Hausdorff Spaces

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Now we define Hausdorff spaces. Before doing so, we first introduce the following.

Convergence of Sequences

Definition 1 For a topological space $X$, a function $\mathbb{N} \rightarrow X$ is called a sequence of points in $X$, and is denoted by $(x_n)_{n\geq 1}$. We say that a sequence $(x_n)_{n\geq 1}$ converges to $x\in X$ if for every neighborhood $U$ of $x$, there exists an $N\in \mathbb{N}$ such that

\[n\geq N\implies x_n\in U\]

holds.

This differs from the usual $\epsilon$-$N$ definition of convergence in calculus only in that an open set $U$ is used in place of an $\epsilon$-ball to represent points close to $x$. Naturally, if a sequence $(x_n)$ converges to $x$, one would like to write $\lim_{n \rightarrow\infty}x_n$, but the following example shows that this notation need not even be well-defined.

Example 2 Let $X$ be any set endowed with the trivial topology, considered as a topological space. Then any sequence in $X$ converges to every point of $X$.

Separation Axioms

The reason phenomena like Example 2 occur is, intuitively, that the topology on $X$ is not strong enough to separate the points of $X$. We may define various separation axioms according to the degree to which points are separated in a topological space, and classify spaces satisfying each of them. To this end, let us fix some terminology.

Two points $x,y$ of a topological space $X$ are distinct if $x\neq y$.
Two points $x,y$ of a topological space $X$ are topologically distinguishable if $\mathcal{N}(x)\neq \mathcal{N}(y)$.¹ (§Open Sets, §§Neighborhood filter)
Two subsets $A,B$ of a topological space $X$ are separated if each has a neighborhood not containing the other.
Two subsets $A,B$ of a topological space $X$ are separated by neighborhoods if they have disjoint neighborhoods.
Two subsets $A,B$ of a topological space $X$ are separated by closed neighborhoods if they have disjoint closed neighborhoods.
Two subsets $A,B$ of a topological space $X$ are separated by continuous functions if there exists a continuous function $f:X \rightarrow \mathbb{R}$ such that $A\subseteq f^{-1}(\{0\})$ and $B\subseteq f^{-1}(\{1\})$.
Two subsets $A,B$ of a topological space $X$ are precisely separated by continuous functions if there exists a continuous function $f:X \rightarrow \mathbb{R}$ such that $A= f^{-1}(\{0\})$ and $B= f^{-1}(\{1\})$.

The conditions above are listed in order of increasing strength. That is, two subsets precisely separated by continuous functions are separated by continuous functions; two subsets separated by continuous functions are separated by closed neighborhoods; two subsets separated by closed neighborhoods are separated by neighborhoods; two subsets separated by neighborhoods are separated; two separated points are topologically distinguishable; and all topologically distinguishable points are distinct.

Now we define the following.

Definition 3 For a topological space $X$, we define the following.

$X$ is a $T_0$-space, or a Kolmogorov space, if any two distinct points are topologically distinguishable.
$X$ is an $R_0$-space if any two topologically distinguishable points are separated.
$X$ is a $T_1$-space, or a Fréchet space, if any two distinct points are separated. Thus, $X$ is $T_1$ if and only if $X$ is $T_0$ and $R_0$.
$X$ is an $R_1$-space if any two topologically distinguishable points are separated by neighborhoods. Hence every $R_1$-space is an $R_0$-space.
$X$ is a $T_2$-space, or a Hausdorff space, if any two distinct points are separated by neighborhoods. Thus, $X$ is $T_2$ if and only if $X$ is $T_0$ and $R_1$, and every $T_2$-space is $T_1$.
$X$ is a $T_{2\frac{1}{2}}$-space, or a Urysohn space, if any two topologically distinguishable points are separated by closed neighborhoods. Hence every $T_{2\frac{1}{2}}$-space is $T_2$.
$X$ is a completely $T_2$-space, or a completely Hausdorff space, if any two distinct points are separated by continuous functions.
$X$ is a regular space if any point $x\in X$ and any closed set $A\subseteq X$ not containing $x$ are always separated by neighborhoods.
$X$ is a $T_3$-space, or a regular Hausdorff space, if $X$ is $T_0$ and regular. Every $T_3$-space is $T_{2\frac{1}{2}}$.
$X$ is a completely regular space if any point $x\in X$ and any closed set $A\subseteq X$ not containing $x$ are always separated by continuous functions.
$X$ is a completely $T_3$-space, or a Tychonoff space, if $X$ is $T_0$ and completely regular. Thus if $X$ is completely $T_3$, then it is completely Hausdorff and regular Hausdorff; it is therefore also called a completely regular Hausdorff space.
$X$ is a normal space if any two disjoint closed subsets are separated by neighborhoods.
$X$ is a normal regular space if $X$ is normal and $R_0$. Thus if $X$ is normal regular, then $X$ is completely regular.
$X$ is a $T_4$-space, or a normal Hausdorff space, if $X$ is $T_1$ and normal. Every $T_1$-space is $R_0$, so every $T_4$-space is normal regular, and hence completely regular. Moreover, every $T_1$-space is a $T_0$-space, so every $T_4$-space is completely $T_3$.
$X$ is a completely normal space if any two separated subsets are separated by neighborhoods. Then every completely normal space is normal.
$X$ is a completely $T_4$-space if $X$ is completely normal and $T_1$. Hence every completely $T_4$-space is $T_4$.
$X$ is a perfectly normal space if any two disjoint closed subsets are precisely separated by continuous functions.
$X$ is a perfectly $T_4$-space if $X$ is perfectly normal and $T_0$.

