Understanding the proof: $M$ is a complete enumerable metric space $\Rightarrow$ $M$ has an isolated point

Question

Theorem 1: $M$ is a complete enumerable metric space $\Rightarrow$ $M$ has an isolated point.

My teacher proved Theorem 1 this way:

$M = \{x_1,x_2,...,x_n,...\}$.

$M = \bigcup_{n \in \mathbb{N}} \{x_n\}$

Baire theorem imply that $\exists n \in \mathbb{N}$ such that $\text{int}(\{x_n\}) \neq \emptyset$.

Then, since $\text{int}\{x_n\} = \{ x_n \}$, $\{ x_n\}$ open $\Rightarrow x_n $ isolated.

$ \square$

Why does Baire theorem imply that? If $M$ were meagre then $\text{int(M)} = \emptyset$. Why is that an absurd?

A set $M$ is meagre if $\exists F_n$ closed sets with $\text{int}(F_n) = \emptyset$ and $M \subseteq \bigcup_{n=1}^{\infty} F_n$.

Baire theorem says $M$ is meagre, then $\text{int}(M) = \emptyset$

Why are the sets $\{x_n\}$ closed?

Please someone explain. Thanks.

Answer 1

The Baire category theorem for complete metric spaces says that every complete metric space is a Baire space; this means that if $X$ is a complete metric space, then $X$ is not the union of countably many meagre subsets. The space $M$ is the union of the countably many singleton sets $\{x_n\}$, and it’s complete, so at least one of the sets $\{x_n\}$ must be non-meagre.

This non-meagre set $\{x_n\}$ is closed because every singleton set $\{x\}$ in a metric space is closed: if you don’t know this already, you should try to prove it. You can do so by showing that if $x\ne y$, then $x\notin\operatorname{cl}\{y\}$. A closed set is meagre if and only if its interior is empty; $\{x_n\}$ is non-meagre, so its interior must be non-empty. The only non-empty subset of $\{x_n\}$ is $\{x_n\}$ itself, so the interior of $\{x_n\}$ must be $\{x_n\}$. Thus, $\{x_n\}$ is an open set, and by definition $x_n$ is an isolated point of $M$.

Answer 2

There are several issues here.

First of all, the Baire category theorem does not say that the interior of a meagre set is empty. For instance, the interior of the whole metric space is always the whole metric space, but the whole metric space may be meager (as it is in this case)!

Rather, BCT says that no complete metric space is a countable union of nowhere dense sets. So since $M=\bigcup \{x_i\}$, we have that some $\{x_i\}$ is not nowhere dense. That is, there is some nonempty open set $U$ such that $\{x_i\}$ is dense in $U$.

We now want to show that this means that $\{x_i\}$ is open. How will we do this? Well, consider the $U$ as above. If $U$ contains some other point $y\not=x_i$, then $\{x_i\}$ is not dense in $U$: $U\cap B_{d(x_i, y)\over 2}(y)$ is an open set containing $y$, contained in $U$, which does not intersect $\{x_i\}$ (why?). So we must have $U=\{x_i\}$, that is, $\{x_i\}$ is open. But "$\{x_i\}$ is open" means exactly that $x_i$ is an isolated point.

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You also ask why each singleton is closed. This isn't important to this proof, but it's a basic fact about metric spaces: any singleton in a metric space is closed, since metric spaces are Hausdorff.