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If a set is endowed with the discrete topology then a subset is compact iff it is finite.

Is the converse true? That is, given a Hausdorff topological space such that every compact subset is finite, then is the topology necessary discrete?

Alex
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2 Answers2

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The converse is not true. Consider the set of continuous functions $[0,1]\to \mathbb{R}$ in the box topology. $X$ is Hausdorff, and a sequence $x_1,x_2,\ldots$ of points of $X$ converges if and only if it is eventually constant by a diagonal argument using the fact that if two continuous functions differ at only finitely many points then they are equal. Therefore any infinite subset of $X$ cannot be compact because there exist sequences of points in the set with no convergent subsequences. $X$ is not discrete because one point sets are not open.

Matt Samuel
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  • Nice example, thanks :). – Alex Nov 26 '14 at 01:45
  • @Alex You're welcome. – Matt Samuel Nov 26 '14 at 01:45
  • @Alex I found a kink in my argument, but I fixed it. A sequence of points in the box topology can converge if all points in the sequence have equal coordinates except in finitely many places. However, this can't happen with continuous functions $[0,1]\to\mathbb{R}$ without them being equal. – Matt Samuel Nov 26 '14 at 02:01
  • Ok, I just "belived" the characterisation of convergent sequence in the Box topology (which turns out to be false as you explained) as I'm used the product topology. But using the continuous functions solved that anyway. – Alex Nov 26 '14 at 02:19
  • @Alex Sorry to mislead you. For a full proof, suppose $f_1,f_2,\ldots$ is a sequence of continuous functions that is not eventually constant. Let $f$ be any continuous function, and for simplicity assume $f$ does not occur in the sequence (otherwise we could just remove it). Then for each $i$ there are uncountably many points where $f$ and $f_i$ differ. Choose distinct $x_i\in [0,1]$ such that $f(x_i)\neq f_i(x_i)$ and take a neighborhood of $f$ that misses all of those values. – Matt Samuel Nov 26 '14 at 02:26
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There is also another example, kind of standard, Arens space.

As yet another example consider the Stone-Cech compactification of the integers $\beta \Bbb N$, and take $x$ in the remainder and consider the space $X=\{x\}\cup \Bbb N$. It is known that every compact subset of $\beta \Bbb N$ mist be either finite, or have cardinality $2^\frak c$, so compact subsets of $X$ must be finite.

These two examples are not compactly generated, where a space $X$ is compactly generated, also called a $k$-space if it satisfies the following condition: A subspace $A$ is closed in $X$ if and only if $A\cap K$ is closed in $K$ for all compact subspaces $K\subseteq X$.

It is clear that if $X$ a is compactly generated Hausdorff and all compact subsets are finite the $X$ has the discrete topology. The class of compactly generated spaces includes all metrizable spaces, as well as all 1st-countable spaces, and all locally compact spaces.

Mirko
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  • Hum interesting, so what if we replace "Hausdorff" with "metrizable" in the condition then? Would it still not work? All of these counterexamples are non-metrizable sapces. – Alex Nov 26 '14 at 01:52
  • @Alex in a non-discrete metrizable space we can find a point that has a nontrivial Cauchy sequence converging to it. Then the union of the sequence and its limit is compact. – Matt Samuel Nov 26 '14 at 02:09
  • A topological space is anticompact if every compact subspace is finite. A first-countable T$_1$-space (more generally, a sequential T$_1$-space) is anticompact if and only if it's discrete. See the answers and comments to this old question. – bof Nov 26 '14 at 03:10