In nowadays, an axiom of separation is a property that satisfies certain topological spaces. We are going to describe these axioms that are labeled as $T_{i}$ but first let’s recall some basic definitions.

## Separeted sets

Let $S$ be a topological space and let $x$ and $y$ be points points such that $x, y \in S$. Let $X, Y$ be neighbourhoods of $x$ and $y$ respectively such that $X, Y \subset S$. Then, we say that:

• $x$ and $y$ are distinguishable, if there a neighborhood $N \subset X$ such that $x \in N$ but $y \notin N$,

• $X$ and $Y$ are separated by a closed neighborhood, if they have disjoints closed neighborhoods (e.g. $X = [0, 1]$ and $Y = [4, 5]$),

• $X$ and $Y$ are separated by a continuous function, if there a map $\phi$ such that $\phi: X \longrightarrow \mathbb{R}$ and the image $\phi(X) = 0$ and $\phi(Y) = 1$,

• $X$ and $Y$ are precisely separated by a continuous function, if the preimage $\phi^{-1}(\{0\}) = X$ and $\phi^{-1}(\{1\}) = Y$.

## Axioms

• $X$ is $T_{0}$ (or Kolmogorov), the simplest axiom which satisfies that for any two distinct points $x$ and $y$ from $S$, then $x$ and $y$ are distinguishable.

• $X$ is $T_{1}$ (or Fréchet), if satisfies $T_{0}$ and if any two dinstinguishable points in $S$ are separated.

• $X$ is $T_{2}$ (or Hausdorff), if satisfies $T_{0}$ and if any two dinstinguishable points in $S$ are separated by neighbourhoods.

• $X$ is $T_{2 \frac{1}{2}}$ (or Urysohn), if any two distinct points in $S$ are separated by closed neighbourhoods. $T_{2 \frac{1}{2}}$ implies $T_{2}$.

• $X$ is $T_{3}$ (or regular Hausdorff), if satisfies $T_{0}$ and if for any $x \in S$ and any closed set $F \subset S$ such that $x \notin F$, then $x$ and $F$ are separated by neighbourhoods.

• $X$ is $T_{3 \frac{1}{2}}$ (or Tychonoff,), if satisfies $T_{0}$ and if for any $x \in S$ and any closed set $F \subset S$ such that $x \notin F$, then $x$ and $F$ are separated by a continuous function. $T_{3 \frac{1}{2}}$ implies $T_{3}$.

• $X$ is $T_{4}$ (or normal Hausdorff), if satisfies $T_{1}$ and if any two disjoints closed subsets of S are separated by neighbourhoods.

• $X$ is $T_{5}$ (or completely normal Hausdorff), if satisfies $T_{1}$ and if for any two separated sets are separated by neighbourhoods.

• $X$ is $T_{6}$ (or perfectly normal Hausdorff), if satisfies $T_{1}$ and if any two disjoints closed sets are precisely separated by a continuous function.

Concluding:

since $T_{i}$ satisfies the properties of $T_{j}$ for a $i > j$.

## Example

Let’s make a very precise example. Let $S = \mathbb{R}^{n}$ be a topological space and let $E, F$ be any two differrent closed disjoints subsets from $\mathbb{R}^{n}$ such that there a function $x \mapsto \frac{d(x, E)}{d(x, E) + d(x, F)}$ where $d(x, E)$ is the distance from a point $x$ to the subset $E$ defined as $\inf_{y \in E}\{d(x, y)\}$. Then,

• $\mathbb{R}^{n}$ is $T_{0}$, if we take any $x \in E$, then $x \notin F$ for any closed $F \subset \mathbb{R}^{n}$ since $E, F$ are disjoints. Then for any $y \in F$, clearly $x, y$ are distinguishable.

• $\mathbb{R}^{n}$ is $T_{1}$, let $X, Y$ be closed subsets from $\mathbb{R}^{n}$ such that $X \supset E$ and $Y \supset F$, then $X, Y$ are disjoints closed neighborhood from $E$ and $F$ respectively. Indeed, for any $x \in X$ and $y \in Y$, then $x$ and $y$ are separared. Since $\mathbb{R}^{n}$ is $T_{0}$, then claim follows.

• $\mathbb{R}^{n}$ is $T_{2}$, as we described previously that for any $x \in X$ an $y \in Y$ are dinstinguishable points in $\mathbb{R}^{n}$ are separated by neighbourhoods $E$ and $F$ respectively, then claim follows.

• $\mathbb{R}^{n}$ is $T_{2 \frac{1}{2}}$, this is a little stronger than $T_{2}$ since follows that for any $x, y \in \mathbb{R}^{n}$ even if they are not dinstinguishable points, they are separated by closed neighbourhoods. Since already our example fulfills all the requirements about this, then claim follows.

• $\mathbb{R}^{n}$ is $T_{3}$, if we taky any $x \in \mathbb{R}^{n}$ and we suppose that $x \notin F$, then $x$ and $F$ are separated by neighbourhoods, this is easy to see if we suppose that $x \in E$, since $E$ and $F$ are disjoints subsets from $\mathbb{R}^{n}$ and since it satisfies $T_{0}$ then claim follows.

• $\mathbb{R}^{n}$ is $T_{3 \frac{1}{2}}$, this is a little stronger than $T_{3}$, since requires that for any $x \in \mathbb{R}^{n}$ is separated by a closed subset $F \subset \mathbb{R}^{n}$, as previously we take $E$ and $F$ and they are closed disjoints subsets, then claim follows.

• $\mathbb{R}^{n}$ is $T_{4}$, let $X$ and $Y$ be two disjoints closed subsets from $\mathbb{R}^{n}$ such that $X \supset E$ and $Y \supset F$, then we can say that $E$ and $F$ are separated by neighbourhoods and since $\mathbb{R}^{n}$ satisfies $T_{1}$, then claim follows.

• $\mathbb{R}^{n}$ is $T_{5}$, this is a little stronger than $T_{4}$, since it satisfies that two any subsets from $\mathbb{R}^{n}$ are separated even if they are not closed. As we have decomposed $\mathbb{R}^{n}$ in closed subsets and already our example fulfills all the requirements about this, then claim follows.

• $\mathbb{R}^{n}$ is $T_{6}$, let $\phi$ be a continuous function such that $\phi: x \longrightarrow \frac{d(x, E)}{d(x, E) + d(x, F)}$ and as we described previously $d(x, E)$ is the distance from a point $x$ to the subset $E$ defined as $\inf_{y \in E}\{d(x, y)\}$, then $\phi$ separates $E$ and $F$ as closed subsets from $\mathbb{R}^{n}$. Note that we never divide by zero, that only could be possible if $d(x, E)$ and $d(x, F)$ are both zero, but this is impossible since $E$ and $F$ are disjoints. Since $\mathbb{R}^{n}$ satisfies $T_{1}$, then claim follows.

This post is inspired by the question asked on MathExchange and the example was inspired by the user Akiva about his answer to the question.