# Is there a set of all topological spaces?

This question is from Willard's General Topology:

Is there a set of all topological spaces?

My try is: Suppose $\mathfrak T$ is set of all topological spaces, then $\mathfrak T$ 'contains' all the sets (i.e., if $S$ is some set, then $\{\varnothing, S\}\in\mathfrak T$). Since Willard assumes that a set cannot be element of itself (exact words in Willard: Russell's paradox can be avoided (in our naive discussion) by agreeing that no aggregate shall be a set which would be an element of itself.) But, if $\mathfrak T$ is a set, then 'set of all sets' is 'subset' of $\mathfrak T$, counter to our agreement.

I find my argument appealing as well as sloppy at the same time :), :(. Have I done something wrong?

• Seems fine to me. If there is a set of all topological spaces, I can give it I dunno the discrete topology and that itself becomes a topological space, hence must be contained it the underlying set of itself. This contradicts Russell. – Balarka Sen Apr 17 '16 at 8:16
• Related: math.stackexchange.com/questions/226413/… (You can use the same idea for groups as for topological space.) – Martin Sleziak Apr 17 '16 at 10:10
• @MartinSleziak, thanks for this amazing link! – Silent Apr 17 '16 at 13:07

Your idea is right, but technically it needs some small adjustment. As a you say it's not $S$ that's in $\mathfrak T$, but $\{\emptyset, S\}$, so "the set of all sets" isn't really going to be a subset, but there's an easy injective map from "the set of all sets" to $\mathfrak T$ so you can easily recreate Russell's paradox.