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Suppose that we have two topological spaces $(X,T_1)$, $(X,T_2)$ such that $T_1\subset T_2$ and $K\subset X$ is compact in $T_1$ topology. Is it also compact in $T_2$ topology ???

I think that the answer is yes, because we can cover $K$ with a finite collection of open sets that belong in $T_1$ topology, but we know that these open sets belong also in $T_2$, so it is going to be compact in $T_2$.

Is this true ??

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  • $\begingroup$ To show compactness in $T_2$, you have to be able to find finite subcovers from any covering by $T_2$-open sets. If you can only handle coverings by $T_1$-open sets, that's not good enough. Remember that compactness is not just about being able to cover $K$ - it's about being able to find a finite subcover from any horrible covering that someone might hand you. $\endgroup$ – Nate Eldredge Feb 23 '18 at 20:58
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    $\begingroup$ More open sets $\rightarrow$ fewer compact sets. This is the whole point behind using weak topologies. $\endgroup$ – Umberto P. Feb 23 '18 at 23:30
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No, but it does go the other way.

Consider an extreme example. Consider the set $\mathbb{N}$ with two different topologies. First, let $T_1 = \{ \emptyset , \mathbb{N} \}$, the "trivial topology", and let $T_2 = \mathcal{P}( \mathbb{N} )$ be the discrete topology on $\mathbb{N}$. Then $\mathbb{N}$ is compact in $T_1$, but not in $T_2$. Specifically, consider the open cover $\mathcal{U} = \{ U_k \}_{k \in \mathbb{N}}$, where $U_k = \{ k \}$. Then $\mathcal{U}$ is an open cover of $\mathbb{N}$ with no finite subcover.

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We can consider $X=[0,1]$, with $T_1$ standard, and $T_2$ discrete and $K=X$. Certainly, $T_1 \subset T_2$, but $[0,1]$ is not compact in the $T_2$ topology.

On the other hand, suppose that $T_1 \subset T_2$, and $K$ is compact wrt $T_2$. Let $U_\alpha$ be some open cover of $K$ in $T_1$, then, since $U_\alpha$ are open in $T_2$ as well, you can find a finite subcover, so it is compact in $T_1$ as well.

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  • $\begingroup$ Thank you very much , but why K is not compact in discrete topology? I can't really see that . $\endgroup$ – Jonathan1234 Feb 23 '18 at 19:56
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    $\begingroup$ @Jonathan1234 just take for your cover the collection of all singletons. You can't find a finite subcollection that works. $\endgroup$ – Andres Mejia Feb 23 '18 at 19:57
  • $\begingroup$ no problem, by the way. $\endgroup$ – Andres Mejia Feb 23 '18 at 19:57
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There is a theorem which you might have met already which answers your question fairly definitively:

If a topology is both compact and Hausdorff, then it is maximal as a compact topology and minimal as a Hausdorff topology. That is any strictly larger topology is not compact and any strictly smaller topology is not Hausdorff.

Moreover, this follows from this theorem:

Suppose $f: X \to Y$ is is a continuous bijection, $X$ is compact and $Y$ is Hausdorff. Then $f$ is a homeomorphism.

Moreover, this follows from two even more basic results:

A compact subset of a Hausdorff space is closed, and a closed subset of a compact space is compact.

and

The continuous image of compact set is compact.

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  • $\begingroup$ I'd never considered that corollary, but it's fascinating no less. $\endgroup$ – AJY Feb 26 '18 at 5:45

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