Countability and Separation axioms

See Munkres (2000) §17, §30, §31, §32.

Separation axioms

Definition 347.1(Definition).

A topological space is called

1. Fréchet ($T_1$) if singleton subsets are closed.
2. Hausdorff ($T_2$) if distinct points have disjoint open neighborhoods.
3. Regular ($T_3+T_1$) if every closed set $C$ and point $p\notin C$ have have disjoint open neighborhoods, and singletons are closed.
4. Normal ($T_4+T_1$) if any two disjoint closed sets have disjoint open neighborhoods, and singletons are closed.
5. Completely normal ($T_5$) if all subspaces are normal.
6. Perfectly normal ($T_6$) if all closed subsets are $G_{\delta }$.

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Clearly, $T_4$+$T_1$ $⟹$ $T_3$+$T_1$ $⟹$ $T_2$ $⟹$ $T_1$.

Example 347.2(Example).

1. The cofinite topology on an infinite set is $T_1$ but not $T_2$.
2. ${\mathbb{R}}_K$ is $T_2$ but not $T_3$.
3. ${\mathbb{R}}_{\ell }$ is $T_4$. ${\mathbb{R}}_{\ell }^2$ is not $T_4$ (but $T_3$, since ${\mathbb{R}}_{\ell }$ is).
4. If $J$ is uncountable, the product space ${\mathbb{R}}^J$ is not normal (and hence not metrizable, by the next example!)
5. Metric spaces are $T_6$.

Fréchet spaces

Proposition 347.3(Proposition).

$X$ is $T_1$ $⟺$ for every pair of distinct points, each has a neighborhood not containing the other point.

Proof.

$(⟹)$ for $x,y\in X$, $\{x{\}}^c$ and $\{y{\}}^c$ work.

$(⟸)$ Let $x\in X$. For every $y\ne x$, let $U_y$ be an open neighborhood of $y$ not containing $x$. Then, $U:={\bigcup }_{y\in X,y\ne x}{U}_y$ is open, and $U^c=\{x\}$.□

Proposition 347.4(Proposition).

Let $X$ be a $T_1$ space, and $A\subseteq X$. Then the point $x$ is a limit point of $A$ iff every neighborhood of $x$ contains infinitely many points of $A$.

Proof.

Let $U_1$ be an open neighborhood of $x$. Since $U_1$ must intersect $A$, we can pick $a_1\in A\cap U_1$. Since $X$ is $T_1$, there exists an open neighborhood $U^{\prime }$ of $x$ which does not contain $a_1$. Let $U_2:=U_1\cap U^{\prime }$. Again, being an open neighborhood of $x$, $U_2$ must contain an element of $A$, say $a_2$. Proceeding in this manner, we see that $U_1$ must contain infinitely many points of $A$.□

Hausdorff, regular and normal spaces

There are several niceties of metric spaces that we take for granted:

1. all singletons are closed;
2. a sequence cannot converge to more than one point.

These are not true for arbitrary topological spaces. As we ascend the ladder from $T_1$ to $T_4$, topological spaces become more well-behaved. As we have seen, merely assuming $T_1$ gives us closed singletons; the uniqueness of limits is attained at $T_2$.

Proposition 347.5(Proposition).

A sequence of points in a $T_2$ space converges to at most one point.

Proof.

Say a sequence $\{x_n\}$ converges to two points, $a$ and $b$. Let $A\ni a$ and $B\ni b$ be disjoint open sets. By definition, there exists $N$ such that for all $n⩾N$, $x_n\in A$ - but this implies $\{x_n\}$ cannot converge to $b$!□

Def 1 can be reformulated in the following useful way:

Proposition 347.6(Proposition).

Let $X$ be a $T_1$ space.

1. $X$ is regular $⟺$ given $x\in X$ and a neighborhood $N\ni x$, there exists a neighborhood $V\ni x$ such that $\overline{V}\subseteq U$.

2. $X$ is normal $⟺$ given a closed set $A\subseteq X$ and an open set $U\supseteq A$, there exists an open set $V\supseteq A$ such that $\overline{V}\subseteq U$.

Proof.

