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 $X$ be any set of some infinite size $\kappa$. A topology on $X$ is a set of subsets of $X$. $X$ has $2^\kappa$ subsets and there are $2^{2^\kappa}$ collections of subsets of $X$. This is an upper bound for the number of topologies on $X$.

Now, choose a point $x_0\in X$ and let $Y=X\setminus\{x_0\}$. Since $X$ is infinite, $Y$ is of size $\kappa$, too. Let $\beta Y$ be the Stone-Čech compactification of the space $Y$ with the discrete topology.
$\beta Y$ can be thought of as the space of all ultrafilters on $Y$, with the ultrafilter generated by a singleton $\{y\}$ identified with $y$. $Y$ is dense in $\beta Y$. The space $\beta Y$ is of size $2^{2^\kappa}$. For each $y\in\beta Y\setminus Y$ let $\tau_y$ be the topology on $X$ that makes the map mapping $x\in Y$ to $x$ and $x_0$ to $y$ into a homeomorphism.

This gives you $2^{2^\kappa}$ different topologies on the set $X$. In the case of $X=\mathbb R$ we get $2^{2^{2^{\aleph_0}}}$ topologies. (Wow!)

Now, you ask about different topologies, and these are different topologies, even pretty good ones, in terms of separation axioms. What about homeomorphism classes of topologies? I am almost certain that you can construct $2^{2^\kappa}$ ultrafilters on $Y$ that give you $2^{2^\kappa}$ pairwise non-homeomorphic topologies. But this needs some more thought.

Ok, I thought about this some more. Let $y,z\in\beta Y\setminus Y$ and let $f:(X,\tau_y)\to(X,\tau_{z})$ be a homeomorphism. For both topologies, $x_0$ is the only non-isolated point of $X$. Hence $f$ restricts to a bijection from $Y$ to $Y$. There are $2^\kappa$ bijections from $Y$ to $Y$.
It follows that for each $y\in\beta Y\setminus Y$ there are at most $2^\kappa$ points $z\in\beta Y\setminus Y$ such that $(X,\tau_y)$ and $(X,\tau_z)$ are homeomorphic.
In other words, in the class of topologies of the form $\tau_y$, the homeomorphism classes are of size at most $2^\kappa$.

But there are $2^{2^\kappa}$ topologies of this form. It follows that there are $2^{2^\kappa}$ pairwise non-homeomorphic topologies on the set $X$.

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.