TITLE: Question about the proof of Rudin's Theorem 2.30
QUESTION [5 upvotes]: The theorem states:
Suppose $Y \subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some subset $G$ of $X$.
I think the proof in the forward direction is relatively clear, however I have some problems relating the backward direction. The proof is relatively quick and goes as (Rudin, pg. 36):
If $G$ is open in X and $E = G \cap Y$, every $p \in E$ has a neighborhood $V_p \subset G$ (open ball $B_{r_p}(p) = \{x \in X: d(p, x) < r_p \}$). Then $V_p \cap Y \subset E$, so that $E$ is open relative to Y.
In order to prove that $E$ itself is and open set in $Y$, wouldn't we want to prove that for each $p \in E$, there is an open ball contained in $Y$. Thus would it work to remedy the proof by taking a ball for each $p$ with the following radius:
$r_p' = \min \{ r_p, \sup_{x \in E} d(p, x) \}$ ?
Then we could guarantee that the ball that is guaranteed by the openness of $G$ will let conclude the openness of $E$ relative to $Y$.
Thank you very much.

REPLY [3 votes]: I initially found this proof very confusing because I was hazy on the meaning of "open relative to." See my response to my own confused question here: Suppose we have an open set $E$ such that $E \subset Y \subset X$ for some metric space $X$. When is $E$ *NOT* open relative to $Y$? Rudin Thm 2.30 to develop a little more intuition on what it means for one set to be open relative to another set.
I didn't find Mikhail D's proof of the reverse direction straightforward, so let me present the way that I thought about this. Hopefully someone else who gets stuck on part 2 of Theorem 2.30 will find it illuminating.
Theorem 2.30 Reverse Direction
Suppose that $E = G \cap Y$ for some $G$ that is open in $X$. We must show that $E$ is open relative to $Y$.
Proof: First, note that for every point $p$, there is a neighborhood $V_p \subset G$ (i.e. there is some $r_p > 0$, such that $\forall q$ where $d(p,q) < r_p$, we have that $q \in G$).
To see why this is true, suppose that it were false. Then there would have to be some $p \in E$ (let's call it $p_0$) such that there is NO $r > 0$ such that $N_r(p) \in G$. Now, because $E \subset Y \cap G$, we know that $E \subset G$. Hence $p_0 \in E \Rightarrow p_0 \in G$. Thus, $G$ includes a point $p_0$ such that $p_0$ is NOT an interior point of $G$. But this contradicts our assumption that $G$ was an open subset of $X$!
Hence, we know that for every point $p \in E$, there is a neighborhood $V_p \subset G$.
Now consider a point $q \in V_p \cap Y$. We know that $q \in V_p \Rightarrow q \in G$ (from our conclusion above). Hence $q \in V_p \cap Y \Rightarrow q \in G \cap Y$ (using our assumption that $E = G \cap Y$). Hence, $V_p \cap Y \subset E$.
But this means that for every point $p \in E$, there exists some $r_p > 0$ (namely, the $r_p$ such that $V_p = \{q | d(p,q) < r_p\} \subset G$), such that for all $q$, if it is the case that $d(p, q) < r_p, q \in Y$, then $q \in V_p \cap Y \Rightarrow q \in G \cap Y \Rightarrow q \in E$, which is precisely the condition for $E$ being open relative to $Y$.