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First Countability and Second Countability of Metric Spaces 📂Topology

First Countability and Second Countability of Metric Spaces

Theorem

Description

After seeing all sorts of abstract spaces in topology, one realizes how convenient and nice metric spaces are.

Proof

[1]

For a metric space (X,d)\left( X , d \right), if we say xXx \in X, {Bd(x,1n)  nN} \left\{ \left. B_{d} \left(x , {{1} \over {n}} \right) \ \right| \ n \in \mathbb{N} \right\} is a countable local base for xx, hence XX is first-countable.

[2]

If a metric space (X,d)\left( X , d \right) has a countable and dense AXA \subset X, then XX is a separable metric space. Since AA is countable, B:={Bd(a,1n)  aA,nN}=aA{Bd(x,1n)  nN} \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. Showing this B\mathscr{B} serves as a basis for XX completes the proof.

For an open set UU of XX, if we say xUx \in U, there exists r>0r>0 that satisfies Bd(x,r)UB_{d} \left( x , r \right) \subset U. Choose nxNn_{x} \in \mathbb{N} such that the reciprocal is less than half of rr, i.e., satisfies 1nx<r2\displaystyle {{1} \over {n_{x}}} < {{r} \over {2}}. Because AA is dense, axABd(x,1nx) a_{x} \in A \cap B_{d} \left( x , {{1} \over {n_{x}}} \right) exists. Then Bd(ax,1nx)B B_{d} \left( a_{x} , {{1} \over {n_{x}}} \right) \in \mathscr{B} and xBd(ax,1nx)Bd(x,r)U x \in B_{d} \left( a_{x} , {{1} \over {n_{x}}} \right) \subset B_{d} \left( x , r \right) \subset U hence U=xUBd(ax,1nx)\displaystyle U = \bigcup_{x \in U} B_{d} \left( a_{x} , {{1} \over {n_{x}}} \right).

Through these two theorems, the following fact can be known.

Corollary

Euclidean space and Hilbert space are second-countable.