## Thoughts about separation

In this and the next couple of posts we will discuss the idea of topological separation. The main idea behind the idea of separation is to give necessary and sufficient conditions for some of the delusions we hold to be true. To motivate the point consider the following example

**Example:** Let be an indiscrete space and , then for any arbitrary we have that .

**Proof:** Since the only neighborhood of is we must merely check that is eventually in . But, this is clear. It follows that .

To the untrained mind this is shocking. This contradicts the tried-and-true fact that in a metric space every convergent sequence has a *unique* limit. It is a violent reminder that in the generality of topological spaces one cannot assume things to behave as expected. So, one might ask “What properties does a metric space have that an indiscrete space does not, which causes the discrepancy?”. Well, that is where separation axioms come in. They discuss necessary and sufficient conditions for certain more well known-properties to occur, the uniqueness of a convergent limit being one of them.

As one might have guessed the “separation axioms” have to do with when certain things may be separated, in the sense they are contained in open sets which in some way serve to separate them from other points, sets, etc. The first of these is the weakest, most spaces have this property:

**Kolomogorov ( ): **A topological space is said to be *Kolomogorov* (or alternatively* )* if given any distinct points there exists an open set which contains one but not the other. Note, this does not say that there exists two open sets, one which contains and not and one which contains and not . This says that only one of those sets must exist.

There aren’t too many theorems pertaining to Kolomogorov spaces, but we list and prove some here:

**Theorem:** Let be a topological space, then is Kolomogorov if and only if any two distinct points have distinct closures.

**Proof:**

: Suppose that is Kolomogorov, then for any distinct there exists some neighborhood which contains one but not the other. Assume WLOG that . It follows that is a closed set containing but not . Thus, and so .

: Let , by assumption we have that

**Lemma:** Let be a topological space and let be distinct. Then, if and then

**Proof:** First note that since and it follows that and so every neighborhood of intersects . A similar argument shows that every neighborhood of intersects

So, let if it is trivially true that . Consequently, assume not, then and so every neighborhood of contains , but since is also a neighborhood of and since every neighborhood of contains we see that . It follows that . Therefore, .

Conversely, an identical argument shows that and so .

The conclusion follows.

With this lemma in mind we see that implies that we may assume WLOG that and so which is an open set containing . The conclusion follows.

**Theorem:** Let be a topological space, then is Kolomogorov if and only if has no indiscrete subspace has exactly two points.

**Proof:**

: Suppose that is Kolomogorov. Let be a subset of . Since is Kolomogorov we have that there exists an open set which contains either but not both. For sake of convenience assume that . Then, forming a subspace out of we see that and so is in the topology of . But, since we see that and so is not a indiscrete subspace.

: Let be distinct. We know that as a subspace of is not indiscrete and so there is some open set in such that . It follows that either or but the other is. The conclusion follows. .

The idea of Kolomogorov spaces is intimately linked with the idea of “topological distinguishability”. The idea is this. Imagine that a topological space is a person, and when presented with presented with two elements of itself it could say either “they’re the same”, “they’re different”, “I’m not sure”. Well, since the topological space only knows open sets the only way it could tell if two elements were distinct is if there existed an open set which contained one but not the other. This is precisely what a Kolomogorov space is.

It might seem pretty amazing that they’re are spaces which are not Kolomogorov, in fact they’re extremely easy to construct.

**Theorem:** Let be an indiscrete topological space such that . Then, is not Kolomogorov.

**Proof:** Let be distinct points of . The only set which contains either of them is . But, this contains both. It follows that is not Kolomogorov. .

**Theorem:** Let be an arbitrary collection of Kolomogorov spaces, then is a Kolomogorov under the product topology.

**Proof:** Let be distinct. It follows that for some . Since is Kolomogorov there exists an open set such that contains one but not the other. Then, where is an open set containing either or but not the other. The conclusion follows.

We next discuss a more non-trivial separation axiom.

**: **Let be a topological space, then is said to be if given any distinct there exists an open set such that and and also that there exists an open set such that but .

So, we see that is just a strengthened version of Kolomogorov. The nice thing about spaces is that fulfill the commonly held belief that points should be closed sets.

**Theorem:** Let be a topological space, then is if and only if points are closed set.

**Proof:**

: Let we prove that is closed by proving that is open. So, let . Since is there exists an open set such that and . But, this implies there exists an open set such that . It follows that is open, and thus closed.

: Let be such that every point is a closed set. Then, be distinct from , but this of course means that . But, by assumption is open, and so it is an open set containing which does not contain . We may find a similar set for by considering .

The conclusion follows

Something that follows from this is that every finite space is discrete.

**Theorem:** Let be a finite space, then is discrete.

**Proof:** We must merely prove that every subsets of is open, or that every subset of is closed (from where it follows that any set’s compliment is open). But, this is clear since given any we have that which is the finite union of closed sets, and thus closed.

But, this ties nicely into an even nicer theorem. But, first we need the obvious (and up until this point incessantly used) theorem.

**Theorem:** Let be a metric space, then is .

**Proof:** Let be distinct. Then clearly satisfy the necessary conditions with . .

From this (and the previous theorem) we see that every finite metrizable space is discrete. Which, although not particularly difficult to prove is quite profound.

Another nice little fact is that :

**Theorem: **Let be a topological space, then is if and only if where is taken to be all the neighborhoods of .

**Proof:**

: Let . Clearly . But, for any we have that there exists some open set such that and . It follows that . The conclusion follows.

: Let be distinct. Since there must exist some open neighborhood of which does not include . A similar argument works for .

The conclusion follows

One might ask “is a hereditary property?”. In other words, if is a space and is a subspace, is ?

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

**Proof: **Let be distinct, then since they must (obviously) be distinct in and so there exists open sets which contain but do not include the other. Clearly then, are the necessary open sets . The conclusion follows.

Another theorem which can be helpful is the following:

**Theorem:** Let be a topological space, and suppose that for every distinct there exists some such that then is .

**Proof:** Let be distinct. By assumption there exists a continuous mapping such that . We may assume WLOG that and so are both open sets in which satisfy the necessary conditions.

A kind of novelty idea is the following

**Theorem: **Let be continuous, indiscrete, and . Then, is constant.

**Proof:** Suppose that for some . Then, since is there exists open sets with the normal properties. But, since we see that and similarly for . But, this contradicts the indiscretness of . The conclusion follows.

Maybe one of the most important fact about spaces comes from the following theorem.

**Theorem:** Let be a space. Then, is a limit point of if and only if every neighborhood of contains infinitely many points of .

**Proof:** Suppose that there existed some neighborhood of which contained only finitely many points . Then, for each there exists corresponding neighborhoods so that and . Clearly then is an neighborhood of which does not contain any points of . The conclusion follows. .

We next talk about the product of spaces.

**Theorem:** Let be an arbitrary collection of spaces, then is under the product topology.

**Proof:** Let be distinct. We must have then that for some . But, since is we know there exists open sets which contain respectively but do not contain the other. Clearly then where , are open sets in which satisfy the necessary conditions. The conclusion follows

We lastly give an example of a Kolomogorov space which isn’t .

**Example:** Let have topology . Then is Kolomogorov but not .

**Proof:** Clearly we must only check that there exists an open set which contains either or but not the other. Clearly is such a set. But, is not since the only open set containing is .

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