# Abstract Nonsense

## Crushing one theorem at a time

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 $X$ be an indiscrete space and $\{x_n\}_{n\in\mathbb{N}}$, then for any arbitrary $y\in X$ we have that $x_n\to y$.

Proof: Since the only neighborhood of $y$ is $X$ we must merely check that $\{x_n\}$ is eventually in $X$. But, this is clear. It follows that $y_n\to x$. $\blacksquare$

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 ( $T_0$): A topological space $X$ is said to be Kolomogorov (or alternatively $T_0$) if given any distinct points $x,y\in X$ 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 $x$ and not $y$ and one which contains $y$ and not $x$. 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 $X$ be a topological space, then $X$ is Kolomogorov if and only if any two distinct points have distinct closures.

Proof:

$\implies$: Suppose that $X$ is Kolomogorov, then for any distinct $x,y\in X$ there exists some neighborhood $O$ which contains one but not the other. Assume WLOG that $x\in O$. It follows that $X-O$ is a closed set containing $y$ but not $x$. Thus, $x\notin \overline{\{y\}}$ and so $\overline{\{x\}}\ne\overline{\{y\}}$.

$\Longleftarrow$: Let $x,y\in X$, by assumption we have that $\overline{\{x\}}\ne\overline{\{y\}}$

Lemma: Let $X$ be a topological space and let $x,y\in X$ be distinct. Then, if $x\in\overline{\{y\}}$ and $y\in\overline{\{x\}}$ then $\overline{\{x\}}=\overline{\{y\}}$

Proof: First note that since $y\in\overline{\{x\}}=\{x\}\cup D\left(\{x\}\right)$ and $x\ne y$ it follows that $y\in D\left(\{y\}\right)$ and so every neighborhood of $y$ intersects $x$.  A similar argument shows that every neighborhood of $x$ intersects $y$

So, let $z\in\overline{\{x\}}=\{x\}\cup D\left(\{x\}\right)$ if $z\in\{x\}\implies z=x$ it is trivially true that $z\in\overline{\{y\}}$. Consequently, assume not, then $z\in D\left(\{x\}\right)$ and so every neighborhood $N$ of $z$ contains $x$, but since $N$ is also a neighborhood of $x$ and since every neighborhood of $x$ contains $y$ we see that $y\in N$. It follows that $z\in\overline{\{y\}}$. Therefore, $\overline{\{x\}}\subseteq\overline{\{y\}}$.

Conversely, an identical argument shows that $z\in\overline{\{y\}}\implies z\in\overline{\{x\}}$ and so $\overline{\{y\}}\subseteq\overline{\{x\}}$.

The conclusion follows. $\blacksquare$

With this lemma in mind we see that $\overline{\{x\}}\ne\overline{\{y\}}$ implies that we may assume WLOG that $x\notin\overline{\{y\}}$ and so $y\notin \left(\overline{\{y\}}\right)'$ which is an open set containing $x$. The conclusion follows.  $\blacksquare$

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

Proof:

$\implies$: Suppose that $X$ is Kolomogorov. Let $\{x,y\}$ be a subset of $X$ . Since $X$ is Kolomogorov we have that there exists an open set $O$ which contains either $x,y$ but not both. For sake of convenience assume that $x\in O$. Then, forming a subspace out of $\{x,y\}$ we see that $\{x,y\}\cap O=O$ and so $O$ is in the topology of $\{x,y\}$. But, since $y\notin O$ we see that $O\ne X$ and so $\{x,y\}$ is not a indiscrete subspace.

$\Longleftarrow$: Let $x,y \in X$ be distinct. We know that $\{x,y\}$ as a subspace of $X$ is not indiscrete and so there is some open set $O\ne\varnothing$ in $X$ such that $\varnothing\subset O\cap\{x,y\}\ne \{x,y\}$. It follows that either $x\notin O$ or $y\notin O$ but the other is. The conclusion follows. $\blacksquare$.

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 $X$ be an indiscrete topological space such that $\text{card }X\geqslant 2$. Then, $X$ is not Kolomogorov.

