# General Topology/Separation

Suppose we have a topological space $X$ and two distinct points $x,y\in X$. Suppose further that using the topological structure, we try to separate these points, that is, to find sets which contain one of the points, but not the other. Usually this is not possible. But the topological spaces where it is possible form important classes of topological spaces, which are then said to satisfy separation axioms. There is an important hierarchy of separation axioms which are numbered T$_0$ until T$_6$, and then there are the R$_0$ and R$_1$ axioms. The T axioms concern separation in the classical sense, whereas the R axioms concern only separation of topologically distinguishable points.
Definition (topological distinguishability):
Let $X$ be a topological space, and let $x,y\in X$. The two points $x$ and $y$ are said to be topologically distinguishable iff there exists an open set $U$ of $X$ that contains one point of $\{x,y\}$, but not the other. Otherwise, $x$ and $y$ are called topologically indistinguishable.
Definition (T$_0$ space):
Let $X$ be a topological space. $X$ is said to be a T$_0$ space if and only if for any two points $x,y\in X$ which are distinct (ie. $x\neq y$) there exists an open set $U$ of $X$ which either contains $x$ or $y$, but not both.
Alternatively, we may define a T$_0$ space to be a space where any two points are topologically distinguishable.
The situation is depicted in the following picture:
Definition (T$_1$ space):
Let $X$ be a topological space. $X$ is said to be a T$_1$ space iff for any two distinct points $x,y\in X$, there exists an open set $U$ of $X$ which does not contain $y$ but does contain $x$.
The situation is depicted in the following picture:
We introduce our first R axiom:
Definition (R$_0$ space):
An R$_0$ space is a topological space $X$ where for any pair of topologically distinguishable points $x,y$, there exists both a neighbourhood $U$ of $x$ not containing $y$ and a neighbourhood $V$ of $y$ not containing $x$.
Proposition (characterisation of T$_1$ spaces):
Let $X$ be a topological space. The following are equivalent:
* $X$ is a T$_1$ space
* All cofinite subsets of $X$ are open
* All finite subsets of $X$ are closed
* The points of $X$ are closed
* $X$ is simultaneously T$_0$ and R$_0$
Proof: If $X$ is a T$_1$ space and $x\in X$, then $X\setminus \{x\}$ is open, because for $y\in X$ we find an open neighbourhood $V_{y}$ of $y$ (which may be chosen the union of all such neighbourhoods in order to avoid the axiom of choice) such that $x\notin V_{y}$, and then
    $X\setminus \{x\}=\bigcup _{y\neq x}V_{y}$.
Then every cofinite subset of $X$ is open as the intersection of finitely many open sets. Since the closed sets are the complements of the open sets, 3. follows. 3. $\Rightarrow$ 4. is trivial. Given distinct points $x,y\in X$, set $U:=X\setminus \{y\}$ and $V:=X\setminus \{x\}$. Then $U$ and $V$ show that $X$ is T$_0$ (any two points are topologically distinguishable) and R$_0$ (if a statement holds for any two points, then in ptic. for topologically distinguishable points). Finally suppose $X$ is simultaneously T$_0$ and R$_0$. Let $x,y$ be any two distinct points. Since $X$ is T$_0$, $x,y$ are topologically distinguishable, and since $X$ is R$_0$, we find neighbourhoods $U$ of $x$ and $V$ of $y$ so that $y\notin U$ and $x\notin V$. $\Box$
Definition (R$_1$ space):
A topological space $X$ is called R$_1$ if and only if for any two topologically distinguishable points $x,y$ of $X$, there exist open neighbourhoods $U$ of $x$ and $V$ of $y$ so that $U\cap V=\emptyset$.
By far the most common class of spaces that satisfy a separation axiom are the Hausdorff spaces:
Definition (Hausdorff space):
Let $X$ be a topological space. $X$ is said to be a Hausdorff space if and only if for any two distinct points $x,y\in X$ there exists two open sets $U$ and $V$ of $X$ so that $U\cap V=\emptyset$, $x\in U$ and $y\in V$.
This situation is depicted in the following picture:
Proposition (separation of finitely many points in Hausdorff spaces):
Let $X$ be a Hausdorff topological space, and let $x_{1},\ldots ,x_{n}$ be distinct points in $X$. Then there exist neighbourhoods $U_{j}$ of $x_{j}$ ($j\in [n]$) such that for $j\neq k$, we have $U_{j}\cap U_{k}=\emptyset$.
