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How many second-countable $T_1$ spaces, up to homeomorphism, are there?

Let $X$ be a second-countable $T_1$ space. Since $X$ is second-countable, there are at most $\beth_1$ open sets in $X$. And since $X$ is $T_1$, every singleton set is closed, so there are at most $\beth_1$ points in $X$.

So at most, there are second-countable $T_1$ spaces as many as topologies on spaces with $\beth_1$ points, which are $2^{2^{\beth_1}} = \beth_3$ many. But what's the exact cardinality?

Dannyu NDos
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    The upper bound can easily be improved to $\beth_2^{\aleph_0}=\beth_2$ since a second-countable topology is determined by giving countably many subsets of the space. Then the question becomes a duplicate of https://math.stackexchange.com/a/2132119/86856 – Eric Wofsey Feb 17 '23 at 03:25

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