I'm reading Analysis on Manifolds by Munkres and in the section Review of Topology Munkres states the following theorem without proof:(adsbygoogle = window.adsbygoogle || []).push({});

Let X be a metric space; let Y be a subspace. A subset A of Y is open in Y if and only if it has the form A = U ∩ Y where U is open in X.

All he has defined is a metric space, subspace, B(y;ε) = {x|d(x,y) < ε} which is the ε neighborhood of y. And he defined open sets to be: A subset U of X is said to be open in X if for each y ∈ U there is a corresponding ε > 0 such that B(y;ε) is contained in U.

I have never taken Topology or even read about it, so I wrote a proof for it which I'm not sure is correct. Here it is:

Since A is open then for x0 ∈ A there exists an ε > 0 such that B(x0;ε) is in A. Further, B(x0;ε) is open (I've proved this already). Now taking the union of all such ε neighborhoods of x's in A also produces and open set (I've also proved this). Therefore, letting U = U B(x; ε) proves this direction as clearly, A = U B(x;ε) ∩ Y.

(⇐)

A = U ∩ Y

This means that any x ∈ A is also in U and Y. Therefore, since x ∈ U, which is open, there exists B(x; ε) in U. But I want the open ball to be in A so letting ε' = ε and then B'(x;ε') = B(x;ε)∩Y. However since B(x;ε) is in U, then B'(x;ε') is in U∩Y which is A.

Am I on the right track? Btw, what if I have a metric X, and a closed set Y in X and then I choose U to be open and not entirely in Y but entirely in X. I define A to be U∩Y, but then A isn't open since it contains part of the boundary of Y. What am I not understanding? And if X is a metric space and Y a subspace, if A is open in Y, does it necessarily imply that A is open in X also?

I'd appreciate the help!

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# Open sets and metric spaces

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