Hausdorff Spaces

One of the notions of particular importance in Definition 3 is that of a Hausdorff space. Here our intuition is well-founded.

Proposition 4 In a Hausdorff space $X$, any sequence $(x_n)$ converges to at most one point.

Proof

Suppose for contradiction that $(x_n)$ converges to two distinct points $x,y$. Then we can choose disjoint open neighborhoods $U,V$ of $x$ and $y$, respectively. From the assumption that $(x_n)$ converges to both $x$ and $y$, there exist $M,N$ such that

\[m\geq M \implies x_m\in U,\qquad n\geq N\implies x_n\in V\]

so letting $K=\max(M,N)$, the term $x_K$ must belong to both $U$ and $V$, a contradiction.

Meanwhile, the following lemma is also useful for showing that $X$ is Hausdorff.

Lemma 5 A topological space $X$ is Hausdorff if and only if the subset

\[\Delta_X=\{(x,x)\mid x\in X\}\]

of $X\times X$ is closed.

Proof

First, assume $X$ is Hausdorff. Then for any $(x,y)\not\in\Delta_X$, we have $x\neq y$, so we can choose disjoint neighborhoods $U,V$ of $x$ and $y$. Then $U\times V$ is an open set containing $(x,y)$ and disjoint from $\Delta_X$.

Conversely, if $\Delta_X$ is a closed subset of $X\times X$, then for any $x,y\in X$ with $x\neq y$, we have $(x,y)\not\in\Delta_X$, so there exists an open neighborhood of $(x,y)$ disjoint from $\Delta_X$, and by considering the base for the product topology, there is a basic open set of the form $U\times V$ contained in it.

More generally, by the same argument one can show that $X$ is Hausdorff if and only if, for any index set $I$, the diagonal

\[\Delta_X=\{(x_i)_{i\in I}:\text{$x_i=x$ for all $i$, where $x\in X$}\}\]

in $X^I=\prod_{i\in I}X$ is closed. From Lemma 5 we obtain the following.

Corollary 6 For continuous functions $f,g:X \rightarrow Y$, if $Y$ is Hausdorff then the set

\[E=\{x\in X\mid f(x)=g(x)\}\]

is a closed subset of $X$.

Proof

Consider the continuous function $x\mapsto (f(x), g(x))$ from $X$ to $Y\times Y$; the given set is the preimage of the closed set $\Delta_Y$ under this map.

Corollary 7 For a continuous function $f:X \rightarrow Y$, if $Y$ is Hausdorff then the set

\[\Gamma(f)=\{(x,f(x))\mid x\in X\}\]

is a closed subset of $X\times Y$.

Proof

Apply Corollary 6 to the two continuous functions

\[(x,y)\mapsto f(x),\quad (x,y)\mapsto y\]

from $X\times Y$ to $Y$.

Subspaces and Products of Hausdorff Spaces

For any Hausdorff space $X$, it is easy to verify that any subspace $A$ of $X$ is also Hausdorff. Indeed, given any $x,y\in A$, if we choose disjoint open neighborhoods $U,V$ of them in $X$, then $U\cap A$ and $V\cap A$ are disjoint open neighborhoods of $x,y$ in $A$. On the other hand, the product of Hausdorff spaces is also Hausdorff.

Proposition 8 For nonempty spaces $X_i$, the product $X=\prod_{i\in I}X_i$ is Hausdorff if and only if each $X_i$ is Hausdorff.

Proof

First, suppose the $X_i$ are Hausdorff and let $x,y\in X$ be given. Then there exists an index $i$ such that $x_i\neq y_i$, and in this $X_i$ we choose open neighborhoods $U,V$ separating $x_i$ and $y_i$. Considering the base for $X=\prod X_i$, the sets whose $i$th factor is $U$ or $V$ and whose remaining $j$th factors are $X_j$ separate $x$ and $y$.

Conversely, if $X$ is Hausdorff, then for arbitrarily chosen elements $x_j\in X_j$, the set

\[\prod_{j\in I} A_j,\qquad A_j=\begin{cases}A_i&i=j\\\{x_j\}&\text{otherwise}\end{cases}\]

is a subset of $X$ homeomorphic to $X_i$.

Quotient Spaces of Hausdorff Spaces

In general, the quotient space $X/R$ of a Hausdorff space $X$ need not be Hausdorff. Moreover, a necessary and sufficient condition for $X/R$ to be Hausdorff is not hard to obtain: since open sets in $X/R$ correspond bijectively to $R$-saturated open sets in $X$, the quotient $X/R$ is Hausdorff if and only if for any $x,y\in X$ belonging to different equivalence classes, there exists an $R$-saturated open set separating them. In particular, the following holds.

Proposition 9 For a continuous function $f:X \rightarrow Y$, if $Y$ is Hausdorff then the quotient space $X/{\sim}$ with respect to the equivalence relation

\[x\sim y\iff f(x)=f(y)\]

is Hausdorff.

This is equivalent to saying that the collections of open sets containing $x$ and $y$ are not the same. ↩

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Convergence of Sequences

Separation Axioms

Hausdorff Spaces

Subspaces and Products of Hausdorff Spaces

Quotient Spaces of Hausdorff Spaces

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