# Defining Open Subsets in Baby Rudin

1. Dec 29, 2011

### gwsinger

The following two definitions are taken directly from Rudin's Principles of Mathematical Analysis.

(1) OPEN SUBSET DEFINITION: If $G$ is an open subset of some metric space $X$, then $G \subset X$ and for any $p \in G$ we can find some $r_{p} > 0$ such that the conditions $d(p,q) < r_p$, $q \in X$ implies $q \in G$.

(2) RELATIVE OPENNESS DEFINITION: Suppose $E \subset Y \subset X$. We say that $E$ is open relative to $Y$ if for each $p \in E$ there is associated an $r_p > 0$ such that $q \in E$ whenever $d(p,q) < r_p$ and $q \in Y$.

1. Couldn't we just say that $E$ is open relative to $Y$ if and only if $E$ is an open subset of $Y$? Rudin never flat out says this but I just wanted to make sure I wasn't missing something.

2. Suppose $X$ has an isolated point $i$ and it happens that $i \in G$. While it's true that this would preclude $X$ from being an open set (since an open set must be comprised solely of internal points and $i$ is an isolated point), wouldn't it still be a possibility that $G$ could be an open subset of $X$? After all, it's (trivially) true that for $i$ we could find some ball $B$ that satisfies the condition set forth in the definition (1) above. If this is true, wouldn't it follow that an open subset is not necessarily itself an open set?

2. Dec 29, 2011

### micromass

True. But you need to make Y a metric space first, but that's not very hard to do.

X is the entire metric space? Then X is open.

The notion of internal point is somewhat confusing I guess. But if i is isolated, then {i} is always open. This means that i is an internal point of every set!!!!

3. Dec 29, 2011

### pwsnafu

Let X be all the real numbers. Let Y be the natural numbers. Let E be the singleton E={0}.
E is a closed set (with respect to X). But it is open relative to Y. Why? Let p be an element of E (there is only one choice). Pick r = 1/2. Suppose n is a natural number satisfying d(p,n) < 1/2. Then the only choice is n = 0, which is in E.

The statement is true if you say "E is an open subset of Y when Y is a metric space by itself". The point is that Y as a subset of (X,d) is different from (Y,d).

Every metric space X is an open subset of X. This is trivial from Def 1: just set G=X and write out the definition in full.

4. Dec 30, 2011

### gwsinger

Thanks for your responses. I had some serious misconceptions about metric spaces that your examples help me clear up. Here are some conclusions (and one question) I've come up with:

(1) Whether a subset $G$ is open is partially a function of its ambient space $X$. Pwsnafu's example shows this for $X = \mathbb{R}$, $Y = \mathbb{N}$, and $E = \{0\}$.

(2) There is more to a metric space than its underlying set. One reason why $E$ can be open relative to $Y$ but not an open subset of $X$ is that the distance function $d_1$ for $<X,d_1>$ can be different than the distance function $d_2$ for $<Y,d_2>$. For example, let $X = \mathbb{R}^2$, let $Y = \mathbb{R}^1$, and let $E = (a,b)$. Then we have $E$ as open relative to $Y$ but not an open subset of $X$ since the two-dimensional distance function in $\mathbb{R}^2$ is different than the one-dimensional distance function in $\mathbb{R}^1$.

But now I have a question of convention: I assume that when we say "$E$ is open relative to $Y$" we use the distance function that corresponds with $Y$ (in this case $d_2$) to determine this openness, and similarly, when we say that some set $G$ is an open subset of $X$, then we will use the corresponding distance function for $X$ (in this case $d_1$). Am I correct on this?

5. Dec 30, 2011

### Fredrik

Staff Emeritus
Yes, but I would prefer to say "open with respect to d2" instead of "open with respect to Y", since there could be other metrics on the set Y. "open with respect to (Y,d2)" would of course make sense too.