Proof: Let $x,y$ be distinct points of $X$. The only set which contains either of them is $X$. But, this contains both. It follows that $X$ is not Kolomogorov. $\blacksquare$.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of Kolomogorov spaces, then $\displaystyle \prod_{j\in\mathcal{J}}X_j$ is a Kolomogorov under the product topology.

Proof: Let $\bold{x},\bold{y}\in X$ be distinct. It follows that $\pi_\ell(\bold{x})\ne\pi_\ell(\bold{y})$ for some $\ell\in\mathcal{J}$. Since $X_\ell$ is Kolomogorov there exists an open set $O$ such that $O$ contains one but not the other. Then, $\displaystyle \prod_{j\in\mathcal{J}}E_j$ where $\displaystyle E_j=\begin{cases} O & \mbox{if} \quad j=\ell \\ X_j & \mbox{if} \quad j\ne \ell\end{cases}$ is an open set containing either $\bold{x}$ or $\bold{y}$ but not the other. The conclusion follows. $\blacksquare$

We next discuss a more non-trivial separation axiom.

$T_1$: Let $X$ be a topological space, then $X$ is said to be $T_1$ if given any distinct $x,y$ there exists an open set $O_x$ such that $x\in O_x$ and $y\notin O_y$ and also that there exists an open set $O_y$ such that $y\in O_y$ but $x\notin O_y$.

So, we see that $T_1$ is just a strengthened version of Kolomogorov. The nice thing about $T_1$ spaces is that fulfill the commonly held belief that points should be closed sets.

Theorem: Let $X$ be a topological space, then $X$ is $T_1$ if and only if points are closed set.

Proof:

$\implies$: Let $\{x\}\subseteq Y$ we prove that $\{x\}$ is closed by proving that $\{x\}'$ is open. So, let $y\in X-\{x\}$. Since $X$ is $T_1$ there exists an open set $O_y$ such that $y\in O_y$ and $x\notin O_y$. But, this implies there exists an open set $O_y$ such that $y\in O_y\subseteq \{x\}'$. It follows that $\{x\}'$ is open, and thus $\{x\}$ closed.

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

The conclusion follows $\blacksquare$

Something that follows from this is that every finite $T_1$ space is discrete.

Theorem: Let $X$ be a finite $T_1$ space, then $X$ is discrete.

Proof: We must merely prove that every subsets of $X$ is open, or that every subset of $X$ is closed (from where it follows that any set’s compliment is open). But, this is clear since given any $E=\{e_1,\cdots,c_n\}$ we have that $\displaystyle E=\bigcup_{j=1}^{m}\{e_j\}$ 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 $X$ be a metric space, then $X$ is $T_1$.

Proof: Let $x,y\in X$ be distinct. Then $B_{\delta}(x),B_{\delta}(y)$ clearly satisfy the necessary conditions with $\displaystyle \delta=\frac{d(x,y)}{2}$. $\blacksquare$.

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 $X$ be a topological space, then $X$ is $T_1$ if and only if $\displaystyle \{x\}=\bigcap_{N\in\mathfrak{N}}N$ where $\mathfrak{N}$ is taken to be all the neighborhoods of $x$.

Proof:

$implies$: Let $x\in X$. Clearly $\displaystyle x\in\bigcap_{N\in\mathfrak{N}}N$. But, for any $y\in X-\{x\}$ we have that there exists some open set $O_x$ such that $x\in O_x$ and $y\notin O_x$. It follows that $\displaystyle y\notin\bigcap_{N\in\mathfrak{N}}N$. The conclusion follows.

$\Longleftarrow$: Let $x,y\in X$ be distinct. Since $\displaystyle \{x\}=\bigcap_{N\in\mathfrak{N}_x}N$ there must exist some open neighborhood of $x$ which does not include $y$. A similar argument works for $y$.

The conclusion follows $\blacksquare$

One might ask “is $T_1$ a hereditary property?”. In other words, if $X$ is a $T_1$ space and $E\subseteq X$ is a subspace, is $E$ $T_1$?

Theorem: Let $X$ be $T_1$ space and $E\subseteq X$ a subspace, then $E$ is $T_1$.

