Question

the second countability axiom, then f (X) satisfies the same axiom. Show that if X has a countable dense subset, every collection of disjoint open sets in X is countable.

Show....

looking for not only a solution but also an explaination.

Answer 1

Suppose that $\{r_n : n\in \mathbb{N}\}$   is a countable dense set in

Then let $\mathbb{U}$ be any family of pairwise disjoint non-empty open sets.

For every
UE U
, there is some
n(C) E N
such that $r_{n(U)}\in U\in\mathbb{U}$

This defines a function: $f: \mathbb{U}\to \mathbb{N}$ defined by If
U V and U, V E U
then either
n(C)メnV ( )
or and then $r_m \in U_{n(U)} \cap U_{n(V)}$ , a contradiction with the disjointness of members of $\mathbb{U}$ . So f is at most 1-1 and so $\mathbb{U}$ is at most countable.