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Prove that there doesn't exist any function $f:\mathbb R \to \mathbb R$ that is continuous only at the rational points. Suggestion: For every $n \in \mathbb N$, consider the set $$U_n=\{x \in \mathbb R : \exists U \subset \mathbb R \,\text{open, with}\, x \in U, {\rm diam}(f(U))<1/n\}.$$

I am supposed to prove this statement using the Baire Category theorem. I am not sure but I think that the suggestion points towards trying to express $\mathbb R$ as the union of the sets $U_n$. If I could prove that any $U_n$ is a nowhere dense set and I affirm $\mathbb R=\bigcup_{n \in \mathbb N} U_n$, since the Baire Category theorem says that the interior of a countable union of nowhere dense sets is empty, I would get to an absurd. I have two problems: what does this have to do with the fact that there can't be any function $f$ continuous only at rational points? How can I assure that every $x \in \mathbb R$ is in some $U_n$? Moreover, is there any non-empty $U_n$?

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    $\begingroup$ I think this may work, or at least may be on the right track: you can show that the points of continuity are those that are in $U_n$ for all positive integers $n$. This is a $G_{\delta}$ set, as the intersection of the open sets $U_n$. This means the complement is an $F_{\sigma}$ , which relates to Baire category. $\endgroup$
    – user99680
    Commented Nov 15, 2013 at 3:50
  • $\begingroup$ It works pretty well, thanks! $\endgroup$
    – user100106
    Commented Nov 15, 2013 at 4:09

2 Answers 2

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You've got it backwards. It's not $U_n$ that will be nowhere dense, but its complement.

  1. Show that $\bigcap_n U_n$ is precisely the set of points at which $f$ is continuous.

  2. Show that $U_n$ is open.

  3. Suppose $f$ is continuous at the rationals. Show $U_n$ is also dense. Hence, $U_n^c$ is closed and nowhere dense.

  4. Using the previous statement and the fact that the rationals are countable, write $\mathbb{R}$ as a countable union of nowhere dense sets, contradicting the Baire category theorem.

Taking complements can be used to restate the Baire category theorem in the following equivalent way: a countable intersection of dense open subsets of $\mathbb{R}$ is dense.

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    $\begingroup$ Your answer was extremely helpful, I was really mixed up and I didn't know how to use the suggestion properly. $\endgroup$
    – user100106
    Commented Nov 15, 2013 at 4:07
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Let $f:\Bbb R\to \Bbb R$ and let $D$ be the set of $x\in \Bbb R$ such that $f$ is discontinuous at $x$.

For $q\in \Bbb Q^+$ let $x\in D(q)$ iff $\sup \{|f(y)-f(z)|:y,z\in U\}>q$ whenever U is open and $x\in U.$

Every $D(q)$ is closed. For if $x'\in \overline {D(q)}$ and $U$ is any open set with $x'\in U$ then there exists $x\in U\cap D(q).$ Now, since $x\in D(q)$ and $U$ is open with $x\in U,$ we have $\sup \{|f(y)-f(z)|: y,z\in U\}>q.$ So $x'\in D(q).$

We have $D=\cup_{q\in \Bbb Q^+}D(q).$ So $D$ is an $F_{\sigma}$ set. So $C=\Bbb R \setminus D$ is a $G_{\delta}$ set.

Suppose $C$ is also dense. Let $C=\cap_{n\in \Bbb N}\,U_n$ where each $U_n$ is open. Each $U_n$ is dense because $\overline U_n\supset \overline C=\Bbb R.$ Let $S$ be any countable set, with $S\subseteq \{s_n:n\in \Bbb N\}.$ Then each $U_n\setminus \{s_n\}$ is dense & open, so by Baire, $C\setminus S\supseteq \cap_{n\in \Bbb N}(U_n \setminus \{s_n\})\ne \emptyset.$ So $C$ is not equal to any countable $S$. In particular $C\ne \Bbb Q.$

Remarks: (1). Regardless of the Continuum Hypothesis, if $C$ is a dense $G_{\delta}$ subset of $\Bbb R$ then $|C|=2^{\aleph_0}=|\Bbb R|$. (2). If $C$ is any $G_{\delta}$ subset of $\Bbb R$ then there exists $f:\Bbb R \to [0,1]$ such that (i) $f(x)=0\iff x\in C,$ and (ii) $f$ is continuous at $x \iff x\in C.$

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  • $\begingroup$ I do not understand the general idea of your proof and why you wrote $D(q)$? $\endgroup$
    – Emptymind
    Commented Mar 16, 2020 at 0:52
  • $\begingroup$ @Emptymind . $D(q)$ is defined in the A. The idea is that each $D(q)$ is closed and that $\Bbb Q^+$ is countable so $D$ is $F_{\sigma}$, so the set of points where $f $ is continuous is a $G_{\delta}$ set ... But $\Bbb Q$ is NOT $G_{\delta}..$ $\endgroup$ Commented Mar 17, 2020 at 5:48

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