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.

  • $\begingroup$ 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 $N_{\epsilon, X}(p) \cap Y \subset E$. $\endgroup$ May 16, 2021 at 21:50
  • $\begingroup$ 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. $\endgroup$ May 16, 2021 at 21:54

2 Answers 2


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


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