$(1)$ Suppose $X$ is regular, and suppose the point $x$ and the neighborhood $U$ of $x$ are given. Let $B=X\setminus U$; then $B$ is a closed set. By hypothesis, there exist disjoint open sets $V$ and $W$ containing $x$ and $B$ respectively. The set $\overline{V}$ is disjoint form $W$, since if $y\in W$, there is an open neighborhood of $y$ contained in $W$. Thus, $\overline{V}$ is disjoint from $B$, and $\overline{V}\subseteq U$.

To prove the converse, suppose the point $x$ and the closed set $B$ not containing $x$ are given. Let $U=X\setminus B$. By hypothesis, there is a neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$. The open sets $V$ and $X\setminus \overline{V}$ are disjoint open sets containing $x$ and $B$, respectively. Thus, $X$ is regular.

$(2)$ is proved similarly.□

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Proposition 347.7(Proposition).

1. A subspace of a $T_2$ space is $T_2$.

2. A product of $T_2$ spaces is $T_2$.

3. A subspace of a $T_3$ space is $T_3$.

4. A product of $T_3$ spaces is $T_3$.

Proof.

$(1)$ and $(3)$ are clear.

$(2)$ Let $\{X_{\alpha }\}$ be a family of Hausdorff spaces. Let $\mathbf{x}=(x_{\alpha })$ and $\mathbf{y}=(y_{\alpha })$ be distinct points of the product space $\prod X_{\alpha }$. Because $\mathbf{x}\ne \mathbf{y}$, there is some index $\beta$ such that $x_{\beta }\ne y_{\beta }$. Choose disjoint open sets $U$ and $V$ in $X_{\beta }$ containing $X_{\beta }$ containing $x_{\beta }$ and $y_{\beta }$ respectively. Then the sets ${\pi }_{\beta }^{-1}(U)$ and ${\pi }_{\beta }^{-1}(V)$ are disjoint open sets in $\prod X_{\alpha }$ containing $\mathbf{x}$ and $\mathbf{y}$ respectively.

$(4)$ Let $\{X_{\alpha }\}$ be a family of regular spaces; let $X=\prod X_{\alpha }$. By $(2)$, $X$ is $T_2$, so $X$ is $T_1$. We use Prp 6. Let $\mathbf{x}\in X$, $U$ be a neighborhood of $\mathbf{x}$ in $X$. Choose a basis element $\prod U_{\alpha }$ about $\mathbf{x}$ contained in $U$. Choose, for each $\alpha$, a neighborhood $V_{\alpha }$ of $x_{\alpha }$ in $X_{\alpha }$ such that ${\overline{V}}_{\alpha }\subseteq U_{\alpha }$; if it happens that $U_{\alpha }=X_{\alpha }$, chose $V_{\alpha }=X_{\alpha }$. Then $V=\prod V_{\alpha }$ is a neighborhood of $x$ in $X$. Since $\overline{V}=\prod {\overline{V}}_{\alpha }$ by Prp 372.22, it follows that $\overline{V}\subseteq \prod U_{\alpha }\subseteq U$.□

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There is no analogue of Prp 7 for normal spaces.

Prp 8 provides three sufficient conditions for a space to be normal:

Proposition 347.8(Proposition).

1. Every regular second countable space is normal.

Proof.

Let $X$ be a regular space with a countable basis $\mathcal{B}$. Let $A$ and $B$ be disjoint closed subsets of $X$. Each point $x\in A$ has a neighborhood $U$ not intersecting $B$. Using regularity, choose a neighborhood $V$ of $x$ whose closure lies in $U$; finally, choose an element of $\mathcal{B}$ containing $x$ and contained in $V$. By choosing such a basis element for each $x\in A$, we construct a countable covering $\{U_n\}$ of $A$ by open sets whose closures do not intersect $B$. Let $\{V_n\}$ be a similar countable covering for $B$. The sets $\bigcup U_n$ and $\bigcup V_n$ are open sets containing $A$ and $B$ respectively, but these may not be disjoint. However, that can be arranged. Let

$$\begin{array}{r}{U}_n^{\prime }=U_n\setminus \bigcup _{i=1}^n{\overline{V}}_i\text{and}{V}_n^{\prime }=V_n\setminus \bigcup _{i=1}^n{\overline{U}}_i.\end{array}$$

Clearly, $U^{\prime }:=\bigcup U_n^{\prime }\supseteq A$, $V^{\prime }:=\bigcup V_n^{\prime }\supseteq B$, and $U^{\prime }\cap V^{\prime }=\varnothing$.□

2. Every metrizable space is normal.

Proof.

