User:Sapiens24/sandbox

Hypolorenose

In order to see this, let $$\mathcal{U}$$ be an open cover of a topological space $$\mathcal{T}$$ which has a countable basis $$\mathcal{B}$$. Then, $U=\bigcup\{ B\in \mathcal{B}\mid B\subseteq U\}$ for all $$U\in \mathcal{U}$$. Especially, the system $\mathcal{B}_{\mathcal{U}}\colon=\bigcup_{U\in \mathcal{U}}\{B\in \mathcal{B} \mid B\subseteq U\}$ is non-empty. As $$\mathcal{U}$$ covers $$\mathcal{T}$$ and as $\bigcup \mathcal{U}=\bigcup_{U\in \mathcal{U}}\left(\bigcup\{B\in \mathcal{B}\mid B\subseteq U\}\right)=\bigcup\left(\bigcup_{U\in \mathcal{U}}\{B\in \mathcal{B}\mid B\subseteq U\}\right)=\bigcup \mathcal{B}_{\mathcal{U}}$, the system $$\mathcal{B}_{\mathcal{U}}$$ is a cover of $$\mathcal{T}$$ as well. Moreover, $$\mathcal{B}_{\mathcal{U}}\subseteq \mathcal{B}$$ implies that $$\mathcal{B}_{\mathcal{U}}$$ is countable. Since for every $$B\in \mathcal{B}_{\mathcal{U}}$$ the set $$\{U\in \mathcal{U}\mid B\subseteq U\}$$ is non-empty by definition of $$\mathcal{B}_{\mathcal U}$$, there exists $u\in \prod_{B\in \mathcal{B}_{\mathcal{U}}}\{U\in \mathcal{U}\mid B\subseteq U\}$ by the axiom of (at least countable) choice. The system $\mathcal{V}\colon=u(\mathcal{B}_{\mathcal U})=\{u(B)\mid B\in \mathcal{B}_{\mathcal{U}}\}$ is a subsystem of $$\mathcal{U}$$ by construction. It is countable because $$\mathcal{B}_{\mathcal{U}}$$ is. And since $$\mathcal{B}_{\mathcal{U}}$$ covers $$\mathcal{T}$$ and since $\bigcup\mathcal{B}_{\mathcal{U}}\subseteq \bigcup \mathcal{V} $, the system $$\mathcal{V}$$ is also a cover of $$\mathcal{T}$$. Hence, $$\mathcal{U}$$ has a countable subcover.