Separable Metric Space is Second-Countable

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Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $M$ be separable.


Then $M$ is second-countable.


Proof

By the definition of separability, we can choose a subset $S \subseteq X$ that is countable and everywhere dense.


Define:

$\BB = \set {\map {B_{1/n} } x: x \in S, \, n \in \N_{>0} }$

where $\map {B_\epsilon } x$ denotes the open $\epsilon$-ball of $x$ in $M$.

We have that Cartesian Product of Countable Sets is Countable.

Hence, by Image of Countable Set under Mapping is Countable, it follows that $\BB$ is countable.


Let $\tau$ denote the topology on $X$ induced by the metric $d$.

It suffices to show that $\BB$ is an analytic basis for $\tau$.


From Open Ball is Open Set, we have that $\BB \subseteq \tau$.


We use Equivalence of Definitions of Analytic Basis.

Let $y \in U \in \tau$.

By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} y \subseteq U$.

By the Archimedean Principle, there exists a natural number $n > \dfrac 2 \epsilon$.

That is:

$\dfrac 2 n < \epsilon$

and so:

$\map {B_{2/n} } y \subseteq \map {B_\epsilon} y$.

From Subset Relation is Transitive, we have $\map {B_{2/n} } y \subseteq U$.

By the definition of everywhere denseness, and by Equivalence of Definitions of Adherent Point, there exists an $x \in S \cap \map {B_{1/n} } y$.

By axiom $\left({M3}\right)$ for a metric, it follows that $y \in \map {B_{1/n} } x$.

For all $z \in \map {B_{1/n} } x$, we have:

\(\displaystyle \map d {z, y}\) \(\le\) \(\displaystyle \map d {z, x} + \map d {x, y}\) Metric Space Axiom $(\text M 2)$
\(\displaystyle \) \(=\) \(\displaystyle \map d {z, x} + \map d {y, x}\) Metric Space Axiom $(\text M 3)$
\(\displaystyle \) \(<\) \(\displaystyle \frac 2 n\)

That is:

$\map {B_{1/n} } x \subseteq \map {B_{2/n} } y$

From Subset Relation is Transitive, we have:

$y \in \map {B_{1/n} } x \subseteq U$


Hence the result.

$\blacksquare$


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