In this question, two potential topologies were proposed for the power set of a set $X$ with a topology $\mathcal T$: one comprised of all sets of subsets of $X$ whose union was $\mathcal T$-open, one comprised of all sets of subsets of $X$ whose intersection was $\mathcal T$-open. I outlined a proof there that neither such construction need be a topology on $\mathcal P(X)$ in general. In fact, depending on how one interprets the intersection of the empty set, each will be a topology on $\mathcal P(X)$ if and only if $\mathcal T$ is the discrete topology on $X,$ in which case both are the discrete topology on $\mathcal P(X).$
This led me to wonder if there are any ways to induce a topology on $\mathcal P(X)$ from a topology on $X$? Some searching shows that one "natural" way to topologize $\mathcal P(X)$ is to give it the topology of pointwise convergence of indicator functions $X\to\{0,1\}.$ This is certainly very nice, but ignores the topology on $X,$ so I'm still curious:
Are there any ways to induce a topology on $\mathcal P(X)$ from any given topology on $X$? Ideally, I would like for different topologies on $X$ to give rise to different (though potentially homeomorphic, of course) topologies on $\mathcal P(X).$
Let a set be open iff it is empty or of the form $\mathcal{P}(X)\backslash\big\{\{x\}:x\in C\big\}$ for some closed set C.