.

Let $U_b=f^{-1}([0, b))$ for b>0. Now the inf of a set is less than b iff some element of the set is less than b, so $U_b=\cup_{t is open.

Let $C=f^{-1}([0, b])$ for some b≥0. Now if t>b and $x\in C,$ then f(x)<t so $x\in V_t \subseteq C_t.$ This means $C\subseteq \cap_{t>b} C_t.$ Conversely, if f(x) > b, then there’re dyadic tu such that f(x) > t > ub. Then $x\not\in V_t$ and so $x\not \in C_u \implies x\not\in \cap_{t>b} C_t.$ Hence $C = \cap_{t>b} C_t$ is closed and $f^{-1}((b, 1])$ is open.

Since [0, b) and (b, 1] form a subbasis for [0, 1], we’re done. ♦

## Urysohn’s Metrisation Theorem

The following result shows a sufficient condition for metrisability.

Definition. A topological space X is second-countable if it has a basis B with countably many elements.

For example, even though set-theoretically R is uncountable, its topology is second-countable since it has a basis comprising of open intervals (ab) for rational a<b.

Urysohn’s Metrisation Theorem. If X is a T3 space which is second-countable, then it’s metrisable.

The proof follows a couple of steps.

Step 1 : T3 + second-countable implies T4.

Let $C, D\subseteq X$ be disjoint closed subsets. For each $x\in C$, since X is T3, we can find an open subset Ux containing x such that its closure is disjoint from D. Since X has a countable basis and $C \supseteq \cup_x U_x,$ we can find a countable subcover

$C\supseteq \cup_{n=1}^\infty U_{x_n},$ with each $\text{cl}(U_{x_n})\cap D = \emptyset.$

Similarly, we’ll cover D via $D\supseteq \cup_{n=1}^\infty V_{y_n}$ with each $\text{cl}(V_{y_n})\cap C =\emptyset.$ Now define:

$U_n' = U_{x_n} -(\text{cl}(V_{y_1}) \cup \ldots \cup \text{cl}(V_{y_n}))$ and $V_n' = V_{y_n} -(\text{cl}(U_{x_1})\cup \ldots \cup \text{cl}(U_{x_n})).$

Each $U_n', V_n'$ is open and $U_m'\cap V_n'=\emptyset.$ Thus $U:=\cup_n U_n'$ and $V:=\cup_n V_n'$ are disjoint open sets containing C and D respectively.

Step 2: Find countably many continuous fnX → [0, 1] such that for any x in X and closed subset C not containing x, there’s an fn such that fn(x)=0, fn(C)={1}.

Pick a countable basis. For any UV in this basis satisfying $U\subseteq \text{cl}(U)\subseteq V,$ Urysohn’s lemma allows us to pick $f: X\to [0, 1]$ such that f(cl(U)) = {0} and f(XV) = {1}. Let’s show that this (countable) collection of f is good enough.

Indeed, for any x in X and closed subset C not containing x, we can pick an open subset V such that $x\in V\subseteq \text{cl}(V)\subseteq X-C.$ Replacing V by a basic open subset, we may assume V lies in the basis. Repeating this process, we can find a basic open subset U such that $x\in U\subseteq \text{cl}(U)\subseteq V.$ Now there’s some f we picked such that f(cl(U)) = {0} and f(XV) = {1}. Hence f(x)=0 and f(C)={1}.

Step 3: Embed X into countably many copies of R.

Let $f_1, f_2, \ldots :X\to [0,1]$ be the continuous functions in step 2. This gives a map $f:X \to [0,1]^\mathbf{N}$ which takes $x\mapsto (f_1(x), f_2(x), \ldots).$ Since $\pi_n\circ f = f_n$ is continuous for each n, by the universal property of productsf is also continuous. We claim that X has the subspace topology from [0, 1]N.

First if xy, then there’s an fnX → [0, 1] such that fn(x)=0 and fn(y)=1. So f(x)≠f(y) and f is injective.

Next we need to show: if $U\subseteq X$ is open, then $U = f^{-1}(V)$ for some open subset $V\subseteq [0,1]^\mathbf{N}.$ We may assume U is non-empty and in our countable basis. For any x in U, pick an fnX → [0, 1] such that fn(x)=0 and fn(X-U)={1}. Then

$x\in f^{-1}(\pi_n^{-1}([0, 1)\ )) = f_n^{-1}([0, 1)) \subseteq U$

and the union of all such $\pi_n^{-1}([0, 1))$ gives an open subset $V\subseteq [0, 1]^\mathbf{N}$ such that $U=f^{-1}(V).$

Thus X has the subspace topology from [0, 1]N. We already saw that the product of countably many metrisable spaces is metrisable, so we’re done. ♦

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