Proof: Let $x,y\in E$ be distinct, then  since they must (obviously) be distinct in $X$ and so there exists open sets $O_x,O_y$ which contain $x,y$ but do not include the other. Clearly then, $O_x\cap E,O_y\cap E$ are the necessary open sets $E$. The conclusion follows. $\blacksquare$

Another theorem which can be helpful is the following:

Theorem: Let $X$ be a topological space, and suppose that for every distinct $x,y\in X$ there exists some $f\in\mathcal{C}\left[X,\mathbb{R}\right]$ such that $f(x)\ne f(y)$ then $X$ is $T_1$.

Proof: Let $x,y\in X$ be distinct. By assumption there exists a continuous mapping $f:X\mapsto \mathbb{R}$ such that $f(x)\ne f(y)$. We may assume WLOG that $f(x) and so $\displaystyle f^{-1}\left(\left((-\infty,\frac{f(x)+f(y)}{2}\right)\right),f^{-1}\left(\left(\frac{f(x)+f(y)}{2},\infty\right)\right)$ are both open sets in $X$ which satisfy the necessary conditions. $\blacksquare$

A kind of novelty idea is the following

Theorem: Let $f:X\mapsto Y$ be continuous, $X$ indiscrete, and $Y$ $T_1$. Then, $f$ is constant.

Proof: Suppose that $f(x)\ne f(y)$ for some $x,y\in X$. Then, since $X$ is $T_1$ there exists open sets $O_x,O_y$ with the normal properties. But, since $f(y)\notin O_x$ we see that $y\notin f^{-1}(O_x)$ and similarly for $f^{-1}(O_y)$. But, this contradicts the indiscretness of $X$. The conclusion follows. $\blacksquare$

Maybe one of the most important fact about $T_1$ spaces comes from the following theorem.

Theorem: Let $X$ be a $T_1$ space. Then, $x$ is a limit point of $E\subseteq X$ if and only if every neighborhood of $x$ contains infinitely many points of $E$.

Proof: Suppose that there existed some neighborhood $N$ of $x$ which contained only finitely many points $e_1,\cdots,e_n$. Then, for each there exists corresponding neighborhoods $O_{x_1},\cdots,O_{x_n}$ so that $x_\in O_{x_k}$ and $e_k\notin O_{x_k}$. Clearly then $N\cap O_{k_1}\cap\cdots\cap O_{k_n}$ is an neighborhood of $x$ which does not contain any points of $E$. The conclusion follows. $\blacksquare$.

We next talk about the product of $T_1$ spaces.

Theorem: Let $\left\{X_j\right\}_{j\in\mathcal{J}}$ be an arbitrary collection of $T_1$ spaces, then $X=\prod_{j\in\mathcal{J}}X_j$ is $T_1$ under the product topology.

Proof: Let $\bold{x},\bold{y}\in X$ be distinct. We must have then that $\pi_\ell(\bold{x})\ne\pi_\ell(\bold{y})$ for some $\ell\in \mathcal{J}$. But, since $X_\ell$ is $T_1$ we know there exists open sets $O_x,O_y$ which contain $x,y$ respectively but do not contain the other. Clearly then $\displaystyle \prod_{j\in\mathcal{J}}E_j\prod_{j\in\mathcal{J}}G_j$ where $\displaystyle E_j=\begin{cases} O_x & \mbox{if} \quad j=\ell\\ O_y & \mbox{if} \quad j\ne \ell\end{cases}$,$\displaystyle G_j=\begin{cases} O_y & \mbox{if} \quad j=\ell\\ O_y & \mbox{if} \quad j\ne \ell\end{cases}$ are open sets in $X$ which satisfy the necessary conditions. The conclusion follows $\blacksquare$

We lastly give an example of a Kolomogorov space which isn’t $T_1$.

Example: Let $X=\{a,b\}$ have topology $\left\{\varnothing,\{a\},X\right\}$. Then $X$ is Kolomogorov but not $T_1$.

Proof: Clearly we must only check that there exists an open set which contains either $a$ or $b$ but not the other. Clearly $\{a\}$ is such a set. But, $X$ is not $T_1$ since the only open set containing $b$ is $X$.