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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.

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    $\begingroup$ Successor ordinals. $\endgroup$
    – Goldstern
    Commented Apr 11, 2013 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
    Commented Apr 11, 2013 at 18:12
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    $\begingroup$ Crossposted at MSE: math.stackexchange.com/questions/358583/… $\endgroup$
    – Zhen Lin
    Commented Apr 11, 2013 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
    Commented Apr 11, 2013 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
    Commented Oct 12, 2014 at 17:04

1 Answer 1

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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.

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  • $\begingroup$ And the empty set is trivially ok. ;) $\endgroup$
    – user56097
    Commented Sep 21, 2017 at 9:51

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