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A coarser topology has more compact sets. Is this claim easy to verify? An if so, how can we prove it?

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This seems to be the same question as this one. – t.b. Apr 29 '12 at 21:01
Well, you are right. Sorry, I did not see that message. Do I have to delete my message? – Oo3 May 1 '12 at 9:26
No, of course not (actually you can't). I think your question is more useful than the other one because it is asking the question without any superfluous clutter that doesn't really add to the question. I was linking to it mainly for the sake of letting you and others know that there was a similar thread. – t.b. May 1 '12 at 9:28
up vote 12 down vote accepted

More accurately, a coarser topology has no fewer compact sets. Consider the discrete and indiscrete topologies on $\{0,1\}$, for instance: the latter is strictly coarser than the former, but they both have exactly the same compact sets, namely, $\varnothing,\{0\},\{1\}$, and $\{0,1\}$.

The claim is trivial to verify when it’s stated properly:

Proposition: Let $\tau$ and $\tau'$ be topologies on a set $X$ such that $\tau'\subseteq\tau$. If $K\subseteq X$ is $\tau$-compact, then $K$ is also $\tau'$-compact. In other words, $\langle X,\tau'\rangle$ has all of the compact sets of $\langle X,\tau\rangle$ and possibly more besides.

To prove this, suppose that $K\subseteq X$ is $\tau$-compact, and let $\mathscr{U}$ be a cover of $K$ by $\tau'$-open sets. Then $\mathscr{U}\subseteq\tau$, so $\mathscr{U}$ is also a $\tau$-open cover of $K$, and as such it has a finite subcover $\mathscr{V}$. Thus, $K$ is $\tau'$-compact. $\dashv$

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Intuitively, reducing the number of open sets in existence reduces the number of open covers for the space, making it harder to find an open cover without a finite subcover. Consider any space $X$ of infinite cardinality: in the trivial topology, every subset is compact. In the discrete topology, only finite subspaces are compact.

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