Proof: We proceed by induction on $n$. The case $n=1$ is trivial. Suppose the claim holds true for $n-1$, and let $x_{1},\ldots ,x_{n}$ be distinct points in $X$. By induction, choose neighbourhoods $V_{1},\ldots ,V_{n-1}$ of $x_{1},\ldots ,x_{n-1}$ respectively such that $V_{j}\cap V_{k}=\emptyset$ for $j\neq k$. Since $X$ is Hausdorff, we find for all $j\in [n-1]$ neighbourhoods $W_{j},U_{j}$ of $x_{j}$ resp. $x_{n}$ such that $W_{j}\cap U_{j}=\emptyset$. Then set $O_{n}:=\bigcap _{j=1}^{n-1}U_{j}$ and for $j\in [n-1]$ set $O_{j}:=V_{j}\cap W_{j}$. Then $O_{1},\ldots ,O_{n}$ is a set of open sets as desired. $\Box$
Hausdorff spaces are located within the T$_0$ until T$_6$ hierarchy:
Definition (T$_2$ space):
A T$_2$ space is a Hausdorff space.
Proposition (R-axiom characterisation of Hausdorff spaces):
Let $X$ be a topological space. $X$ is Hausdorff if and only if $X$ is T$_0$ and R$_1$.
Proof: If $X$ is Hausdorff, it is certainly T$_0$ and R$_1$. On the other hand, if it is T$_0$, all points are topologically indistinguishable, so that for any two points $x,y\in X$ we find $U,V$ open with $x\in U$, $y\in V$ and $U\cap V=\emptyset$. $\Box$
Proposition (diagonal criterion for being Hausdorff):
Let $X$ be a topological space, and let $\Delta \subseteq X\times X,\Delta :=\{(x,x)|x\in X\}$ be the diagonal. Then $X$ is Hausdorff if and only if $\Delta \subseteq X\times X$ is closed, where we consider $X\times X$ as a topological space with the product topology.
Proof: Let $x,y\in X$ be arbitrary. $X$ being Hausdorff means that there exist open neighbourhoods $y\in U_{y}$ and $x\in U_{x}$ so that $U_{x}\cap U_{y}=\emptyset$. If that is always the case, then whenever $(x,y)$ is not in $\Delta$ and $U_{x},U_{y}$ are so chosen, then $U_{x}\times U_{y}$ is a neighbourhood of $(x,y)$ in $X\times X\setminus \Delta$, so that the diagonal is closed as the complement of an open set. Conversely, if $\Delta$ is closed, let $x,y\in X$ be arbitrary and not equal to each other. Then $(x,y)\notin \Delta$, and since $X\times X\setminus \Delta$ is open, by definition of the product topology we find open sets $U_{y}$ and $U_{x}$ of $X$ so that $(x,y)\in U_{x}\times U_{y}$ and $U_{x}\times U_{y}\in X\times X\setminus \Delta$. From the first of these conditions we deduce that $x\in U_{x}$ and $y\in U_{y}$, and from the second we deduce that $U_{x}\cap U_{y}\neq \emptyset$, since any $z\in U_{x}\cap U_{y}$ would yield $(z,z)\in U_{x}\times U_{y}$. $\Box$
Definition (Urysohn space):
A topological space $X$ is called a Urysohn space iff for all distinct points $x,y\in X$ we find closed neighbourhoods $A,B$ of $x$ resp. $y$ so that $A\cap B=\emptyset$.
Proposition (Urysohn spaces are Hausdorff):
Let $X$ be a Urysohn space. Then $X$ is Hausdorff.
Proof: Let $x,y\in X$ so that $x\neq y$, and pick closed neighbourhoods $A,B$ of $x$ resp. $y$ so that $A\cap B=\emptyset$. Then by def. of a neighbourhood of a point, pick $U\subseteq A$ and $V\subseteq B$ open s.t. $x\in U$, $y\in V$. Then $U,V$ are neighbourhoods of $x$ resp. $y$ that satisfy the requirements given by the Hausdorff condition. $\Box$
Definition (T$_{2½}$ space):
A topological space $X$ is a T$_{2½}$ space iff it is a Urysohn space.
