What is the cardinality of the set of all topologies on R\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 κ. A topology on X is a set of subsets of X. X has 2κ subsets and there are 22κ collections of subsets of X. This is an upper bound for the number of topologies on X.

Now, choose a point x0X and let Y=X{x0}. Since X is infinite, Y is of size κ, too. Let βY be the Stone-Čech compactification of the space Y with the discrete topology.
βY 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 βY. The space βY is of size 22κ. For each yβYY let τy be the topology on X that makes the map mapping xY to x and x0 to y into a homeomorphism.

This gives you 22κ different topologies on the set X.
In the case of X=R we get 2220 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 22κ ultrafilters on Y
that give you 22κ pairwise non-homeomorphic topologies.
But this needs some more thought.

Ok, I thought about this some more. Let y,zβYY and let f:(X,τy)(X,τz) be a homeomorphism.
For both topologies, x0 is the only non-isolated point of X.
Hence f restricts to a bijection from Y to Y.
There are 2κ bijections from Y to Y.
It follows that for each yβYY there are at most 2κ points zβYY such that (X,τy) and (X,τz) are homeomorphic.
In other words, in the class of topologies of the form τy,
the homeomorphism classes are of size at most 2κ.

But there are 22κ topologies of this form. It follows that there are 22κ pairwise non-homeomorphic topologies on the set X.

Source : Link , Question Author : Alon Amit , Answer Author : Stefan Hamcke

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