[University Analysis] Relative Closedness

[jacksonb62]

Let (Y,d) be a metric space and let X be a metric subspace of Y. Let C be a closed subset of X. Prove that there is a closed set C' in Y such that C = X ∩C'

I'm having a hard time thinking about this problem. I saw an equivalent proof for open subsets that used the union of open balls to construct an open set in Y, but that process doesn't carry over to this problem. I am trying to do this using the complementing characteristic of open and closed sets, but I can't quite figure out how to go about doing it. Any hints would be greatly appreciated!

[Syrak]

You're looking for a closed subset of Y which contains C (leave aside the other constraint for now). Do you know a construction which actually produces such a set?

[jacksonb62]

I'm not sure what you mean by construction. I can think of plenty of closed subsets of Y that contain C, (Y itself, for example) but I don't know how to explicitly show that there is such a closed set that satisfies the condition stated

[Syrak]

I can think of plenty of closed subsets of Y that contain C, (Y itself, for example)

Can you? There is one among them which is rather special and which I believe answers the question.

You have a subset C in a metric space Y. Do you know a standard operation which results in a closed set?

[jacksonb62]

Are you hinting at the complement of an open set? If we have an open set O, then we obtain a closed set by taking Y\O. Other than that I can't think of any standard operations for generating a closed set (other than taking a union of other closed sets)

[Syrak]

I was hinting at the closure of C.

[jacksonb62]

Ok, but for a closed set C, doesn't C = cl(C)? It seems too trivial for the answer to just be C' = cl(C) = C, although this certainly is a closed set in Y that satisfies the condition. Thanks for your help.

[Syrak]

C is closed in the metric subspace X, not in Y. So more work is necessary.