Proposition

Let $X$ be a second countable topological space. Show that every collection of disjoint open subsets of $X$ is countable.

Solution

Let $B = \{ B_1, B_2, \cdots \}$ be a countable basis of $X$. Without loss of generality, assume each $B_i$ is nonempty.

It suffices to show that every collection of disjoint, nonempty open subsets of $X$ is countable.

Let $\{ U_{\alpha} \} \subset \mathcal{T}$ be given where $I$ is the index set. Assume $U_{\alpha} \ne \emptyset$ for each $\alpha \in I$. Suppose that $U_{\alpha} \cap U_{\beta} = \emptyset$ if $\alpha \ne \beta$. We will show that $I$ is countable.

Let $f: I \rightarrow \mathbb{N}$ be defined such that \(B_{f(\alpha)} \subset U_{\alpha}\) for each $\alpha \in I$. Each \(U_{\alpha}\) is a nonempty, open subset of $X$, so it is the union of at least one basis element. Therefore, such a function must exist.

Let $\alpha, \beta \in I$. If $f(\alpha) = f(\beta)$, then $B_{f(\alpha)} \subset U_{\alpha} \cap U_{\beta}$. Since each basis element is nonempty, this is only possible if $\alpha = \beta$.

Therefore, $f$ is an injective map from $I$ into $\mathbb{N}$, so $I$ is countable.