Open Covers in a Second Countable space reduce to a Countable Subcover

Last updated Nov 1, 2022

Let $X$ be a Topological Space that is Second Countable. Let $S \subset X$ and let $\mathcal{U} := {U_\alpha}_{\alpha \in I}$ be an Open Cover of $S$. Then there exists a Countable Open Subcover $\mathcal{V} \subset \mathcal{U}$ of $S$.

We first prove the result for $X$, then carry it over to $S$ through the Subspace Topology. Let $\mathcal{U}:= {U_\alpha}{\alpha \in I}$ be an Open Cover for $X$. Since $X$ is Second Countable, there exists a Countable Topological Basis $\mathcal{B}$ of $X$. For each $\alpha \in I$, for each $x \in U\alpha$, let $B_{\alpha}^{(x)} \in \mathcal{B}$ be such that $x \in B_{\alpha}^{(x)} \subset U_\alpha$. Now consider $$\mathcal{B}’ := {B_\alpha^{(x)} : \alpha \in I, x \in U_{\alpha}}\subset \mathcal{B}.$$First note that $\mathcal{B}’$ is an Open Cover for $X$, since $\forall x \in X$, there is some $\alpha \in I$ so that $x \in U_\alpha$. Also, it is Countable since $\mathcal{B}$ is Countable.

We use $\mathcal{B}’$ to construct our Countable Open Subcover $\mathcal{V}$. For each $B \in \mathcal{B}’$, choose $\beta \in I$ so that $U_{\beta} \supset B$. Such a $\beta$ exists by construction of $\mathcal{B}’$. Denote this sub-Index Set $J \subset I$ and set $\mathcal{V} := {U_{\beta}}_{\beta \in J}$. Then, for all $x \in X$

1. There exists $B \in \mathcal{B}’$ so that $x \in B$, since $\mathcal{B}’$ is an Open Cover of $X$.
2. There exists $U \in \mathcal{V}$ so that $U \supset B$, by construction of $\mathcal{V}$. Thus $x \in U$.

Therefore, $\mathcal{V} \subset \mathcal{U}$ is an Open Subcover. Since we only made one choice per $B \in \mathcal{B}’$, $\mathcal{V} \preccurlyeq \mathcal{B}’$ and $\mathcal{V}$ is Countable.

Now we carry the result over to an $S \subset X$. Suppose $\mathcal{U}$ is an Open Cover of $S$. Endow $S$ with the Subspace Topology. Since Subspace Topology inherits Second Countability, and $S \cap \mathcal{U}$ is an Open Cover in $S$, we can reduce to a Countable Open Subcover $\mathcal{V}{S}$. Then, for each $V{S}\in \mathcal{V}{S} \subset S \cap \mathcal{U}$ there exists a $V \in \mathcal{U}$ so that $V{S} = S \cap V$. Thus $\mathcal{V}{S}$ induces a $\mathcal{V} \subset \mathcal{U}$ so that $$S = \bigcup\limits{V \in \mathcal{V}} V \cap S \subset \bigcup\limits_{V \in \mathcal{V}}V$$ so $\mathcal{V}$ is an Open Subcover of $S$ in $X$. Since $\mathcal{V}_{S}$ was Countable, $\mathcal{V}$ is as well. $\blacksquare$

• Surjection
• Set Cardinality