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With a complete metric space $X$ and a set $F$ of continuous functions $f: X\to \mathbb{R}$ such that $\forall a\in X\colon \quad F_a=\left \{ f(a)\mid f\in F \right \}$ is bounded, I want to prove that

There exists a non-empty open subset $U\subseteq X$ and $M>0$ such that for all $x\in U$ and $f\in F$, $\left | f(x) \right | \leq M$

The functions $f$, are called in this context "Uniformly bounded" in $U$.

From a hint that the professor gave, Baire's category theorem can be used to solve this, but I haven't found even an approach that I think will lead to solving this. Can you give me any suggestions?

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You know that since $X$ is complete, then if you have $X=\cup_{i\in\mathbb{N}} X_i$ with closed $X_i$ there is some $X_i$ with non empty interior. Then set $X_i:=\{a\in X: \lvert f(a)\rvert\leq i \text{ for any }f\in F\}$, by the hypothesis you made from a certain point on $X_i$ is non empty. $X_i$ are obviously closed sets by continuity of f and by stability under arbitrary intersections of closed set. Hence one of the $X_i$ has non empty inerior.

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