Which way does the inclusion go? Lemma Let $\mathcal{B}$ and $\mathcal{B'}$ be bases for topologies $\mathcal{T}$ and $\mathcal{T'}$, respectively, on $X$. Then the following are equivalent:


*

*$\mathcal{T'}$ is finer than $\mathcal{T}$.

*For each $x\in X$ and each basis element $B\in \mathcal{B}$ containing $x$, there is a basis element $B'\in \mathcal{B'}$ such that $x\in B' \subset B$.


Why don't we write "for each $x\in X$ and each basis element $B\in \mathcal{B}$ containing $x$, there is a basis element $B'\in \mathcal{B'}$ such that $x\in B' \supset B$." (isn't that also true?) instead? I can see that the original statement is true but it seems very counterintuitive. 

The idea is that we need every $\mathcal{T}$-open set to be $\mathcal{T}'$-open. Since $\mathcal{B}$ is a basis for $\mathcal{T}$, then every $\mathcal{T}$-open set is a union of $\mathcal{B}$-elements (and every union of $\mathcal{B}$-elements is $\mathcal{T}$-open), so it suffices that every $\mathcal{B}$-element is $\mathcal{T}'$-open. Since $\mathcal{B}'$ is a basis for $\mathcal{T}'$, then we must show that every $\mathcal{B}$-element is a union of $\mathcal{B}'$-elements, which is what the Lemma shows.
