Pointwise boundedness of continuous functions on a complete metric space implies uniform boundedness

Question

I am in the process of proving the following:

Let $\mathcal{F}$ be a family of continuous real-valued functions on a complete metric space $X$ that is pointwise bounded in the sense that for each $x \in X$, there is a constant $M_{x}$ for which $|f(x)|\leq M_{x}$ for all $f \in \mathcal{F}$.

Then, there is a nonempty open subset $\mathcal{O}$ of $X$ on which $F$ is uniformly bounded in the sense that there is a constant $M$ for which $|f|\leq M$ on $\mathcal{O}$ for all $f \in \mathcal{F}$.

So far, I have done the following:

Define the sets $E_{n} = \{ x \in X: |f(x)|\leq n, \forall f \in \mathcal{F}\} = \cap_{f \in \mathcal{F}}\{x \in X: |f(x)|\leq n\} = \cap_{f \in \mathcal{F}}f^{-1}[-n,n]$, which is closed in $X$, since $\mathcal{F}$ is a family of continuous functions.

Since $\mathcal{F}$ is pointwise bounded, then $\exists x \in E_{n}$ whenever $|f(x)|\leq M_{x} \leq n$, which implies that $X = \cup_{n=1}^{\infty}E_{n}$.

Since $X$ is nonempty, it has nonempty interior, and since $X = \cup_{n=1}^{\infty}E_{n}$, by a previous result, at least one of the $E_{n}$'s, say $E_{n_{0}}$ has nonempty interior; i.e., for $x_{0} \in E_{n_{0}}$, $\exists r>0$ such that $B_{0}(x_{0},r)\subset E_{n_{0}}=\overline{E_{n_{0}}}$.

From here, I have no idea where to go in order to get the uniform boundedness. Please help. I am completely and utterly clueless as to how to proceed.

Answer 1

You are done. All that remains is that you recognise that you are done, and sum it up in a concluding sentence.

You have defined

$$E_n = \bigcap_{f \in \mathcal{F}} f^{-1}([-n,n]),$$

so directly by definition, the family $\mathcal{F}$ is uniformly bounded on each $E_n$ - and on all subsets of $E_n$ - with a uniform bound of $n$, any $M \geqslant n$ is a fortiori a uniform bound of $\mathcal{F}$ on a subset of $E_n$.

You have used Baire's theorem to conclude

at least one of the $E_n$'s, say $E_{n_0}$ has nonempty interior;

so that means $\overset{\Large\circ}{E}_{n_0}$, or $\operatorname{int}(E_{n_0})$, or whatever your preferred notation is, is a nonempty open subset of $X$.

And since $\overset{\Large\circ}{E}_{n_0}$ is a subset of $E_{n_0}$, we know that $\mathcal{F}$ is uniformly bounded by $n_0$ on $\overset{\Large\circ}{E}_{n_0}$.

So we can take $\mathcal{O} = \overset{\Large\circ}{E}_{n_0}$ and $M = n_0$.

We can also take any nonempty open subset of $\overset{\Large\circ}{E}_{n_0}$ and any $M \geqslant n_0$.

Or, since we have the inclusion $E_n \subseteq E_{n+1}$ for all $n$, it follows that if $\overset{\Large\circ}{E}_{n_0} \neq \varnothing$, then $\overset{\Large\circ}{E}_{n} \neq \varnothing$ for all $n \geqslant n_0$, so we can take a nonempty open subset of $E_{n_1}$ for some $n_1 > n_0$ and an $M \geqslant M_1$. We have lots of options to choose from.