Let $X$ be metrizable with metric $d$. Suppose $A$ and $B$ are two closed disjoint subsets of $X$. For each $a\in A$, choose ${ϵ}_a$ so that the ball $B(\alpha ,{ϵ}_a)$ does not intersect $B$. Similarly, for each $b\in B$, choose ${ϵ}_b$ so that the ball $B(b,{ϵ}_b)$ does not intersect $A$. Define

$$U:=\bigcup _{a\in A}B(a,{ϵ}_a/2)\text{and}V=\bigcup _{b\in B}B(b,{ϵ}_b/2).$$

The sets $U$ and $V$ are open sets containing $A$ and $B$ respectively, and are disjoint, as can be shown by a routine application of the triangle inequality.□

3. Every compact Hausdorff space is normal.

Proof.

First, we show that a compact Hausdorff space is regular. Let $A$ be closed and $x\notin A$. For each $p\in A$, let $U_p$ and $V_p$ be disjoint open neighborhoods of $p$ and $x$ respectively. Since $A$ is compact, there exist $p_1,\dots ,p_n\in A$ such that $U_{p_1}\cup \cdots \cup U_{p_n}\supseteq A$. Clearly, $V_{p_1}\cap \cdots \cap V_{p_n}\ni x$ and is disjoint from $U_{p_1}\cup \cdots \cup U_{p_n}$. Repeat the same argument with two closed sets to show that the space is normal.□

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Proposition 347.9(Proposition).

Every well-ordered set $X$ is normal in the order topology.

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Countability axioms

Definition 347.10(Definition).

A topological space $X$ is said to be

1. First countable if for each $x\in X$ there is a countable collection $\{U_n{\}}_{n\in \mathbb{Z}}$ of neighborhoods of $x$ such that any neighborhood $U$ of $x$ contains at least one of the sets $U_n$.
2. Second countable if $X$ has a countable basis.

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Remark 347.11(Remark).

1. Clearly, second countability implies first countability.
2. A metrizable space always satisfies the first countability axiom, but the converse is not true.
3. If $X$ is second countable, then any discrete subspace $A$ of $X$ must be countable.

Both countability axioms are well behaved with respect to the operations of taking subspaces and countable products:

Proposition 347.12(Proposition).

1. A subspace of a (first/second) countable space is (first/second) countable.
2. A countable product of (first/second) countable spaces is (first/second) countable.

In a first countable space, convergent sequence are adequate to detect limit points of sets and to check continuity of functions; see Prp 372.31.

Two important consequences of the second countability axiom relate to notions we have encountered before in the context of metric spaces.

Proposition 347.13(Munkres (2000) 30.3).

Let $X$ be a second countable space. Then,

1. Every open covering of $X$ contains a countable subcover. (Thm 178.6)
2. $X$ is separable. (Thm 178.4, $⟸$)

1. A space for which every open cover contains a countable subcover is called a Lindelöf space.
2. The Lindelöf property and separability are weaker in general than the second countability axiom.
3. separability and first countability together do not imply second countable either; the proof of (Thm 178.4, $⟹$) does not go through when the metric space hypothesis is replaced by first countability.
4. However, both properties are equivalent to second countability for metrizable spaces. Given a Lindelöf metric space, the countable union ${\bigcup }_{r\in {\mathbb{Q}}_{>0}}{B}_r$ where $B_r$ is a countable subcover of the collection of all balls of radius $r$ is a countable basis.

Example 347.14(Example).

1. The space ${\mathbb{R}}_{\ell }$ is first countable, separable, Lindelöf, but not second countable.
2. The product of two Lindelöf spaces need not be Lindelöf: ${\mathbb{R}}_{\ell }^2$ is not.
3. A subspace of a Lindelöf space need not be Lindelöf. The ordered square (being compact) is Lindelöf, but the subspace $[0,1]\times (0,1)$ is not.

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References

Munkres, J. R. (2000). Topology (2nd ed). Prentice Hall, Inc.