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Given a set $A$ with $n$ elements, the finest topology is powerset $2^A$ and the coarsest is the trivial topology $\{ \emptyset, A\}$. And it seems to count the exact number of different topologies in between is an open problem.

But maybe to fulfill my curiously, do we know if the number of different topologies is finite?

peng yu
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    Every topology is an element of $\mathcal P(\mathcal P(X))$, where $X$ is the set on which the topology is defined. If $X$ is finite, $\mathcal P(\mathcal P(X))$ is too, so the answer is yes. – Rushabh Mehta Mar 17 '21 at 16:32

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Yes, the number of different topologies on a set of cardinality $n$ is at most $2^{2^n}$, the number of sets of subsets of a set of cardinality $n$ (since every topology is a set of subsets of the set of points).

Note, incidentally, that there are two possible ways to interpret "the number of topologies on a set with $n$ elements." We could count the literal number of topologies on such a set, or the number of topologies up to homeomorphism. The latter is vastly smaller than the former in general; for example, there are $355$ different topologies on a $4$-element set but only $33$ different homeomorphism types. The former is, to my understanding, significantly easier to count than the latter; for example, there is a simple formula for the "log-asymptotic" behavior of the former, but to the best of my knowledge no such result is known for the latter.

Noah Schweber
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