# Rudin Theorem 2.30: relative openness

I do not fully understand Rudin's Theorem 2.30. The context, which is not immediately clear from the problem, is a metric space $$(X,d)$$ and a subset $$Y \subset X$$. 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 open subset $$G$$ of $$X$$.

I've replicated Rudin's proof's verbatim below, and will add my questions after.

Suppose $$E$$ is open relative to $$Y$$. To each $$p \in E$$ there is a positive number $$r_p$$ such that that the conditions $$d(p,q) < r_p$$, $$q \in Y$$ imply that $$q \in E$$. Let $$V_p$$ be the set of all $$q \in X$$ such that $$d(p,q) < r_p$$, and define $$G = \bigcup\limits_{p \in E} V_p$$. Then $$G$$ is an open subset of $$X$$, by Theorems 2.19 and 2.24. Since $$p \in V_p$$ for all $$p \in E$$, it is clear that $$E \subset G \cap Y$$. By our choice of $$V_p$$, we have $$V_p \cap Y \subset E$$ for every $$p \in E$$, so that $$G \cap Y \subset E$$. Thus $$E = G \cap Y$$, and one half of the theorem is proved. Conversely, if $$G$$ is open in $$X$$ and $$E = G \cap Y$$, every $$p \in E$$ has a neighborhood $$V_p \subset G$$. Then $$V_p \cap Y \subset E$$, so that $$E$$ is open relative to $$Y$$.

First, I'm not sure that I understand fully what 'open relative to $$Y$$' means. The definition of open is: $$W \subset X$$ ($$X$$ a metric space) is open if for every $$u \in W$$, there exists $$\epsilon > 0$$ such that $$N_{\epsilon} (u) \subset W$$. My understanding of 'relative openness' is as follows. If we have $$W \subset U \subset X$$, and for every $$u \in W$$, there exists $$\epsilon > 0$$ such that $$N_{\epsilon} (u) \subset W \subset U$$, so we s ay $$N_{\epsilon} (u)$$ is "open relative to $$W$$" or open in $$W$$, which I believe are equivalent notions. Is that correct? (If not, what I'm about to say almost will be f alse.)

The first assumption is that $$E$$ is open relative to $$Y$$, so given $$p \in E$$, there is $$r_p > 0$$ so that $$N_{r_p} (p) \subset Y \subset X$$. We call $$V_p = N_{r_p} (p)$$, which is a neighborhood and, by Theorem 2.19, open. The problem is: the theorem only says open, but open relative to what? I assume we say open relative to $$Y$$ since $$V_p \subset W$$. Then $$G$$ is a union of open sets and therefore open by Theorem 2.24, but relative to what? I would assume $$Y$$ again, but Rudin asserts that $$G$$ is an open subset of $$X$$. This isa. problem, because Rudin warns right before stating this theorem that we may have a set that is open relative to $$Y$$ but not an open subset of $$X$$. So this leads me to believe that I'm misinterpreting this, and perhaps $$V_p$$ is open in $$X$$ only.

I think I can follow the remainder of the theorem, provided that I can get this fact straight, as much of it depends on understanding the concept of relative openness.

• Your neighborhoods should either depend on the space or should be intersected with it. For $E \subset Y \subset X$, $E$ is open relative to $Y$ iff for all $p \in E$, there is $\epsilon > 0$ s.t. the $p$-neighborhood $$N_{\epsilon, Y}(p) := \{q \in Y : d(p, q) < \epsilon\}$$ of $Y$ satisfies $N_{\epsilon, Y}(p) \subset E$. This is equivalent to: for all $p \in E$, there is $\epsilon > 0$ s.t. the $p$-neighborhood $N_{\epsilon, X}(p) := \{q \in X : d(p, q) < \epsilon\}$ of $X$ satisfies $N_{\epsilon, X}(p) \cap Y \subset E$. May 16, 2021 at 21:50
• When Rudin considers neighborhoods like $N_{\epsilon}(p)$ without explicit dependence on the space, he is usually considering it within the largest containing space which is $X$ here. May 16, 2021 at 21:54

Concerning the concept of “open relative to”, it's the other way around: if $$X$$ is a metric space, if $$U\subset X$$ and if $$W\subset U$$, $$W$$ is open relative to $$U$$ if, for every $$x\in W$$, there is some open disk $$D_\varepsilon(x)$$ such that $$D_\varepsilon(x)\cap U\subset W$$.

For instance, $$[0,1)$$ is not an open subset of $$\Bbb R$$. But it is open relative to $$[0,\infty)$$: if $$x\in[0,1)$$, take $$\varepsilon>0$$ such that $$(x-\varepsilon,x+\varepsilon)\subset(-1,1)$$ (it exists, since $$(-1,1)$$ is an open subset of $$\Bbb R$$) and then $$(x-\varepsilon,x+\varepsilon)\cap[0,\infty)\subset[0,1)$$.

Let $$X$$ be a metric space with $$E \subset X$$.

If for every point $$p$$ in $$E$$ you can find a radius $$r$$ s.t. $$N_r(p)$$ fits inside $$E$$ without any of its elements falling out of $$E$$, then $$E$$ is open. For example, consider $$[-3, 3].$$ Now $$N_1(3) = (3 - 1, 3 + 1) = (2, 4)$$ which means $$[-3, 3]$$ is not open because for $$r = 1$$, we have $$4 \not \in [-3, 3].$$ We have $$N_r(3) \not \subset [-3, 3]$$ for any $$r > 0$$.

For another example, consider $$F = (-8, 8)$$ with $$N_2(3) = (1, 5) \subset F$$. This shows $$3$$ is an interior point of $$F$$. If $$F$$ contains all its interior points, then $$F$$ is open. In other words, if for any $$p \in F$$, we can find $$r > 0$$ s.t. $$N_r(p) \subset F$$, then $$F$$ is open. Just take $$r = \min\{|-8, -p|, |8 - p|\}$$. Then $$N_r(p) \subset F$$ meaning $$F$$ is open.

Note $$N_2(3) \cap \mathbb R \subset F$$. Now, if for any point $$p \in F$$, we can find some $$r > 0$$ s.t. $$N_r(p) \cap \mathbb R \subset F$$, then $$F$$ is open relative to $$\mathbb R$$. In other words, show that $$q \in \mathbb R$$ and $$q \in N_r(p) \iff d(p, q) < r$$ implies $$q \in F.$$

Here below is the definition of relative openness:

A set $$E$$ is open relative to $$Y$$ if, for each $$p \in E$$, there is an $$r > 0$$ such that $$q \in Y$$ and $$d(p, q) < r \implies q \in E$$. In other words, for every $$p \in E, \ N_r(p) \cap Y \subset E$$ for some $$r > 0$$.

Given the definition above, check out this thread below that deals with the proof in the direction you find difficult.

Show relative opennes wrt set implies intersection