## 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 are disjoint non-empty closed subsets, then we can find open subsets U, V of X such that:

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 Then the distance from x to C:is positive.

**Proof of Lemma**.

Since and *X*–*C* is open, there is an ε>0 such that It follows that any *y* in *C* satisfies *d*(*x*, *y*) ≥ ε so ♦

Now we’re ready to prove the theorem.

**Proof of Theorem**.

For each , we have *d*(*x*, *D*) > 0. Take the union of all over and call this set *U*; this is an open set containing *C*. Likewise, take the union of all over all 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 and for some This gives:

But by definition *d*(*x*, *y*) ≥ *d*(*x*, *D*) and *d*(*y*, *x*) ≥ *d*(*y*, *C*), 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 we can find an open subset U of X containing x but not y.T2: for any distinct points we can find open subsets U, V of X, such thatT3: T1 and regular (X isregularif, for any point and closed subset C of X not containing x, we can find open subsets U, V of X, such that ).T4: T1 and normal (X isnormalif, for any disjoint closed subsets we can find open subsets U, V of X such that ).

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. Hence *X*–*C* is open.

(←) If *x*, *y* 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 there’s an open subset *V* of *X* satisfying

Now suppose *X* is locally compact and Hausdorff. Let *U* be an open subset of *X* containing *x*. By theorem 1 here, *x* 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 *C*, *D* be disjoint closed subsets of *X*. For each since *X* is T3, we can pick open subsets:

such that

Now the collection of {*V _{y}*} covers

*D*. Since

*D*is a closed subset of compact

*X*, it’s compact as well, so there’s a finite subcover Let:

Since *U _{y}* and

*V*are disjoint for each

_{y}*y*, so are

*U*and

*V*. We get: and ♦

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 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 Denote which is closed in *X*. Since

we see that *C* doesn’t contain *x*. Since *X* is regular, we can enclose *x*, *C* in disjoint open subsets *U*, *V* of *X*. Then and 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 is a collection of topological spaces which satisfy T1 (resp. T2, T3), then so is their product

**Proof**.

(T1). Let If each *X _{i}* is T1, then the singleton set is closed. Since a product of closed subsets is closed, it follows that is also closed. Hence,

*X*is T1.

(T3) Let and be a closed subset of *X* not containing **x**. Then *X*–*C* is an open subset of *X* containing **x** so it contains a basic open subset:

for some open

where equality holds for Then for each *i*. For each since *X _{i}* is regular, there’s an open subset

*V*of

_{i}*X*such that For the remaining we set

_{i}*V*:=

_{i}*X*.

_{i}Now is open and:

♦

**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 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

Armed with this, first define and For the first step, let be open and be closed such that

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

For each 0 ≤ *r*/2^{k }< 1, given the inclusion find open and closed such that:

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

Now define the function:

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

Let for *b*>0. Now the inf of a set is less than *b* iff some element of the set is less than *b*, so is open.

Let for some *b*≥0. Now if *t*>*b* and then *f*(*x*)<*t* so This means Conversely, if *f*(*x*) > *b*, then there’re dyadic *t*, *u* such that *f*(*x*) > *t* > *u* > *b*. Then and so Hence is closed and 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 issecond-countableif 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 (*a*, *b*) 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 be disjoint closed subsets. For each , since *X* is T3, we can find an open subset *U _{x}* containing

*x*such that its closure is disjoint from

*D*. Since

*X*has a countable basis and we can find a countable subcover

with each

Similarly, we’ll cover *D* via with each Now define:

and

Each is open and Thus and are disjoint open sets containing *C* and *D* respectively.

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

Pick a countable basis. For any *U*, *V* in this basis satisfying Urysohn’s lemma allows us to pick such that *f*(cl(*U*)) = {0} and *f*(*X*–*V*) = {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 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 Now there’s some *f* we picked such that *f*(cl(*U*)) = {0} and *f*(*X*–*V*) = {1}. Hence *f*(*x*)=0 and *f*(*C*)={1}.

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

Let be the continuous functions in step 2. This gives a map which takes Since is continuous for each *n*, by the universal property of products, *f* is also continuous. We claim that *X* has the subspace topology from [0, 1]^{N}.

First if *x*≠*y*, then there’s an *f _{n}*:

*X*→ [0, 1] such that

*f*(

_{n}*x*)=0 and

*f*(

_{n}*y*)=1. So

*f*(

*x*)≠

*f*(

*y*) and

*f*is injective.

Next we need to show: if is open, then for some open subset We may assume *U* is non-empty and in our countable basis. For any *x* in *U*, pick an *f _{n}*:

*X*→ [0, 1] such that

*f*(

_{n}*x*)=0 and

*f*(

_{n}*X-U*)={1}. Then

and the union of all such gives an open subset such that

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. ♦