LEC CAL1 9

Separable metric spaces

Definition 178.1(Definition).

A topological space $X$ is said to be separable provided there is a countable subset of $X$ that is dense in $X$.

Warning

The notion of separable for a topological space with a dense countable subset is distinct from that of connectedness and separations.

Theorem 178.2(Theorem).

A compact metric space is separable.

Proof.

Let$X$ be a compact metric space. Then, $X$ is totally bounded. For each $n\in \mathbb{N}$, let $B_n$ be a finite collection of balls of radius $1/n$ which covers $X$. Let $D=\{x|B_{1/n}(x)\in B_n\text{ for some }n\}$. Then $D$ is countable and dense.□

Second-countable spaces

Definition 178.3(Definition).

A topological space $X$ is second-countable if there is a countable collection $\{O_n{\}}_{n=1}^{\infty }$ of open subsets of $X$ called a basis such that any open subset of $X$ is the union of a subcollection of $\{O_n{\}}_{n=1}^{\infty }$.

Theorem 178.4(Theorem).

A metric space is separable iff it is second-countable.

Proof.

Suppose$X$ is separable. Let $D$ be a countable dense subset of $X$. Assume $D$ is countably infinite, and let $(x_n)$ be an enumeration of $D$. Then, $\mathcal{A}=\{B_{1/m}(x_n){\}}_{n,m\in \mathbb{N}}$ is a countable collection of open subsets of $X$. We claim that every open subset of $X$ is the union of a subcollection of $\mathcal{A}$. Let $\mathcal{O}$ be an open subset of $X$. Let $x\in \mathcal{O}$. Since $\mathcal{O}$ is open, $B_{2/m}(x)\subseteq \mathcal{O}$ for some $m$. Since $D$ is dense in $X$, we can find $x_n$ such that $x_n\in B_{1/m}(x)$. Now, observe that $x\in B_{1/m}(x_n)\subseteq \mathcal{O}$. Thus, every $x\in \mathcal{O}$ is in some member of $\mathcal{A}$ contained in $\mathcal{O}$.

To prove the converse, suppose there is a countable collection $\{{\mathcal{O}}_n{\}}_{n=1}^{\infty }$ of open sets such that any open subset of $X$ is the union of a subcollection of $\{{\mathcal{O}}_n{\}}_{n=1}^{\infty }$. For each $n$, pick a point in ${\mathcal{O}}_n$ and call it $x_n$. Then the set $\{x_n{\}}_{n=1}^{\infty }$ is countable and dense since every nonempty open subset of $X$ is the union of a subcollection of $\{{\mathcal{O}}_n{\}}_{n=1}^{\infty }$ and therefore contains points in $\{x_n{\}}_{n=1}^{\infty }$.□

Theorem 178.5(Theorem).

Every subset of a separable metric space is separable.

Proof.

Let$E$ be a subset of a separable metric space $X$. From the previous theorem, we know that $X$ has a countable base $\mathcal{A}=\{{\mathcal{O}}_n{\}}_{n=1}^{\infty }$. Now, every element in $\mathcal{B}=\{{\mathcal{O}}_n\cap E{\}}_{n=1}^{\infty }$ is open in $E$ due to subspace topology, and $\mathcal{B}$ is countable. Also, any open set $\mathcal{O}$ in $E$ can be expressed as ${\mathcal{O}}^{\prime }\cap E$, where ${\mathcal{O}}^{\prime }$ is open in $X$. Since ${\mathcal{O}}^{\prime }$ can be expressed as a countable union of elements in $\mathcal{A}$, it follows that $\mathcal{O}$ can be expressed as a countable union of elements in $\mathcal{B}$.□

Lindelöf covering theorem

Theorem 178.6(Theorem).

Let $X$ be a separable metric space, and let $A\subseteq X$. Let $F$ be an open covering of $A$. Then, there exists a countable subcollection of $F$ which also covers $A$.

Proof.

Since$X$ is separable, it has a countable basis $\mathcal{A}=\{{\mathcal{O}}_n{\}}_{n=1}^{\infty }$. Express each $f\in F$ as a union of elements in $\mathcal{A}$. Then, $\bigcup f$ is a subcollection $\mathcal{B}\subseteq \mathcal{A}$. For each $b\in \mathcal{B}$, pick $f\in F$ such that $f\supseteq b$. The collection of all these $f$ gives a countable subcollection of $F$ which covers $A$.□

Succinctly, “every open cover in a separable metric space has a countable subcover”.

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Baire category theorem

Definition 178.7(Definition).

We call a subset $A$ of a metric space $X$ dense in $X$ if every nonempty open subset of $X$ contains a point in $A$.

We call a subset $A$ of a metric space $X$ hollow in $X$ if $A$ has an empty interior (taken in $X$).

Lemma 178.8(Lemma).

A set is dense if and only if its complement is hollow.

Theorem 178.9(Baire category theorem).

Let $X$ be a complete metric space.

1. Let $\{{\mathcal{O}}_n{\}}_{n=1}^{\infty }$ be a countable collection of open dense subsets of $X$. Then the intersection ${\bigcap }_{n=1}^{\infty }{\mathcal{O}}_n$ is dense in $X$.
2. Let $\{{\mathcal{F}}_n{\}}_{n=1}^{\infty }$ be a countable collection of closed hollow subsets of $X$. Then the union ${\bigcup }_{n=1}^{\infty }{F}_n$ is hollow in $X$.