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 $E \neq G$?

  • 1
    $\begingroup$ 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. $\endgroup$ Sep 28, 2012 at 18:41

2 Answers 2

  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 $G=(-1,1)$.

  • $\begingroup$ Thanks for such a clear and quick response! $\endgroup$
    – Student
    Sep 28, 2012 at 18:45
  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 $\subset\atop\ne$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.