proof of alternative characterization of filter

First, suppose that $𝐅$ is a filter. We shall show that, forany two elements $A$ and $B$ of $𝐅$, it is the case that $A\cap B\in 𝐅$ if and only if $A\in 𝐅$ and $B\in 𝐅$.

By the definition of filter, if $A\in 𝐅$ and $B\in 𝐅$ then $A\cap B\in 𝐅$. Since $A\supseteq A\cap B$ and $𝐅$ is a filter, $A\cap B\in 𝐅$ implies $A\in 𝐅$. Likewise, $A\cap B\in 𝐅$ implies $B\in 𝐅$.

Next, we shall show that any proper subset $𝐅$ of the powerset of $X$ such that $A\cap B\in 𝐅$ if and only if $A\in 𝐅$ and $B\in 𝐅$ is a filter.

If the empty set were to belong to $𝐅$ then for any $A\subset X$, we would have $A\cap \mathrm{\varnothing }=\mathrm{\varnothing }\in 𝐅$. This would imply that every subset of $X$ belongs to$𝐅$, contrary to our hypothesis that $𝐅$ is a propersubset of the power set of $X$.

If $A\subseteq B\subseteq X$ and $A\in 𝐅$, then $A\cap B=A\in 𝐅$. By our hypothesis, $B\in 𝐅$.

The third defining property of a filter — If $A\in 𝐅$ and$B\in 𝐅$ then $A\cap B\in 𝐅$ — is part of ourhypothesis.