# Question about the proof of Rudin's Theorem 2.30

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.

• 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? – DanielWainfleet Nov 24 '15 at 23:54
• Isn't it $V_p\cap Y$ exactly the ball in $Y$? – Patricio Nov 24 '15 at 23:56
• @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. – user118837 Nov 25 '15 at 0:01
• @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 $V_p \subset of Y$. – user118837 Nov 25 '15 at 0:04
• 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 $Y$. – Patricio Nov 25 '15 at 0:13

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

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

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 $$Y$$.

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

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 $X$