Suppose (X,τ) is a compact topological space and C ⊆ X is closed. Show that C is compact in (X,τ)

So far I have come up with:

Let P be an open cover of C. Thus C is closed, then its complement, P', is open.

Now, P ∪ { C ' } is an open cover of X and, therefore, of C. Since X is compact, P ∪ { C '} contains a finite subcollection P ' that also covers X and, therefore, covers C.

If C ' is not in P', then P' is a finite subcollection of P that covers C. Otherwise, if we remove C' from P', we get a finite subcollection of P that covers C.

So, every open cover of C contains a finite subcover. It follows C is compact

I'm assuming I will need to show more proof as to how you would reach this conclusion

  • $\begingroup$ If you’re supposed to show that $C$ is a compact subset of $X$, then your proof is just fine as it is. If you’re supposed to show that $C$ is a compact subspace of $X$, then, as Mike Miller notes in his answer, you have to start with a cover of $C$ by sets that are (relatively) open in $C$ and expand them to open sets in $X$ before applying your argument. $\endgroup$ – Brian M. Scott Dec 16 '14 at 22:22

You have to be careful when constructing the open cover of $X$ - the elements of the open cover of $C$ don't have to be open in $X$.

To do this, remember the definition of the subspace topology - $U$ is open in $C$ iff there's an $U' \subset X$ such that $U' \cap C = U$. Now pick these open sets and cover $X$ with them. Continue your proof from there - it shouldn't take more than minor edits to make it work.

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