Suppose we have an open set $E$ such that $E \subset Y \subset X$ for some metric space $X$. When is $E$ *NOT* open relative to $Y$? Rudin Thm 2.30

Question

Update: There are no counterexamples. This question stemmed from my confusion over the definition of "open relative to." Sorry if it is muddled. If you are similarly confused over the meaning, you might find it enlightening to read my response below as well as my proof of the second part of Theorem 2.30 posted here: Question about the proof of Rudin's Theorem 2.30

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What would be a concrete example where $E$ is not open relative to $Y$?

A few definitions from Rudin:

Suppose $E \subset Y \subset X$ where $X$ is a metric space. To say that $E$ is an open subset of $X$ means that to each point $p \in E$ there is associated a positive number $r$ such that the conditions $d(p,q) < r, q \in X$ imply that $q \in E$.

Additionally, $E$ is open relative to $Y$ if to each $p \in E$ there is associated an $r > 0$ such that $q \in E$ whenever $d(p,q) < r$ and $q \in Y$.

Answer 1

In the question in your title, you start by saying

Suppose we have an open set $E$ such that $E \subset Y \subset X$ for some metric space $X$.

This is slightly ambiguous since it is not clear whether you mean that $E$ is given to be an open subset of $X$ or an open subset of $Y$. I assume you mean the former, so it is better to say

Suppose we have an open set $E \subset X$ such that $E \subset Y \subset X$ for some metric space $X$.

Now, since $E$ is an open subset of $X$, for every $p \in E$ there exists $r > 0$ such that $d(p,q) < r$ for $q \in X$ implies that $q \in E$. So, the same value of $r$ works to show that $E$ is relatively open in $Y$, because if $d(p,q) < r$ for $q \in Y \subset X$, then $q \in E$ by the previous statement, and $E \subset Y$ so $q \in Y$. Thus, there is no example of the kind that you seek.

This does not make Theorem 2.30 superfluous, however.

Some open subsets of $Y$ could indeed be nothing but open subsets of $X$ that happen to be contained in $Y$. I assume this is where your intuition is taking you when you said in the comments

Doesn't $E = Y \cap G \implies E \subset Y$. So wouldn't the proof of the theorem be completely trivial and superfluous if that were always true?

But the fact is there are other subsets of $Y$ that are not of this kind. For instance, let $X = \Bbb{R}$, $Y = [0,1]$ and $E = [0,1)$. Clearly $E$ is not an open set in $X$, but it is open relative to $Y$ because $E = Y \cap (-1,1)$. So, the answer to your question

Perhaps the theorem does not assume that $E$ is an open set at all??

is yes, $E$ is not assumed to be an open set in $\mathbf{X}$ in the hypotheses of the theorem. Let me again emphasise that you must specify which is the ambient space when you say something is or isn't an open set.

Answer 2

If $E \subseteq Y \subseteq X$ where $E$ is open relative to $X$, then it is also open relative to $Y$. Indeed, check that the open sets in $Y$ are those of the form $Y\cap G$ where $G$ is open in $X$. Clearly, $Y \cap E = E$ satisfies this.

For the rest, if this doesn't answer your question let me know. I find it unclear what you are exactly asking.

Answer 3

This question stemmed from some confusion over my understanding of Theorem 2.30 from Principles of Mathematical Analysis and the meaning of "open relative to."

Here are a few considerations that helped me understand this better:

(1.) A closed set (that is, closed in the metric space $X$) can be open relative to another set (i.e. $E$ is open relative to $Y$ does NOT imply that $E$ is open in $X$). For example, every set is open relative to itself. Suppose $E = Y \subset X$. Then for every point $p$ in $E$, we can choose any arbitrary $r > 0$ and find that $d(p,q) < r, q \in Y \Rightarrow q \in E$, because $q \in Y \Rightarrow q \in E$ regardless of our choice of $r$. (If $P \rightarrow Q$, then $P \wedge Z \rightarrow Q$)

(2.) Similarly, I believe a set $E$ is always open relative to any proper subset of itself (but correct me if I'm wrong), again, because $q \in Y \Rightarrow q \in E.$ $\forall q \in Y$.

This leaves us with the more interesting case where $E$ is a proper subset of $Y$. For example, suppose that $Y$ is the (non-open) K-cell such that $Y = \{(x_1, x_2) | 0 < x_1 \leq 2, -2 \leq x_2 \leq 2\}$, and $G = \{(x_1, x_2) | \sqrt{x_1^2 + x_2^2} < 1\}$, and $E = Y \cap G$. Theorem 2.30 proves (among other things) that $E$ is open relative to $Y$.

See my proof of the backwards direction of 2.30 here: Question about the proof of Rudin's Theorem 2.30