[Munkres Topology] 30. The Countablility Axioms

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Definition. A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\mathcal{B}$ of neighborhoods of $x$ such that each neighborhood of $x$ contains at least one of the elements of $\mathcal{B}$. A space that has a countable basis at each of its points is said to satisfy the first countability axiom, or to be first-countable.

Theorem 30.1. Let $X$ be a topological space.

(a) Let $A$ be a subset of $X$. If there is a sequence of points of $A$ converging to $x$, then $x\in \bar{A}$; the converse holds if $X$ is first-countable.

(b) Let $f:X\rightarrow Y$. If $f$ is continuous, then for every convergent sequence $x_n\rightarrow x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. The converse holds if $X$ is first-countable.

Proof. Chap 21.

Definition. If a space $X$ has a countable basis for its topology, then $X$ is said to satisfy the second countability axiom, or to be second-countable.

Example that has metric but not second-countable: $\mathbb{R}^\omega$.

Theorem 30.2. A subspace of a first-countable space if first-countable, and a countable product of first-countable spaces is first-countable. A subspace of a second-countable space is second-countable, and a countable product of second-countable spaces is second-countable.

Proof. [subspace] Consider the second countability axiom. If $\mathcal{B}$ is a countable basis for $X$, then $\{ B\cap A | B \in \mathcal{B}\}$ is a countable basis for the subspace  $A$ of $X$. [product] If $\mathcal{B}_i$ is a countable basis for the space $X_i$, then the collection of all products $\prod U_i$, where $U_i \in \mathcal{B}_i$ for finitely many values of $i$ and $U_i = X_i$ for all other values of $i$, is a countable basis for $\prod X_i$ (remember projection).

The proof for the first countability axiom is similar.

Definition. A subset $A$ of a space $X$ is said to be dense in $X$ if $\bar{A}=X$.

Theorem 30.3. Suppose that $X$ has a countable basis. Then:

(a) Every open covering of $X$ contains a countable subcollaction covering $X$.

(b) There exists a countable subset of $X$ that is dense in $X$.

Proof. Let $\{B_n\}$ be a countable basis for $X$.

(a) Let $\mathcal{A}$ be an open covering of $X$. For each positive integer $n$ for which it is possible, choose an element $A_n$ of $\mathcal{A}$ containing the basis element $B_n$. The collection $\mathcal{A}'$ of the sets $A_n$ is countable, since it is indexed with a subset $J$ of the positive integers. Furthermore, it covers $X$: Given a point $x\in X$, we can choose an element $A$ of $\mathcal{A}$ containing $x$. Since $A$ is open, there is a basis element $B_n$ such that $x\in B_n\subset A$. Because $B_n$ lies in an element of $\mathcal{A}$, the index $n$ belongs to the set $J$, so $A_n$ is defined; since $A_n$ contains $B_n$, it contains $x$. Thus $\mathcal{A}'$ is a countable subcollection of $\mathcal{A}$ that covers $X$.

(b) From each nonempty basis element $B_n$, choose a point $x_n$. Let $D$ be the set consisting of the points $x_n$. Then $D$ is dense in $X$: Given any point $x$ of $X$, every basis element containing $x$ intersects $D$, so $x$ belongs to $\bar{D}$.

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