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Mar 25, 2020 at 13:48 answer added Maxime Ramzi timeline score: 2
Mar 25, 2020 at 13:38 comment added YCor @MaximeRamzi thanks! (of course you mean iff it's empty or in the ultrafilter). BTW the number of metrizable topologies is $=2^{|X|}$: the upper bound just consists in counting the number of possible distances. Also if you fix $x_0$ in $X$ and consider, for a nonprincipal ultrafilter $\xi$ on $X-\{x_0\}$, open subsets as finite subsets of $X-\{x_0\}$ and subsets containing $x_0$ and intersecting $X-\{x_0\}$ in an element of $\xi$, one gets $2^{2^{|X|}}$ Hausdorff topologies.
Mar 25, 2020 at 13:32 comment added Maxime Ramzi @YCor : the number of ultrafilters on $X$ is $2^{2^{|X|}}$, and there is an injection $\beta X\to T_X$ ( a subset is open iff it is in the ultrafilter)
Mar 24, 2020 at 11:43 history became hot network question
Mar 24, 2020 at 9:24 comment added YCor For every subset $Y$ of $X$ with $|Y|\le |X|$, there exists a metrizable topology on $X$ for which the set of accumulation points is exactly $Y$. Hence $X$ has $\ge 2^{|X|}$ topologies. This is enough to answer the question. [Each topology being an element of $2^{2^X}$, the upper bound is $2^{2^{|X|}}$, so it would be more interesting to ask whether the number of topologies on $X$ is $2^{2^{|X|}}$. For instance I remember once checking that the number of group topologies on the symmetric group $S_\omega$ is $2^{2^{2^{\aleph_0}}}$.]
Mar 24, 2020 at 8:54 answer added Peter LeFanu Lumsdaine timeline score: 6
Mar 24, 2020 at 8:29 answer added Wlod AA timeline score: 4
Mar 24, 2020 at 4:37 answer added Keith Kearnes timeline score: 11
Mar 24, 2020 at 3:43 history edited user153451 CC BY-SA 4.0
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Mar 24, 2020 at 3:36 history asked user153451 CC BY-SA 4.0