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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.

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  • $\begingroup$ You should say some OPEN G, subset of X. Since the subspace topology on Y is defined this way, your Q is equivalent to the def'n. Does the book have a different way of defining the subspace topology? $\endgroup$ – DanielWainfleet Nov 24 '15 at 23:54
  • $\begingroup$ Isn't it $V_p\cap Y$ exactly the ball in $Y$? $\endgroup$ – Patricio Nov 24 '15 at 23:56
  • $\begingroup$ @user254665 As far as I know, he does not define as subspace topology. Thus I was assuming if $(X, d)$ is taken to be a metric space, then so is $(Y, d)$ for $Y \subset X$ with the same definition of the metric. $\endgroup$ – user118837 Nov 25 '15 at 0:01
  • $\begingroup$ @Patricio As far as I understand it, it does not have to be the case. Since $V_p \subset G$ guaranteed by openness of $G$ does not necessarily imply $V_p \subset of Y$. $\endgroup$ – user118837 Nov 25 '15 at 0:04
  • $\begingroup$ That's for sure, what I'm saying is that $V_p\cap Y$ is the ball in Y, so you have a ball, namely $V_p\cap Y$, contained in $Y$. $\endgroup$ – Patricio Nov 25 '15 at 0:13
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This two-liner proof actually means quite a bit. I am pasting the whole proof from my notes.

In this theorem, we analyze a metric space $X$ with its [non-empty] subsets, $Y, E$.

Abstract

Proof in two parts, forward and reverse directions. Forward direction presents a direct proof by construction, reverse direction is a proof by contradiction.

Forward direction

$Y \in X$. $E \subset Y$, $E$ is open relative to $Y$. We have to prove there exists an open set $G \subset X$ such that $E = Y \cap G$.

We explicitly construct such a set by doing the following.

Since $E$ is open to $Y$, then for each point $p \in E$ there exists $r > 0$ and $\mathcal{N}(p)$ such that all points of $E$ in this neighborhood are points of $E$. So we take a union of all such neighborhoods,

$$ G = \bigcup_{p \in E} \mathcal{N}(p) $$

$G$ is open since a union of open sets is open (Rudin 2.24a).

$G$ contains all points of $p \in E$, so $E \subset G$. Since by assumption $E \subset Y$, we have

$$ E = G \cap Y \quad \text{as required.} $$

Reverse direction

Again, we have a metric space $X$, with $E \subset Y$, $Y \subset X$. For this $E$ there exists an open set $G \subset X$ such that

$$ E = Y \cap G \quad (1) $$

We have to prove $E$ is open relative to $Y$. That is, for all $p \in E$ there exists $r>0$ such that

$$ \underbrace{d(p,q) < r \quad \text{for all $q \in Y$ } }_\text{call this 'P'} \quad \text{implies} \quad \underbrace{q \in E}_\text{call this 'Q'} $$

We prove by contradiction. Suppose the assumptions are true, but $E$ is not open relative to $Y$. The logical structure of such statement is this

\begin{align} &\text{Not}(\forall p \in E : \exists r > 0 : \text{P implies Q}) = \nonumber \\ &\exists p \in E : \text{Not}(\exists r : \text{P implies Q}) = \nonumber \\ &\exists p \in E : \forall r : \text{Not}(\text{P implies Q}) = \nonumber \\ &\exists p \in E : \forall r : (\neg Q \text{ and } P) = \nonumber \\ &\underbrace{ \exists p \in E }_\text{(2)} : \forall r : P \text{ and Not} (Q) \nonumber \end{align}

The last line reads as follows. There exists a point $p \in E$ such that whatever $r > 0$ its neighborhood $\mathcal{N}(p)$ 'captures' at least one 'weird' point of Y that does not belong to $E$ ($q' \in Y $, $q' \notin E$) (3).

We have by (2) $p \in E$, and by (1) $E = Y \cap G$, $p \in G$, so $p$ is an interior point of an open subset $G \subset X$. Therefore we can choose $$ r > 0 : \mathcal{N}(p) \subset G \quad \text{(4)} $$

and we will still have a 'weird' point $q' \in \mathcal{N}(p)$ in it. That is, even if we 'squeeze' $\mathcal{N}(p)$ to fit into $G$, it will still contain a point $q' \notin E$.

As soon as we 'fit' $\mathcal{N}(p)$ into $G$, we blow up our initial assumption $E = Y \cap G$ since we have deduced that by (4) $$ q' \in Y, \quad q' \in \mathcal{N}(p) \subset G, \quad \text{so} \quad q' \in G, \text{ or if combined } q' \in Y \cap G $$

But by hypothesis, by (3) and (1) $$ q' \in Y, \quad q' \notin E, \quad E = Y \cap G, \quad \text{so} \quad q' \notin Y \cap G, \quad \text{a contradiction.} $$

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  • $\begingroup$ E is a subset of G, and E is a subset of Y. Why can we get E=G intersection Y? $\endgroup$ – Yao Zhao Nov 3 '20 at 17:06
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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 $Y$.

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This question is a duplicate in disguise (with glasses). Although it was asked before the question,

Restrict a metric, gives same topology as subspace topology from larger space X,

I would vote to close this one out and redirect to the 'guts of the question'.

Also, the wording for this question should state

$\qquad $ for some subset OPEN set $G$ of $X$

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