# Compactness of a closed subspace

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.

• Don't you need to consider an open cover of $C$ (w.r.t. the topology $\tau_C$) in order to prove $C$ compact? – Lord Shark the Unknown Jan 25 at 18:18
• Yes, sorry I made a mistake. I meant : $\mathcal{V} = (C \cap U_{\alpha})$. – deeppinkwater Jan 25 at 18:22
• No, you need to start with an arbitrary open cover of $C$. – Ted Shifrin Jan 25 at 18:25
• 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 $X$. – Rhys Steele Jan 25 at 18:26

## 2 Answers

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 $$[0,1]$$.

• Thank you for your answer and this example. – deeppinkwater Jan 25 at 18:34

It is wrong because you did not take an open cover of $$C$$. Instead, you only considered open covers of a certain type.