The problem is the following:

Let $X$ be a Baire Space and $f : X \to \mathbb{R}$ continuous. Prove that every non empty open set of $X$ has a non empty open subset where $f$ is bounded.

So, let $A \subseteq X, A \neq \emptyset$ be an open set. Take $p \in A$ and look at $f(p) \in \mathbb{R}$. Since $f$ is continuous, there is an open neighborhood $V \subseteq X$ of $p$, such that $f(V) \subseteq B_1(f(p))=(f(p)-1, f(p)+1)$.

Since $A, V$ are open, $A \cap V$ is also open, $A \cap V \subseteq A,$ and $f(A \cap V) \subseteq f(V) \subseteq (f(p) -1, f(p) +1)$, so $f$ is bounded in $A \cap V$. In particular $$f(p)-2 < f(x) < f(p) + 2, \,\,\, \forall x \in A \cap V .$$

What did I do wrong? I never used the fact that $X$ is a Baire Space, so I'm pretty sure I did something wrong, but I can't see it.


Your proof looks fine.

I think whoever wrote the problem was thinking of a proof based on showing that some $f^{-1}([-a,a])$ had nonempty interior, but missed that the desired result holds more generally for simpler reasons.

  • $\begingroup$ Thanks! That was quite a mistake. $\endgroup$ – Luis Vera Sep 3 '16 at 19:05

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