## Thoughts about separation (Hausdorff)

In this post we go one step further and discuss Hausdorff spaces, a stronger property than either Kolomogorov or .

**Hausdorff (): **Let be a topological space, then we call *Hausdorff* (or ) if given any distinct there exists open sets such that and

Hausdorff spaces are the first real interesting spaces since we can say many important and profound things concerning mappings, etc.

It is clear first-off that every Hausdorff space is . We begin by discussing some alternative definitions for when a space is Hausdorff.

**Theorem: **Let be a topological space, then is Hausdorff if and only if

**Proof:**

: Clearly . Now, let be distinct from . By assumption there exists neighborhoods such that . It is relatively easy to prove (see Random Questions 1) that since that . It follows then that . Since was arbitrary the conclusion follows.

: Conversely, let be distinct. Since we see there exists some neighborhood of such that . It follows that are sets that contain respectively but not the other.

The conclusion follows.

This next theorem ends up being very important later on.

**Theorem:** Let be a topological space. Define to be . Then is Hausdorff if and only if is closed in .

**Proof:**

: Suppose that is Hausdorff, then we prove that is closed by proving its compliment is open. So, let then . And since is Hausdorff there exists there exists disjoint open sets which contain and respectively. Clearly but since we see that and so . The conclusion follows.

: Conversely, suppose that is closed. Then, is open, and so given any we see that . Clearly this implies though that (otherwise if they shared a point then ) from where it follows that are disjoint open sets containing respectively. The conclusion follows.

Now we begin by showing that Haudorff spaces react well with compact subspaces.

**Theorem:** Let be a Hausdorff space and compact, then for any there exists an open sets such that and .

**Proof:** For each there exists disjoint corresponding sets . Clearly then is an open cover for . But, by assumption it must have a finite subcover . It clearly follows that and and that these two open sets are disjoint. The conclusion follows. .

**Corollary:** A compact subspace of a Hausdorff space must be closed.

Another very nice property is the next.

**Theorem:** Let be a Hausdorff space and let be compact subspaces. Then there exists disjoint open sets such that .

**Proof:** By the previous theorem for each there exists some which are open, disjoint, and contain respectively. It is clear that is an open cover for latex and so by it’s compactness there exists a finite subcover and so and are the necessary open sets.

An obvious question is “Is the product of Hausdorff spaces Hausdorff?”. The answer is…:

**Theorem:** Let be a non-empty collection of Hausdorff spaces then is Hausdorff under the product topology.

**Proof:**Let be distinct. Then, it follows that for some . But, since is Hausdorff there exists disjoint open sets such that . It clearly follows that where , are disjoint open sets containing respectively. The conclusion follows.

Hausdorffness is also hereditary (in the sense that subspaces inherit it).

**Theorem:** Let be a Hausdorff space and a subspace. Then, is Hausdorff.

**Proof: **Let be distinct, then clearly they are distinct in . Thus, by assumption we have that there exists disjoint open sets such that are in them respectively. Clearly then which are open in and latex . The conclusion follows.

Now, a nice theorem which wraps up the motivation of the very first question in the last thread is the following:

**Theorem:** Let be a topological space, then if is Hausdorff the limit of a sequence is unique.

**Proof:**

: Suppose that be a sequence in such that . Now, let by assumption there exists disjoint open sets which contain respectively. By assumption though we have that all but finitely many of latex are in and so at most finitely many are in . Thus, . The conclusion follows.

We now discuss some nice qualities of Hausdorff spaces and maps.

**Theorem: **Let be continuous and suppose that is Hausdorff, then is closed.

**Proof:** By an earlier discussion we know that given by is continuous. But, it is relatively easy to see that and since is Hausdorff we know that is closed we know by ‘s continuity that is closed. the conclusion follows.

We know discuss quite a fruitful subject. But, first a defition

**Kernel:** Let be a mapping from topological spaces . Define .

*Remark:* For those familiar with set theory this can be interpreted as the quotient set where

**Theorem:** If is continuous and Hausdorff, then is closed.

**Proof: **Once again, by previous argument we know that given by is continuous. But, it is relatively easy to see that . Now, since is Hausdorff we know that is closed and the argument is finished by noting ‘s continuity.

