Separation and countability

A space is $T_1$ if and only if its points are closed, and Hausdorff if distinct points have disjoint neighbourhoods. A $T_1$ space is regular if a point and a closed set not containing it have disjoint neighbourhoods, and normal if disjoint closed sets have disjoint neighbourhoods. Urysohn’s lemma: if $X$ is normal and $A,B\subseteq X$ are disjoint closed sets, there is a continuous $f\colon X\to[0,1]$ with $f=0$ on $A$ and $f=1$ on $B$. A space is first countable if every point has a countable neighbourhood basis, second countable if its topology has a countable basis, and separable if it has a countable dense subset; a second countable space is first countable and separable.

Hausdorff space (Tag D-ZFRV4)

A topological space $X$ is Hausdorff, or $T_2$, if for all $p, q \in X$ with $p\neq q$ there exist disjoint open sets $U, V\subseteq X$ with $p\in U$ and $q\in V$.

In a Hausdorff space, every sequence converges to at most one point [@Mun00].

What does it mean for a space to be Hausdorff?

Separation axioms $T_0$ through $T_4$ (Tag D-OM7TD)

Let $X$ be a topological space. Two subsets $S, T\subseteq X$ are separated by neighborhoods if there exist disjoint open sets $U\supseteq S$ and $V\supseteq T$.

$X$ is $T_0$ if for all $x_1\neq x_2$ in $X$ some open set contains exactly one of $x_1,x_2$.
$X$ is $T_1$ if for all $x_1\neq x_2$ in $X$ there is an open set containing $x_1$ but not $x_2$, and an open set containing $x_2$ but not $x_1$.
$X$ is $T_2$ if any two distinct points are separated by neighborhoods; this is the Hausdorff condition.
$X$ is $T_{2.5}$ if any two distinct points $x_1,x_2$ have open neighborhoods $U_1\ni x_1$ and $U_2\ni x_2$ with $\overline{U_1}\cap\overline{U_2}=\emptyset$.
$X$ is $T_3$ if it is $T_0$ and regular: for every $x\in X$ and every closed $F\subseteq X$ with $x\notin F$, the sets $\ts{x}$ and $F$ are separated by neighborhoods.
$X$ is $T_{3.5}$ if it is $T_0$ and completely regular: for every $x\in X$ and every closed $F\subseteq X$ with $x\notin F$ there is a continuous $f\colon X\to[0,1]$ with $f(x)=0$ and $f(F)\subseteq\ts{1}$.
$X$ is $T_4$ if it is $T_1$ and normal: any two disjoint closed subsets of $X$ are separated by neighborhoods.

A space is $T_1$ if and only if every one-point subset is closed.

A space that is not $T_0$: let $\mathcal L^2(\RR)$ be the set of measurable $f\colon\RR\to\CC$ with $\int_\RR\abs{f}^2<\infty$, topologized by the pseudometric $d(f,g)=\qty{\int_\RR\abs{f-g}^2}^{1/2}$. If $f\neq g$ agree almost everywhere, then $d(f,g)=0$, so every open set containing one of them contains the other.

A space that is $T_0$ but not $T_1$: for a commutative ring $R$ with a prime ideal that is not maximal, $\Spec R$ with the Zariski topology. The closure of the point $\mathfrak p$ is $V(\mathfrak p)=\ts{\mathfrak q\in\Spec R \st \mathfrak q\supseteq\mathfrak p}$, so a prime that is not maximal is not a closed point; distinct primes have distinct closures, which gives $T_0$.

Regular space (Tag D-EPTMG)

A topological space $X$ is regular if for every closed set $F\subseteq X$ and every point $x\in X\sm F$ there are disjoint open sets $U, V\subseteq X$ with $x\in U$ and $F\subseteq V$.

Normal space (Tag D-YEQC3)

A topological space $X$ is normal if for all disjoint closed subsets $A, B\subseteq X$ there exist disjoint open sets $U, V\subseteq X$ with $A\subseteq U$ and $B\subseteq V$.

What does it mean for a space to be normal?

Points are closed in $T_1$ and Hausdorff spaces (Tag PR-R72XL)

$T_1$ makes points closed; Hausdorff implies $T_1$.Suggest a slogan

Let $X$ be a topological space in which every one-point set is closed, that is, a $T_1$ space. Then every finite subset of $X$ is closed. Every Hausdorff space is $T_1$, so in a Hausdorff space every finite subset, in particular every singleton, is closed [@Mun00].

Urysohn's lemma (Tag FT-52GNK)

Normality is exactly the ability to separate disjoint closed sets by a continuous $[0,1]$-valued function.Suggest a slogan

Let $X$ be a topological space. Then $X$ is normal if and only if for every pair of disjoint closed subsets $A, B \subseteq X$ there is a continuous function $f\colon X\to [0,1]$ with $f(a) = 0$ for all $a\in A$ and $f(b) = 1$ for all $b\in B$ [@Mun00].

The forward direction is Urysohn’s lemma; the converse follows by taking the disjoint open sets $f^{-1}([0,1/2))$ and $f^{-1}((1/2,1])$.

State Urysohn's lemma.

Which separation property does Urysohn's lemma characterise, and by what function?

Neighborhood basis (Tag D-EMJTU)

Let $X$ be a topological space and $p\in X$. A neighborhood basis at $p$ is a collection $\mathcal B_p$ of neighborhoods of $p$ such that every neighborhood $N$ of $p$ contains some $B\in\mathcal B_p$.

What is a neighborhood basis at a point $x$?

What collection of open sets forms a neighborhood basis at $x$?

First countable space (Tag D-E2NWN)

A topological space $X$ is first countable if every point $x\in X$ has a countable neighborhood basis.

What does it mean for a space to be first countable?

Second-countable space (Tag D-23NTI)

A topological space $X$ is second-countable if there exists a countable basis for its topology.

Dense subspace (Tag D-KJBAK)

Let $X$ be a topological space. A subset $Q\subseteq X$ is dense in $X$ if every neighborhood of every point $x\in X$ intersects $Q$.

A subset $Q\subseteq X$ is dense in $X$ if and only if its closure is $\cl_X(Q) = X$.

See [@Mun00].

In terms of closure, when is $A \subset X$ dense in $X$?

Separable space (Tag D-3O6QH)

A topological space $X$ is separable if there exists a countable subset $Q\subseteq X$ that is dense in $X$.

What does it mean for a space to be separable?

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