Rudin Theorem 2.23 Rudin's Theorem 2.23 states that a set $E$ is open if and only if its complement is closed.
Both directions of Rudin's proof involving choosing an $x$, either in $E$ or in $E^c$. But there is no requirement that $E$ or $E^c$ be non-empty. We don't have, "by definition," that $\emptyset$ and $X$, the entire metric space, are both open/closed, though I know it to be true. Am I correct that, for full rigor, we have to rule these out?
EDIT: The proof is below for context.

First, suppose $E^c$ is closed. Choose $x \in E$. Then $x \not \in E^c$, and $x$ is not a limit point of $E^c$. Hence there exists a neighborhood $N$ of $x$ such that $E^c \cap N$ is empty, that is, $N \subset E$. Thus $x$ is an interior point of $E$, and $E$ is open. Next, suppose $E$ is open. Let $x$ be a limit point of $E^c$. Then every neighborhood of $x$ contains a point of $E^c$, so that $x$ is not an interior point of $E$. Since $E$ is open, this means that $x \in E^c$. It follows that $E^c$ is closed.

 A: No, Rudin’s proof is fine as is, though I understand your confusion.
This is definitely a case where Rudin is trying to prove (in the first case) that $E$ is open, which means “For all $x\in E$ blah blah blah.” that definition works when $E$ is empty, too. To prove that, you assume $x$ is any element of $E.$ The proof still works if $E$ is empty. Same for the proof that $E^c$ is closed.
The definitions for closed and open sets of the form “$\forall x\in S\dots$“ are “vacuously true” for empty sets. But you can still prove them by assuming that $x\in S$ and proving the condition we require for $x.$
Rudin could be slightly more clear, since the book is intended for readers who are new to proofs, but this is a pretty common proof techniquue. To prove that $E$ is open, you want to show every point of $E$ has the property, so you assume you have such a point.
Another way to think of it is rephrase Rudin’s proof as a proof by contradiction. If $E$ is not open, then there is some $x\in E$ where the property doesn’t hold. Assuming such $x$ exists reaches a contradiction. So no such $x$ exists, and $E$ is open.
A: For both the entire space $X$ and the empty set $\emptyset$, truthfulness are asserted by checking the definition. In the case of the set being empty, we say that the property is vacuously true.
