Let $(X, d)$ be a separable metric space that the whole family of subsets of x do not empty. open and disjoint two to two is at most countable

Question

let $(X, d)$ be a separable metric space that the whole family of subsets of $X$ do not empty. open and disjoint two to two is at most countable. deduce that the set of isolated points of $X$ is at most countable.

Answer 1

If I understand the question correctly you are asking for a proof of the following fact: let $(X,d)$ be a separable metric space. Show that we cannot have an uncountable family $\{A_i:i\in I\}$ of non empty open subsets of $X$ such that the sets in the family are pairwise disjoint. Here is a proof: let $\{x_1,x_2,..\}$ be a dense sequence in $X$. For each $i\in I$ pick an integer $n(i)$ such that $x_{n(i)} \in A_i$. This is possible because $A_i$ is a non -empty open set, so it must contain at least one element from the dense set $\{x_1,x_2,..\}$. The map $i \to n(i)$ is then a one-to-one map from $I$ into $\mathbb N$ so $I$ is at most countable.