What is the cardinality of the set of all topologies on $\mathbb{R}$? This was asked on Quora. I thought about it a little bit but didn't make much progress beyond some obvious upper and lower bounds. The answer probably depends on AC and perhaps also GCH or other axioms. A quick search also failed to provide answers.

Let me give a slightly simplified version of Stefan Geschke's argument.  Let $X$ be an infinite set.  As in his argument, the key fact we use is that there are $2^{2^{|X|}}$ ultrafilters on $X$.  Now given any ultrafilter $F$ on $X$ (or actually just any filter), $F\cup\{\emptyset\}$ is a topology on $X$: the topology axioms easily follow from the filter axioms.  So there are $2^{2^{|X|}}$ topologies on $X$.
Now if $T$ is a topology on $X$ and $f:X\to X$ is a bijection, there is exactly one topology $T'$ on $X$ such that $f$ is a homeomorphism from $(X,T)$ to $(X,T')$ (namely $T'=\{f(U):U\in T\}$).  In particular, since there are only $2^{|X|}$ bijections $X\to X$, there are only at most $2^{|X|}$ topologies $T'$ such that $(X,T)$ is homeomorphic to $(X,T')$.
So we have $2^{2^{|X|}}$ topologies on $X$, and each homeomorphism class of them has at most $2^{|X|}$ elements.  Since $2^{2^{|X|}}>2^{|X|}$, this can only happen if there are $2^{2^{|X|}}$ different homeomorphism classes.
