Let X be a second countable topological space. Show that every collection of disjoint open subsets of X is countable.
Proof: Let $X$ be a second countable space. Choose ${\{B_{i}\}{_i\in \mathbb{N}} }$ to be our countable basis for the topology on $X$. Let $C$ be any collection of disjoint open subsets of $X$. Our goal is to show that $C$ is countable. More formally, let $C=\{U_{i} \in \{B_{i}\}\subseteq X | \: i \in \mathbb{N}, \: U_{i} \bigcap U_{j} = \emptyset \}$. Now $C$ is countable since for every $U_{i} \in \{B_{i}\} $ there is a correspondence between $i \in \{1, …, n\}$ for $n \in \mathbb{N}$. Thus it follows that $C$ is countable.
Is the construction of the set $C$ precise or is their a better way.
Best Answer
In your construction of $C$, you're assuming it's countable (its elements are indexed by a countable set). What you should do is start with an arbitary collection $C=\{U_\alpha\}$ of pairwise disjoint open subsets of $X$, then show that $C$ is countable.
Hint: Each $U_\alpha$ contains some $B_\alpha$, and since $U_\alpha\displaystyle\cap U_\beta=\varnothing$ if $\alpha\neq\beta$, we know that $B_\alpha\displaystyle\cap B_\beta=\varnothing$ if $\alpha\neq\beta$.