Show that a separable metric space is secondcountableSolutio

Show that a separable metric space is second-countable.

Solution

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

Define:

B = {B_1/n(x) : x S , n N > 0 }

where B(x) denotes the open -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 B is countable.

Let denote the topology on X induced by the metric d.

It suffices to show that B is an analytic basis for .

From Open Ball is Open Set, we have that B .

We use Equivalence of Definitions of Analytic Basis.

Let y U .

By the definition of an open set, there exists a strictly positive real number such that B(y) U.

By the Archimedean Principle, there exists a natural number n > 2.

That is, (2/n) < , and so B_2/n(y) B(y).

From Subset Relation is Transitive, we have B_2/n(y) U.

By the definition of everywhere denseness, and by Equivalence of Definitions of Adherent Point, there exists an x S B_1/n(y).

By axiom (M3) for a metric, it follows that yB1/n(x).

For all z B_1/n(x), we have:

That is, B_1/n(x) B_2/n(y).

From Subset Relation is Transitive, we have y B_1/n(x) U.

d(z,y)

d(z,x) + d(x,y)

by axiom (M2) for a metric

=

d(z,x) + d(y,x)

by axiom (M3) for a metric

<

2/n