For every set $X$ and every topology $\tau$ over $X$ we have that $\tau$ contains the trivial topology $\{ X, \emptyset\}$, which is compact, and is contained in the discrete topology $\{ S: S \subseteq X\}$, which is Hausdorff. I was wondering if there is any topology on X "between" the trivial and the discrete such that it has both properties.

It seems that there is such a topology for specific sets, such as the natural numbers, but I haven't found any result for arbitrary $X$. I don't know if any additional condition must be established on $X$ for the result to hold, or if it isn't possible in general.


closed as too localized by Asaf Karagila, Bill Johnson, Andreas Blass, Martin Brandenburg, Goldstern Apr 11 '13 at 18:48

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    $\begingroup$ Successor ordinals. $\endgroup$ – Goldstern Apr 11 '13 at 18:09
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    $\begingroup$ Or one point compactifications of discrete spaces, in case you want to avoid the axiom of choice. $\endgroup$ – Ramiro de la Vega Apr 11 '13 at 18:12
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    $\begingroup$ Crossposted at MSE: math.stackexchange.com/questions/358583/… $\endgroup$ – Zhen Lin Apr 11 '13 at 18:28
  • $\begingroup$ A silly solution: Pick one element x. Declare every finite set NOT containing x to be open. Declare the COMPLEMENT of every finite set NOT containing x to be open. This is a compact Hausdorff topology. $\endgroup$ – Alexander Woo Apr 11 '13 at 18:36
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    $\begingroup$ @AlexanderWoo Not silly at all, although I think it's the same as the one-point compactification mentioned by Ramiro. Same applies to Peter's answer of course. $\endgroup$ – Todd Trimble Oct 12 '14 at 17:04

Choose $x_0\in X$ and declare as open neighborhoods of $x_0$ the subsets which contain $x_0$ and all but finitely many of the points of $X$. Declare all other points of $X$ as open. This is Hausdorff and compact.

  • $\begingroup$ And the empty set is trivially ok. ;) $\endgroup$ – user56097 Sep 21 '17 at 9:51

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