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I'm taking all of this to be that \(\displaystyle E \subseteq A \subseteq X\) and that X is simply \(\displaystyle \mathbb{R} ^2\) (with it's usual topology.)Let X be the plane with the usual topology while A = E = the x-axis. Then \(\displaystyle Int_A E = A\) while \(\displaystyle Int_X E = \emptyset\), so that the former cannot be obtained by intersecting the latter with A. Itisalways true, however, that \(\displaystyle Int_A E \supset A \cap Int_X E\)

I understand the final equation in the above. I haven't proven it in general but it easily satisfies the mental image that I have. (I probably should make sure to prove it, but this has little to do with the question I'm going to ask.)

So, the question I'm going to ask... Why is \(\displaystyle Int_X E = \varnothing\)? Every open interval in E can be found to be the intersection of an open set in X and E.

Note: My text defines

Thanks!If X is a topological space and \(\displaystyle E \subset X\) then \(\displaystyle Int _X E = \cup \{G \in X | \text{G is open and } G \subset E \}\)

-Dan