# 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 Feb 23 '17 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 $d(p,q) < -5 billion \not \implies p \in [0,1]$. – fleablood Feb 23 '17 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. – fleablood Feb 23 '17 at 2:51
• Yes, the box was an embarrassing mistake on my part. I fixed it. – Curious Math Student Feb 23 '17 at 2:55