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First and Second Countability of Metric Spaces 📂Topology

First and Second Countability of Metric Spaces

Theorem

Explanation

After seeing all sorts of abstract spaces in topology, one comes to realize how convenient and nice a space a metric space is.

Proof

[1]

For a metric space $\left( X , d \right)$, if we let $x \in X$, then $$ \left\{ \left. B_{d} \left(x , {{1} \over {n}} \right) \ \right| \ n \in \mathbb{N} \right\} $$ is a countable local basis at $x$, so $X$ is first-countable.

[2]

For a metric space $\left( X , d \right)$, suppose there exists a countable and dense $A \subset X$; then $X$ is a separable metric space. Since $A$ is countable, $$ \mathscr{B} := \left\{ \left. B_{d} \left(a , {{1} \over {n}} \right) \ \right| \ a \in A, n \in \mathbb{N} \right\} = \bigcup_{ a \in A } \left\{ \left. B_{d} \left(x , {{1} \over {n}} \right) \ \right| \ n \in \mathbb{N} \right\} $$ is also countable. If we show that this $\mathscr{B}$ is a basis of $X$, the proof is complete.

For an open set $U$ of $X$, if we let $x \in U$, then there exists $r>0$ satisfying $B_{d} \left( x , r \right) \subset U$. Let us take $n_{x} \in \mathbb{N}$ whose reciprocal is smaller than half of $r$, that is, satisfying $\displaystyle {{1} \over {n_{x}}} < {{r} \over {2}}$. Since $A$ is dense, $$ a_{x} \in A \cap B_{d} \left( x , {{1} \over {n_{x}}} \right) $$ there exists an $a_{x}$ satisfying this. Then $$ B_{d} \left( a_{x} , {{1} \over {n_{x}}} \right) \in \mathscr{B} $$ and $$ x \in B_{d} \left( a_{x} , {{1} \over {n_{x}}} \right) \subset B_{d} \left( x , r \right) \subset U $$ so $\displaystyle U = \bigcup_{x \in U} B_{d} \left( a_{x} , {{1} \over {n_{x}}} \right)$.

Through these two theorems, we can learn the following fact.

Corollary

Euclidean space and Hilbert space are second-countable.