Questions on proof of theorem 2.30 in Baby Rudin.

Before I ask questions about the proof I am having trouble differentiating between these two definitions that are related to the proof.

Def. 1. $E$ is open in $X$ means to each point $p \in E\ \exists r>0$ such that $$d(p,q)<r,\ q \in X \Rightarrow q \in E.$$

Def. 2. $E$ open relative to $Y$ if to each point $p \in E\ \ \exists r>0$ such that $q \in E$ whenever $$d(p,q)<r\ \&\ q \in Y$$

To me, both of these definitions seem to be the same (and trivial, to be honest). So what is the "big" difference?

And now, this is the proof my professor gave in class which is more or less the same as Rudin's. I am looking for clarification on some parts (to be honest, most parts) of the proof. My comments and questions in bold refer to the statement above.

$\textbf{Theorem}$. Suppose $Y \subseteq X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y \cap G$ for some open subset $G$ of $X$.

$Proof.$ $(\Rightarrow)$ Suppose $E$ is open relative to $Y$. To each $p \in E,\ \exists r_p >0$ such that

$$d(p,q) < r_p,\ q \in Y \Rightarrow q \in E.$$

Let $V_p$ be the set of all $q \in X$ such that $d(p,q) < r_p$ and define $G = \cup _{p \in E}\ V_p$.

Why do this? Just because is works?

Then $G$ is an open subset of $X$.

$G$ is an open subset of $X$ because arbitrary union of open sets is open. Right?

Since $p \in V_p\ \forall p \in E$, it is clear that $E \subseteq G \cap Y.$

This is not "clear" like my professor states, please explain.

By our choice of $V_p$, we have $V_p \cap E \subseteq E$ for every $p \in E$ so that $G \cap Y \subseteq E$.

This makes sense in my head if I think in terms of sets in general. It's always true that the intersection of two sets is a subset of each of the sets, so it's no different in this case. BUT the words "by our choice" is bothering me, should it? Lastly, I don't understand why the intersection of $G$ and $Y$ would be a subset of $E$.

Hence, $E = G \cap Y$.

$(\Leftarrow)$ Conversely, if $G$ is open in $X$ and $E = G \cap Y$, every $p \in E$ has a neighborhood $V_p \subseteq G$. Then,

$$V_p \cap Y \subseteq E$$

so that $E$ is open relative to $Y$. $□$

Thank you all in advance for the help.

