An exercise on Baire Category Theorem

Question

Let $(X,\rho)$ be a complete metric space. Let $\Phi$ be a non-empty set of continuous real valued functions on $X$. Define $g : X \to \mathbb R \cup \infty$ by $g(x)=\sup\{f(x):f\in\Phi\}$. Suppose that $g(x)<\infty$ for each $x \in X$. Let $G$ be the set of all $x$ such that there is a neighborhood $V$ of $x$ in $X$ such that $g$ is bounded above on $V$. Prove that $G$ is open and dense in $X$.

Answer 1

First the fact that $G$ is open. Take $x\in G$. There exists a neighborhood $V$ of $x$ and a real $M$ such that $g(y)\leq M$ for all $y\in V$. There exists an open subset $U$ of $V$ containing $x$. Then every $z\in U$ is in $G$, hence $U\subset G$, and we are done.

We now show that $G$ is dense (this imply $G$ not empty). Let $U$ a not empty subset of $X$. Recall that as an open subset of a Baire space (as a complete metric space is a Baire space), $U$ is also a Baire space. For $n\in \mathbb{N}$, put $F_n=\{x\in U, g(x)\leq n\}$. Then it is clear that $\cup F_n=U$. We show that $F_n$ is closed in $U$. For that let $x_k\in F_n$, $x_k\to x\in U$. We have $g(x_k)\leq n$, hence for $f\in \Phi$, $f(x_k)\leq n$ . As $f$ is continous, this imply that $f(x)\leq n$, and then that $g(x)\leq n$, hence $x\in F_n$. By Baire's theorem, there exist an $m$ such that the interior $W$ of $F_m$ is not empty. Clearly, $W\subset G$. Hence $U\cap G$ is not empty and this show that $G$ is dense.