**Theorem:** Let be a open surjection and closed, then is Hausdorff.

**Proof:** Define by . Now, it was proved earlier that the product of open sujrective maps is openand surjective and since is open it follows that is open and surjective. So consider the following lemma

**Lemma:** .

**Proof:** Let , then since for some . But, this implies that and so .

Conversely, let then , but since is surjective we know that for some . But this implies that and thus thus .

The conclusion follows.

The theorem now follows since is closed we see that is open and since is open and we see that is open. Thus, closed. The conclusion follows from an earlier problem.

From this we can easily prove a specific case of a future theorem.

**Theorem:** Let be two topological spaces. Then if is Hausdorff under the product topology then so are and .

**Proof:** We know that is a surjective open mapping. It just remains to show that is closed. But, . Note though that since we must merely prove that the set is closed in . But it is clear that this is merely . But, the product of finitely many closed sets is closed and so it follows that is closed in . But, from the previous observation it follows that is closed in . The conclusion follows.

The exact same process works for .

**Corollary:** It follows by induction and the homeomorphic-associative properties of the product topology that is Hausdorff implies that is Hausdorff.

We can combine these two theorems concerning kernels to get.

**Theorem:** Let be a quotient map (continuous open surjection) then is Hausdorff if and only if is closed.

**Proof:** Follows from the combination of earlier theorems.

For the next theorem we need another definition.

**Agreement set:** Let be two mappings, then we define the *agreement set* to be

**Theorem:** Let be two continuous mappings and Hausdorff, then is closed.

**Proof:**

**Lemma:** Define by then is continuous.

**Proof:** It is readily verified that from where the conclusion follows immediately. .

Using this lemma we must merely note that from where it follows from ‘s continuity and being closed (since is Hausdorff) that is closed. .

Using this we can prove a very powerful theorem.

**Theorem:** Let be two continuous mappings and be Hausdorff. Then if is dense then .

**Proof:** By the previous problem we know that is closed and thus . The conclusion follows.

Also, the next theorem is nice.

**Latex:** Let be continuous and Hausdorff, then the set of fixed points of is closed.

**Proof:** Clearly the set of fixed points of is but these are both continuous mappings. The conclusion follows .

*Remark:* It follows from this that if you have a non-identity continuous function into a Hausdorff and you have some point that there exists some neighborhood of such that .

We now discuss a partially discussed converse to the fact that the product of Hausdorff spaces is Hausdorff, namely:

**Theorem:** Let be a non-empty collection of Hausdorff spaces, then if is Hausdorff then so is for each .

**Proof:** Let be distinct. So, let be a fixed but arbitrary set of points. It follows that are distinct points in where and . Thus, by ‘s assumed Hausdorffness there exists disjoint open sets such that . It is clear then that are open sets containing respectively. Now, suppose that then we can see that where where where is as in . This of course contradicts that . It follows that and the conclusion follows.

Before we move onto the next thing we need another definition.

**Perfect map:** Let be continuous closed and surjective. Then if is compact for every we call a *perfect map.*

**Theorem:** Let be perfect, then if is Hausdorff so is .

**Proof:** Let be distinct, then are disjoint compact subsets of . It follows from ‘s Hausdorffness that there exists disjoint open sets which contain respectively. Clearly then are closed sets containing respectively. By virtue that is perfect we see that are closed sets containing respectively. Clearly then are open sets containing respectively.

Now, let . Then and so and thus and thus . Using the exact same argument we see that . It follows that . The conclusion follows.

We lastly show that not every space is Hausdorff.

**Example:** Let be given the cofinite topology (sets are open if their compliments are finite). Then clearly given any distinct we see that are open sets which contain and don’t contain respectively. But, it is not possible to have disjoint open sets. To see this, let be open, then is finite and so any disjoint set must be finite. But, this implies that is infinite contradicting ‘s openness.

[…] An (-dimensional continuous) atlas for consists of a set of (-dimensional) charts such that . A Hausdorff second countable (i.e. having a countable basis) topological space which admits a continuous […]

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[…] let, for functions , be the agreement set . Note that is closed, indeed since is Hausdorff we know that the diagonal is closed in . Now, by the definition of the product topology, the map given […]

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