Question about the proof of Rudin's Theorem 2.30

Question

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.

Answer 1

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) $$ q' \in Y, \quad q' \notin E, \quad E = Y \cap G, \quad \text{so} \quad q' \notin Y \cap G, \quad \text{a contradiction.} $$

Answer 2

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 $Y$.

Answer 3

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 $X$

Answer 4

I have found Rudin to be very condensed in his proofs, and it requires reading really carefully what he's writing (and sometimes he compresses his proofs so much that details are lost).

Here's my attempt at unpacking it piece by piece. Not that it's a full proof, just trying to explain what Rudin is getting at, hopefully:

If $G$ is open in $X$

$G$ being open in $X$ means that for each point $g \in G$ there's $r > 0$ such that $d(g,q) < r$ for $q \in X$. (1)

and $E=G\cap Y$, every $p\in E$ has a neighborhood $V_p \subset G$

$E = G \cap Y$ means that $E$ is a subset of $G$, and since (1) applies for all points in $G$, it also applies to all $p \in E$. Let's label every such neighborhood $V_p$.

Then $V_p \cap Y \subset E$, so that $E$ is open relative to $Y$

Now, we have:

• $E = G \cap Y$
• $V_p \subset G$ for all $p \in E$

That means $V_p \cap Y \subset E$. What this tells us, is that there's a nonempty intersection of $Y$ with the neighborhood of each $p \in E$, so there must be such a neighborhood of $p$ in $Y$ as well, eventually proving that $E$ is open relative to $Y$.