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If we take X=R and Y=Z and E=N i.e. {1,2,3,4,5.........} then since for this case E is open in Y (as Y is itself an entire metric space) however there does not exist any open set G in X for this particular set. then how E is open in Y.

Pl clarify

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    These X, Y, and$E$are the letters from baby Rudin, though he does not specifically call them the reals, naturals, etc. This is think is a problem created to understand the theorem 2.30. Pl clarify:= Please Clarify.2015-10-17

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Take $G = (0,\infty)$.

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If you want to be fancy, let $U=\bigcup_{n\in\Bbb Z^+}\left(n-\frac12,n+\frac12\right)\;;$ then $U$ is an open subset of $\Bbb R$, and $U\cap\Bbb Z=\Bbb Z^+$. But there’s no need to go to so much trouble, as Sean’s answer shows. For that matter, you could observe that $\Bbb Z_{\le0}=\Bbb Z\setminus\Bbb Z^+$ is a closed set in $\Bbb R$, so $\Bbb R\setminus\Bbb Z_{\le0}$ is an open set in $\Bbb R$ whose intersection with $\Bbb Z$ is clearly $\Bbb Z^+$.