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If in a topological space only finite subsets are compact sets, is it then the discrete topological space?

Thank you.

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    $\begingroup$ A finite indiscrete space is a counterexample. Do you want to add the assumption that the space is infinite? $\endgroup$ Oct 7, 2013 at 5:57
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    $\begingroup$ @JonasMeyer Infinite isn't enough; the sum of an infinite discrete space and a finite indiscrete space is a counterexample. I believe I've heard spaces with the property "all compact sets are finite" called "anticompact" spaces. $\endgroup$
    – bof
    Oct 7, 2013 at 6:06

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A topological space is anticompact if all compact sets are finite. In this answer Stefan H pointed out that an uncountable set with the cocountable topology is an example of an anticompact $T_1$-space which is not discrete. The real line, with the topology generated by the usual open sets and the cocountable sets, is an example of an anticompact Hausdorff space which is not discrete.

The set $\mathbb N$ of natural numbers, with the topology consisting of the sets not containing $1$ and the sets of asymptotic density $1$, is an example of a countable zero-dimensional Hausdorff space which is anticompact but not discrete. This is the Appert space.

P.S. In the last example, instead of taking the sets of asymptotic density $1$ as neighborhoods of the point $1$, we could take the sets $U$ such that $\sum_{n\notin U}\frac1n\lt\infty$; maybe this example is slightly easier or more familiar.

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The examples given by bof are quite nice.

On the other hand, for $T_1$ and first countable spaces, being "anticompact" is equivalent to being discrete. (If you require compact spaces to be Hausdorff, as Bourbaki does, then "$T_1$" should be replaced by "Hausdorff").

Indeed, assume that $X$ is a $T_1$, first-countable and anticompact space. To show that $X$ is discrete, it is enough to show that any point $x\in X$ has a finite neighbourhood (then, since $X$ is $T_1$, it follows that in fact every point is isolated).

Assume that some point $x\in X$ has the property that every neighbourhood of $x$ is infinite. Let $(V_n)_{n\in\mathbb N}$ be a countable basis of neighbourhoods for $x$. Since each $V_n$ is infinite, one can construct a sequence $(x_n)\subset X$ such that the $x_n$ are pairwise distinct and $x_n\in V_n$ for all $n$. Then the sequence $(x_n)$ converges to $x$, so $K=\{ x\}\cup\{ x_n;\;n\in\mathbb N\}$ is compact. This contradicts the anticompactness of $X$.

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    $\begingroup$ More generally, we can replace "first countable" with "sequential": every anticompact sequential $T_1$-space is discrete. $\endgroup$
    – bof
    Oct 7, 2013 at 23:52
  • $\begingroup$ @bof Thanks for the typo, and the second comment! $\endgroup$
    – Etienne
    Oct 8, 2013 at 5:29
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Just for the record, here is a proof of the proposition mentioned by bof in his comment to the answer by Etienne.

Proposition: Every anticompact sequential $T_1$ space is discrete.

Proof: Given an arbitrary point $x$ in an anticompact sequential $T_1$ space $X$, the complement $X\setminus\{x\}$ is sequentially closed. Otherwise there would be a sequence $(x_n)$ converging to $x$ with each $x_n\ne x$. By the $T_1$ property, such a sequence must contain infinitely many different points, so the set of all those points together with $x$ would form a compact infinite set, contradicting the anticompactness property. Therefore, by the sequential property, $X\setminus\{x\}$ is closed and its complement $\{x\}$ is open.

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