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Let $(X, \tau)$ be a Baire space, $I$ an index set and for each $x \in X$, let the set $\{f_i(x) : i \in I\}$ be bounded above, where each mapping $f_i : (X,\tau) \to \mathbb{R}$ is lower semicontinuous. Using the Baire Category Theorem prove that there exists an open subset $O$ of $(X,\tau)$ such that the set $\{f_i(x) : x \in O, i \in I\}$ is bounded above.

My proof.

  1. Let $X_n = \bigcap_{i \in I}\big[f_i^{-1}( (-\infty; n])\big]$. Since $f_i$ is lower semicontinuous, $f_i^{-1}( (-\infty; n])$ is closed in $X$ and $X_n$ is closed as an infinite intersection of closed sets. Do we need to tell here that for some $n$, $X_n$ will not be an empty set, because $\{f_i(x) : i \in I\}$ is bounded above?

  2. $\bigcup_{n=1}^{\infty}(-\infty; n] = \mathbb{R}$.

  3. $f_i^{-1}(\mathbb{R}) = X$, for any $f_i$.
  4. From 2 and 3 it follows that $\bigcup_{n=1}^{\infty} X_n = X$ - I'm not sure if it is correct.
  5. Since $X$ is Baire space, then one of $X_n$ is not nowhere dense. Also $X_n$ is closed. Hence $\operatorname{Int}(\bar{X_n}) = \operatorname{Int}(X_n) \neq \emptyset$.
  6. Let $O = \operatorname{Int}(X_n)$ be an open set.
  7. Then $O \subseteq X_n \subseteq f_i^{-1}((-\infty; n])$
  8. Then $f_i(O) \subseteq (-\infty; n]$
  9. And $\bigcup_{i \in I}f_i(O) = \{f_i(x) : x \in O, i \in I\} \subseteq (-\infty; n]$

Complete the proof.

Could you please verify my proof? Thank you!

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The idea is essentially OK, but could be formulated a bit better, my attempt:

So given is: $$\forall x \in X: \exists n_x : \forall i \in I: f_i(x) \le n_x$$

This is just a formal restatement that all sets $\{f_i(x): i \in I\}$ are bounded above.

So again reformulating:

$$\forall x \in X: \exists n_x (\in \mathbb{N}): x \in \bigcap_{i \in I} (f_i)^{-1}[(-\infty, n_x]]$$ or using your definition of $X_n$:

$$\forall x \in X: \exists n_x: x \in X_{n_x}$$

which immediately shows $$X = \bigcup_n X_n$$

without other computations (your 2 and 3 are irrelevant), and each $X_n$ is closed by the $f_i$ all being lsc, and intersections of closed sets being closed, as you stated.

Note that $X_n \subseteq X_{n+1}$ for all $n$, so we have an increasing family.

By the fact that $X$ is Baire we know that $\exists N: \operatorname{int}(X_N) \neq \emptyset$.

Now if $x \in O, i \in I$ we know that $x \in X_N \subseteq f_i^{-1}[(-\infty, N]]$ so that $f_i(x) \le N$. This shows that $N$ is an upperbound for the set $\{f_i(x): x \in O, i \in I\}$, as required.

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  • $\begingroup$ why do we need to note that $X_n$ is an increasing family? $\endgroup$ – Andreo Apr 29 '18 at 20:37
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    $\begingroup$ @Andreo It's not really important but it helps to build a picture, I think. $\endgroup$ – Henno Brandsma Apr 29 '18 at 21:44

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