# 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.

• I believe this can be formulated in $\beth$ numbers, regardless to GCH. Sep 19, 2011 at 7:15
• This question math.stackexchange.com/questions/34838 seems to be related. (It is about number of topologies on a countable infinite set, but some of the answers might be adapted to the more general settings, I guess.) Sep 19, 2011 at 12:52
• @Martin Sleziak: My answer below generalizes one of the answers in the countable case, namely the one using ultrafilters. Sep 19, 2011 at 16:17

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 $X$.

• Welcome Stefan! Sep 19, 2011 at 8:14
• So there are $\beth_3$ many topologies on $\mathbb R$? Sep 19, 2011 at 13:44
• @Asaf: Yes, if I have not miscounted the number of times the exponential gets iterated. The real point here is that there is the maximal possible (wrt. the easily proved upper bound) number of pairwise non-homeomorphic topologies on any infinite set. If you want to use the $\beth$-notation in an even fancier way, an infinite set of size $\kappa$ carries $\beth_2(\kappa)$ many pw non-homeomorphic topologies. Sep 19, 2011 at 14:32
• I thought so, it was my hunch when I was trying to write an answer of my own. Either way, some entrance you made on this site! :-) Sep 19, 2011 at 14:42
• The second half of this answer can be simplified: Given a topology on a set $X$ of size $\kappa$, there are at most $2^\kappa$ other topologies on $X$ that are homeomorphic to it, since each one must be the pushforward of your topology under some bijection $X\to X$. So as soon as you know there are $2^{2^\kappa}$ different topologies (regardless of what they look like), you know there must be $2^{2^\kappa}$ different homeomorphism classes. Jul 30, 2016 at 6:12

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.