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Given $S^3$ the three dimensional sphere and it's usual Euclidean topology, call it $\tau$, consider $U:=\{(X \setminus A) \mid A \in \tau \}$. Does $U$ form a topology on $S^3$?

My guess is no. First I noticed that $U$ should form a topology on finite sets, but on infinite sets there could be a problem..since $S^3$ is homeomorphic to $R^3$, and the complements on the open sets in $R^3$ are the closed sets, $U$ shouldn't form a topology because one could take an countable union the closed sets such that it is not closed. I'm wondering if this is correct and even if so how I could formalize the idea?

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    $\begingroup$ $S^3$ is not homeomorphic to $\mathbb{R}^3$, but $S^3\setminus\{p\}$ is. $\endgroup$ – Michael Albanese Dec 17 '14 at 22:16
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    $\begingroup$ Infinite intersection of open sets doesn't need to be open. I think, by the way, that "complement" is the right word in english. $\endgroup$ – Peter Franek Dec 17 '14 at 22:19
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As Michael Albanese points out in the comments, there is a flaw in your argument.

Notice that singletons are open in $U$. Since any subset of $S^3$ is a union of singletons, all what you need to do is find a subset that is not open in $U$. This should be straightforward.

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