A Topological Basis can be thinned out by a smaller one
# Statement
Let $X$ be a Topological Basis and let $\mathcal{M}$ be a Topological Basis. Then if $\mathcal{B}$ is another Topological Basis for $X$, there exists $\mathcal{B}’ \subset \mathcal{B}$ so that
- $\mathcal{B}’$ is a Topological Basis for $X$.
- $|\mathcal{B}’| \leq \max\limits(|\mathcal{M}|, \aleph_{0})$
# Proof
Construct Index Set $I = {(M_{1}, M_{2}) : M_{1}, M_{2} \in \mathcal{M}, \exists B \in \mathcal{B} \text{ so that } M_{1} \subset B \subset M_{2}}$. Using Axiom of Choice, choose $B_{\alpha}$ for $\alpha \in I$ so that $$\alpha_{1} \subset B_{\alpha}\subset \alpha_{2}$$ and call this Set $\mathcal{B}’$. Since $I \subset \mathcal{M} \times \mathcal{M}$, we have $|\mathcal{B}’| = |I| \leq |\mathcal{M}|^{2} \leq \max\limits(|\mathcal{M}|, \aleph_{0})$.
Now we must show $\mathcal{B}’$ is a Topological Basis. Let $U \subset X$ be Open. Then, because $\mathcal{M}$ is a Topological Basis, there exists ${M_x}{x \in U} \subset \mathcal{M}$ s.t. $\bigcup\limits{x \in U} M_{x} = U$. Because each $M_x$ is Open, there exists $B \in \mathcal{B}$ so that $x \in B \subset M_{x}$. Since $B$ is Open, there is a $M’ \in \mathcal{M}$ so that $x \in M’ \subset B$. Thus $\alpha := (M’, M_{x}) \in I$ so there is a $B_{\alpha} \in \mathcal{B}’$ so that $x \in M’ \subset B_{\alpha} \subset M_x$. Denote this $B_{\alpha}=: B_{x}$. Then $$\bigcup\limits_{x \in U} B_{x} = U$$ because for each $x \in U$, $x \in B_{x}$ and $B_{x} \subset M_{x} \subset U$. Thus $\mathcal{B}’$ is a Topological Basis. $\blacksquare$
# Encounters
# Remarks
- This could be tightend up even further to assert that $\mathcal{B}’$ will be a finite if $\mathcal{M}$ is.