There are also stronger separation conditions. They are usually formulated in regard to closed sets, and hence they don't really fit in with the T-axiom hierarchy (you'll see what I mean), but combining them with a small T axiom (namely T$_0$ or T$_1$), we produce corresponding axioms in the T-axiom hierarchy.
Definition (regular space):
Let $X$ be a topological space. $X$ is regular iff for every closed set $A\subseteq X$ and every point $x\in X\setminus A$, there exist open sets $U,V\subseteq X$ so that $x\in U$, $A\subseteq V$ and $U\cap V=\emptyset$.
Proposition (regular spaces are characterised by the existence of closed neighbourhood fundamental systems):
Let $X$ be a topological space. $X$ is regular if and only if for all $x\in X$ there exists a fundamental system of $N(x)$ consisting of closed sets.
Proof: Suppose first that each $N(x)$ ($x\in X$) has a fundamental system of closed sets. Let $A\subseteq X$ be any closed set and let $x\in X\setminus A$. Now $U:=X\setminus A$ is an open neighbourhood of $x$. Since $x$ has a fundamental system consisting of closed neighbourhoods, pick $B\subseteq U$ closed so that $x\in B$. Since $B$ is a neighbourhood of $x$, pick $V\subseteq B$ open so that $x\in U$. Then $V$ and $X\setminus B$ do the trick in the definition of a regular space.
Suppose now that $X$ is regular, let $x\in X$, and let $U$ be any open neighbourhood of $x$. Claim that $U$ contains a closed neighbourhood of $x$. Indeed, set $A:=X\setminus U$. By regularity, pick a neighbourhood $V$ of $x$ and an open set $W$ containing $A$ such that $V\cap W=\emptyset$. Then $B:=X\setminus W$ is the closed neighbourhood as desired, since it contains the open neighbourhood $V$ of $x$. $\Box$
Definition (T$_3$ space):
A T$_3$ space is a topological space which is T$_0$ and regular.
Proposition (T$_3$ spaces are Urysohn):
Let $X$ be a T$_3$ space. Then $X$ is a Urysohn space.
Proof: Since $X$ is regular, there exists for each point a closed fundamental system of the neighbourhood system. Hence, let $x,y\in X$ be any two distinct points. It suffices to prove that they are separated by disjoint open neighbourhoods. By the T$_0$ property wlog. choose an open neighbourhood $U$ of $x$ so that $y\notin U$. Set $A:=X\setminus U$, which is closed and contains $y$. Since $X$ is regular, we may pick open disjoint sets $V$ and $W$ such that $x\in V$, $A\subseteq W$ and $V\cap W=\emptyset$. $\Box$
Definition (completely regular space):
Let $X$ be a topological space. $X$ is a completely regular space iff for all closed sets $A\subseteq X$ and points $x\in X\setminus A$ we can find a continuous function $f:X\to [0,1]$ such that $f(x)=0$ and $\forall z\in A:f(z)=1$.
Proposition (completely regular spaces are regular):
Let $X$ by a completely regular space. Then $X$ is regular.
Proof: Let $x\in X$ and let $A\subset X$ be closed so that $x\notin A$. Since $X$ is completely regular, pick a continuous function $f:X\to [0,1]$ so that $f(x)=0$ and $f(z)=1$ for $z\in A$. Then set $U:=f^{-1}([0,1/3))$ and $V:=f^{-1}((2/3,1])$ and observe that $U$ and $V$ are open by the definition of the subspace topology of $[0,1]$, that $U\cap V=f^{-1}([0,1/3)\cap (2/3,1])=\emptyset$ and certainly $x\in U$, $A\subseteq V$, so that $X$ is regular. $\Box$
Definition (T$_{3½}$ space):
A T$_{3½}$ space is a topological space that is both T$_0$ and completely regular.
Proposition (T$_{3½}$ spaces are T$_3$):
Let $X$ be a T$_{3½}$ space. Then $X$ is T$_3$.
Proof: It is T$_0$ by definition and regular since completely regular spaces are regular. $\Box$
Definition (normal space):
A normal space is a topological space so that for all closed disjoint subsets $A,B\subset X$, $A\cap B=\emptyset$, there exist open sets $U,V$ such that $A\subseteq U$, $B\subseteq V$ and $U\cap V=\emptyset$.
