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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?

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    $\begingroup$ 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. $\endgroup$ – Balarka Sen Apr 17 '16 at 8:16
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    $\begingroup$ Related: math.stackexchange.com/questions/226413/… (You can use the same idea for groups as for topological space.) $\endgroup$ – Martin Sleziak Apr 17 '16 at 10:10
  • $\begingroup$ @MartinSleziak, thanks for this amazing link! $\endgroup$ – Silent Apr 17 '16 at 13:07
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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.

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  • $\begingroup$ So, in fact I can say that there isn't any set of 'all trivial topological spaces', right? Thank you so much. $\endgroup$ – Silent Apr 17 '16 at 9:12
  • $\begingroup$ Yes, that would correct. $\endgroup$ – Henrik supports the community Apr 17 '16 at 9:35

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