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I have a couple questions regarding the following proof:

2.30 Theorem: Suppse $Y \subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset $G$ of $X$.

Proof: Suppose $E$ is open relative to $Y$. To each $p \in E$ there is a positive number $r_p$ such that the conditions $d(p, q) < r_p, q \in Y$ imply that $q \in E$. Let $V_p$ be the set of all $q \in X$ such that $d(p, q) < r_p$, and define $G = \bigcup_{p \in E} V_p$. Then $G$ is an open subset of $X$.

Since $p \in V_p$ for all $p \in E$, it is clear that $E \subset G \cap Y$.

And the proof goes on...

My questions are the following:

  1. Is the statement "$p \in V_p$ for all $p \in E$" true because if I have a point $p$, there is a point nearby $p'$ such that $p \in V_{p'}$?

  2. I see why $E \subset G \cap Y$. But why is $E \neq G$?

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    $\begingroup$ I assume $X$ is here a metric space and $Y$ a subspace with iunduced metric? For otherwise the statement of the theorem looks like the definition of relative open set. $\endgroup$ – Hagen von Eitzen Sep 28 '12 at 18:41
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  1. The statement is true because $V_p$ is defined as the set of $q\in X$ such that $d(p,q)<r_p$, and if we let $q=p$ we get $d(p,q)=0$ which is certainly less than $r_p$.

  2. It is possible that $E=G$, but this is not a problem. For an example where $E\neq G$, consider $X=\mathbb R$ and $Y=E=\{0\}$, and take $r_0=1$ so $G=(-1,1)$.

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  • $\begingroup$ Thanks for such a clear and quick response! $\endgroup$ – Student Sep 28 '12 at 18:45
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  1. $p\in V_p$ is true because $V_p$ is the set of points at distance $<r_p$ from $p$, and this includes $p$.

  2. It is not claimed that $E\ne G$. (I assume $\subset$ is used as $\subseteq$, not as $\subset\atop\ne$)

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