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I was looking at the following proposition :

"Every closed subspace of a compact topological space is compact"

and I am wondering why the following proof is not good :

Let $(X, \tau)$ be a compact topological space and $C \subset X$ be a closed set in $(X, \tau)$ endowed with the induced topology $\tau_{C}$.

Let $A$ be an arbitrary set. For any open cover $\mathcal{U} = (U_{\alpha})_{\alpha \in A}$ of $(X, \tau)$, $\mathcal{V} = (C \cap U_{\alpha})_{\alpha \in A} = (V_{\alpha})_{\alpha \in A}$ is an open cover of $(C, \tau_{C})$.

Since $(X, \tau)$ is compact, for any open cover $(U_{\alpha})_{\alpha \in A}$ of $(X, \tau)$, there exists a finite subcover $(U_{\alpha_{i}})_{i = 1, ..., n}$.

So $(V_{\alpha_{i}})_{i = 1, ..., n}$ is a finite subcover for $(C, \tau_{C})$. $\square$

In that proof, we don't need the fact that $C$ is closed, but I cannot manage to convince myself that it's wrong (because if I replace $C$ by an open set, I think it works also and it is obviously a false statment). No need to give me the usual proof (https://proofwiki.org/wiki/Closed_Subspace_of_Compact_Space_is_Compact) that I understand, I just don't get why the one above is not good.

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    $\begingroup$ Don't you need to consider an open cover of $C$ (w.r.t. the topology $\tau_C$) in order to prove $C$ compact? $\endgroup$ – Lord Shark the Unknown Jan 25 at 18:18
  • $\begingroup$ Yes, sorry I made a mistake. I meant : $\mathcal{V} = (C \cap U_{\alpha})$. $\endgroup$ – deeppinkwater Jan 25 at 18:22
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    $\begingroup$ No, you need to start with an arbitrary open cover of $C$. $\endgroup$ – Ted Shifrin Jan 25 at 18:25
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    $\begingroup$ Not all open covers of $C$ are of the form $\{C \cap U_{\alpha}\}_{\alpha \in I}$ where $\{U_{\alpha}\}_{\alpha \in I}$ is an open cover of $X$. $\endgroup$ – Rhys Steele Jan 25 at 18:26
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It is not a good proof because you didn't prove that any open cover of $C$ has a finite subcover. You showed that if you take an open cover of $X$ and intersect all the sets in it with $C$ then you have a finite subcover. But what if you build an open cover of $C$ in a different way? For example, if you take the set $(0,1)$ then the open cover $\{(\frac{1}{n},1)\}_{n=1}^\infty$ has no finite subcover, despite $(0,1)$ being a subspace of $[0,1]$.

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  • $\begingroup$ Thank you for your answer and this example. $\endgroup$ – deeppinkwater Jan 25 at 18:34
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It is wrong because you did not take an open cover of $C$. Instead, you only considered open covers of a certain type.

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