• Your first definition isn't stated correctly. It should read, to each point $p \in E$ there exists an $r>0$ such that $q \in E$ whenever $d(p,q)<r$, which equivalent to saying each point $p \in E$ has around it some open ball $B(p,r)$ such that $B(p,r) \subset E$. Your second definition is analogous, it says $E \subset Y$ is open in $Y$ if each point $p \in E$ has around it some open ball $B(p,r)$ such that $B(p,r) \cap Y \subset E$. A set of the form $B(p,r) \cap Y$ is otherwise known as an open ball in $Y$
– joeb
Commented Feb 23, 2017 at 2:45
• That boxed statement $q \in E \implies q \in E$ was obviously written incorrecctly. That is always true but it not always true that every set is open. Take the set [0,1] not open for every $p \in [0,1]$ there is an $r = -5 billion$ so that if $d(p,q) < -5 billion$ then then if $q \in E \implies q \in E$. So $[0,1]$ is open? That make no sense. Instead the definition is if for every $p \in E$ there is a $r$ so that $d(p,q) < r \implies q \in E$. If $d(p,q) < -5 billion \not \implies p \in [0,1]$. Commented Feb 23, 2017 at 2:47
• Open in universal space X, and open relative to Y are very similar. Open in universal space mean for every point p in E, there is same distance r so that all the points that are within r distance of p, all those points are in E. Open relative to Y just means that instead of looking at all the points in the "universe" we are only looking at all the points in Y. If all the points within r distance of p that are in Y are also in E then E is open in Y. Commented Feb 23, 2017 at 2:51
• Yes, the box was an embarrassing mistake on my part. I fixed it. Commented Feb 23, 2017 at 2:55
• Example: E= (0,538] is not open in the "universe" of all real numbers because if p = 538 any r, d(538,q) < r does not mean q \in E. If Y = [3,240] then E is open relative to Y because for every p in E there is an r so that if d(p,q) < r and q in Y, then the q is also in E. If d(p, q) < 1/2 AND if q \in Y$the$q \in E$. Commented Feb 23, 2017 at 2:58 3 Answers Let$B(p,r)=\{q \in X : d(p,q) < r\}$which is an open ball of radius$r$in$X$.$E$open in$X$means that for any point$p$in$E$we can find an open ball small enough so that every point (of$X$) in the ball is also in$E$, i.e.,$B(p,r)\subset E$. On the other hand, open relative to$Y$means that we only care about points in the ball that are in$Y$, in other words for any point$p$in$E$we may find an open ball containing$p$that is small enough that every point (of$Y$) in the ball is also in$E$. So we are considering$B(p,r)\cap Y \subset E$. On to the proof. What is happening here is that we are going to cover$E$in small open balls$V_p=B(p,r_p)$. This cover is given by$G=\bigcup_{p\in E} V_p$and is (indeed) open as it is a union of open sets. As it is a cover,$E \subset G$. (Not to beat a dead horse, but if this still isn't entirely clear,$p \in V_p$, so$E = \bigcup_{p\in E} \{p\} \subset \bigcup_{p\in E} V_p = G$.) Now, in the hypothesis we have$E \subset Y$and because$E\subset G$, $$E = E\cap E = (E \cap Y) \cap (E \cap G) \subset Y \cap G$$ This fact is really just the obvious fact that if two sets contain a common subset then their intersection also contains this subset. ["By choice"] We constructed each$r_p$from the definition of open relativeness, hence$V_p \cap Y = B(p,r_p) \cap Y\subset E$. Therefore $$G\cap Y = \left(\bigcup_{p\in E}V_p\right) \cap Y =\bigcup_{p\in E}(V_p \cap Y) \subset \bigcup_{p\in E} E = E$$ Don't get too worked up about Rudin saying "by choice", it was really just his way of skipping the explanation I just gave. • I apologize for the late response, I've been busy with classes. Anyway, I looked through your explanation and it makes more sense than Rudin (and my professors)!! I will let it sit in my head for a few days and get back to you. In the meantime, thank you very much! Commented Feb 28, 2017 at 4:48 • This is an excellent explanation. Commented Aug 29, 2017 at 14:01 Consider$E = [0, 2]$then$E$is not open. Consider$p = 0$then there is no$r$so that$d(0,q) < r \implies q \in E$because if$q = 0-r/2$then$q \not \in E$but$d(0, q) = r/2 < r$. But consider$E = [0,2]$and$Y = (1/2, 3/4)$. Consider$p = 0$then for any$r$if$d(0,q) < r$AND if$q\in Y$then$q \in E$. Our counter example of$q = 0-r/2$won't do anymore because$q= 0 -r/2 \not \in Y$. Indeed as$Y \subset E$so if$q \in Y$then$q \in E$so$E$must be open in$Y$. Basically taking$E$open in$Y$just means to restrict our "space" to$Y$. Take$E = (0, 100]$and$Y= [-1, 7]$.$E$is open relative to$Y$. For any point in$p \in E$then for any$r > 0$, if$d(p,q) < r$then$q \in Y$. We don't have to worry about$p = 100$because that's completely beyond any consideration of$Y$--- if$d(100, q) < 93$then$q \not \in Y$so$(d(100, q) < 93$and$q \in Y) \implies q \in E$(also$(d(100,q) < 93$and$q \in Y) \implies \text {I'm a green unicorn}\$).

This theorem also appears in the book Espaços Métricos by Prof. Elon Lages Lima (Example 9 of Chapter 3, Pg. 75). Unfortunately, it is only available in Portuguese.

I find his proof clearer. I've tried to adapt it here.

Proof:

Assume the metric $$d: X \times X \to \mathbb{R}$$. An open ball in $$X$$ with center in $$p$$ and radius $$r>0$$ is $$B(p,r) =\left\{x \in X: d(p,x) < r\right\}$$.

$$E$$ is open in $$Y$$ if and only if for every $$p\in E$$ there is $$r_p > 0$$ such that the open ball in $$Y$$, that is $$B(p,r_p)\cap Y$$, is included in $$E$$.

This means that $$E = \bigcup_{p\in E} \left( B(p, r_p) \cap Y \right) = \left( \bigcup_{p\in E} B(p, r_p) \right) \cap Y = G \cap Y \, ,$$ where $$G$$ is a subset of $$X$$ by the definition of the open ball in X given above.

$$\square$$