Topological separation axioms

Motivation

The indiscrete topology and discrete topology are not particularly useful, as first countable and second countable topological spaces show by trying to restrict the cardinality of the topology we may derive some interesting properties that are not true for topological spaces in general. Cardinality arguments are "weak" in the sense of they don't really provide a useful constraint on the topology. Consider for example the topology $\mathcal{J}$ defined as:

• $\mathcal{J}:=\{\mathcal{K}\}\cup\{ (1,+\infty),\ [0,+\infty)\}$ where $\mathcal{K}$ is some topology on $[0,1]\subset\mathbb{R}$ (power-set / discrete topology being a good example)

We can satisfy any cardinality argument with this (depending on the $\mathcal{K}$) but for anything outside $[0,1]$ we have no information. This is what I mean by cardinality arguments are weak. They don't govern the space.

However, suppose that a topological space, $(X,\mathcal{ J })$ is Hausdorff say, now we have a "strong" property (or requirement for the space), because it applies to ALL points equally well.

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TODO: Flesh out

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Overview

(Here $(X,\mathcal{ J })$ is a topological space) Warning:Some authors alter the terms slightly, see the warnings at the bottom

Space	Meaning	Comment
$T_1$	$\forall x,y\in X\exists U\in\mathcal{J}[x\ne y\implies (y\in U\wedge x\notin U)]$
$T_2$	$\forall x,y\in X\exists U,V\in\mathcal{J}[x\ne y\implies(U\cap V=\emptyset\wedge x\in U\wedge y\in V)]$	AKA: Hausdorff space. Every T2 space is a T1 space.
$T_3$	A regular $T_1$ space.	Every T3 space is a T2 space
$T_4$	A normal $T_1$ space.	Every T4 space is a T3 space, All metric spaces are T4 spaces

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TODO: Picture

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The $T_i$ notation exists because the German word for "separation axiom" is "Trennungsaxiome"

Recall

Regular topological space

A topological space, $(X,\mathcal{ J })$ is regular if:

• $\forall E\in C(\mathcal{J})\ \forall x\in X-E\ \exists U,V\in\mathcal{J}[U\cap V=\emptyset\implies(E\subset U\wedge x\in V)]$ - (here $C(\mathcal{J})$ denotes the closed sets of the topology $\mathcal{J}$)

Warning:Note that it is $E\subset U$ not $\subseteq$, the author () like me is pedantic about this, so it must matter

Normal topological space

A topological space, $(X,\mathcal{ J })$, is said to be normal if:

• $\forall E,F\in C(\mathcal{J})\ \exists U,V\in\mathcal{J}[E\cap F=\emptyset\implies(U\cap V=\emptyset\wedge E\subseteq U\wedge F\subseteq V)]$ - (here $C(\mathcal{J})$ denotes the collection of closed sets of the topology, $\mathcal{J}$)

Warnings

According to some authors define:

• $T_3$ - a regular topological space
• $T_4$ - a normal topological space

This rarely leads to problems as a lot of the topological spaces of interest are Hausdorff spaces ($T_2$) and on such spaces the definitions coincide.

We shall use the definitions as given above, because we already have a term for "normal topological spaces" - that term is "normal topological space"

See also

• Urysohn's lemma - for disjoint closed sets of a normal topological space, $(X,\mathcal{ J })$, there exists a continuous function, $f:X\rightarrow [0,1]\subset\mathbb{R}$ such that $f$ is $0$ on the first closed set and $f$ is $1$ on the other.