Definition (T$_4$ space):
A T$_4$ space is a topological space that is both T$_1$ and normal.
The situation is depicted in the following picture:
Theorem (Urysohn's lemma):
Let $X$ be a normal space and let $A,B\subseteq X$ be two closed disjoint subspaces of $X$. Then we can find a continuous function $f:X\to \mathbb {R}$ such that $f(x)=0$ for all $x\in A$ and $f(y)=1$ for all $y\in B$.
Proof: We instead prove that if we are given an open set $U$ of $X$ and a closed set $A\subseteq U$, then we can find a continuous function $f:X\to \mathbb {R}$ such that $f$ is zero inside $A$ and $1$ outside $U$; the theorem will then follow upon defining $U:=X\setminus A$ and applying our auxiliary result.
Indeed, note that since $X$ is normal, we find an open and a closed set (which we suggestively denote by $U_{1/3}$ and $A_{2/3}$) such that $A\subseteq U_{1/3}\subseteq A_{2/3}\subseteq U$. We now repeat this procedure; that is, for every $k\in \mathbb {N}$, we construct a sequence
    $A\subseteq U_{1/3^{k}}\subseteq A_{2/3^{k}}\subseteq U_{1/3^{k-1}}\subseteq A_{4/3^{k}}\subseteq \cdots \subseteq U$
from normality and the sequence that we had gotten in the $k-1$ step (this requires the axiom of dependent choice). Then we define
    $f(x):=\sup\{n/3^{k}|k\in \mathbb {N} ,0\leq n\leq 3^{k},x\notin S_{n/3^{k}}\}$,
where we define $S_{n/3^{k}}$ to equal $U_{n/3^{k}}$ or $A_{n/3^{k}}$, depending on whether $n$ is even or odd, and $A_{0}:=A$ and $U_{1}:=U$. We then have $f(x)=0$ for $x\in A$ and $f(x)=1$ for $x\notin U$, and further $f$ is continuous; indeed, it is continuous at each point, since if we set $y_{0}:=f(x_{0})\in [0,1]$ for a fixed $x_{0}\in X$ and fix an $\epsilon >0$, then we choose $k$ sufficiently large so that $1/3^{k-1}<\epsilon$ and then $n\in \mathbb {N}$ so that $x\in S_{n/3^{k}}\setminus S_{(n-1)/3^{k}}$ (if $x\in A$, then $U_{1/3^{k}}$ is a neighbourhood that will map wholly to $[0,\epsilon )$) and then either $n$ is even, so that $U_{(n+1)/3^{k}}\setminus A_{(n-2)/3^{k}}$ maps to $[0,\epsilon )$ (note $n\geq 2$ if $n\neq 0$ and even), or $n$ is odd and $U_{n/3^{k}}\setminus A_{(n-1)/3^{k}}$ is a neighbourhood of the desired form. $\Box$
Proposition (T$_4$ spaces are T$_{3½}$):
Let $X$ be a T$_4$ space. Then $X$ is a T$_{3½}$ space.
Proof: By Urysohn's lemma, it suffices to note that all points of $X$ are closed. But this is the case, since $X$ is T$_1$. $\Box$
Theorem (Tietze–Urysohn theorem):
Let $X$ be a normal space, let $A\subset X$ be closed, and let $f:A\to \mathbb {R}$ be a continuous function. Then there exists a continuous function $F:X\to Y$ such that $F\upharpoonright _{A}=f$.
Proof: $\Box$

## Exercises

* Let $X$ be a finite set. Prove that there exists exactly one Hausdorff topology on $X$ and find out what it is.
* Let $X$ be a topological space. Prove that the sets $U\subseteq X$ such that $X\setminus U$ is finite, when the empty set is adjoined to it, form a topology on $X$ (this topology is called the co-finite topology). Further prove that if $X$ is infinite, then together with this topology it is $T_{1}$, but not Hausdorff.
* Let $X$ be a topological space that is equipped with the initial topology stemming from a family of functions $f_{\alpha }:X\to X_{\alpha }$. Prove that $X$ is a T$_0$ space if and only if for each $x\neq y$ in $X$, there exists an $\alpha$ such that $f_{\alpha }(x)$ and $f_{\alpha }(y)$ are topologically distinguishable.