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Let $f: \mathbb Q \to \mathbb R$ be a continuous function.

An extension of $f$ is a function $\tilde f: \mathbb R \to \mathbb R$ such that $\tilde f = f$ on $\mathbb Q$.

We say an extension $\tilde f$ of $f$ is maximally continuous if for any other extension $g$ of $f$, we have that if $g$ is continuous at $x \in \mathbb R$, then so is $\tilde f$.

Question: For any continuous function $f: \mathbb Q \to \mathbb R$, does there exist a maximally continuous extension of $f$?

Remark: One can always obtain an extension that is continuous at every point in $\mathbb Q$, see for example, the answer in this post by Fedor Petrov.

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    $\begingroup$ Can't you apply Zorn's lemma to the pairs $(A,g)$, where $\mathbb{Q}\subset A\subset\mathbb{R}$ and $g$ is an extension of $f$ continuous on $A$? $\endgroup$
    – abx
    Dec 1, 2021 at 6:37
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    $\begingroup$ There doesn’t seem to be a natural choice of an upper bound though. $\endgroup$
    – Nate River
    Dec 1, 2021 at 9:44

1 Answer 1

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There even exists a largest set $X$ to which $f$ can be continuously extended.

The trick is the following result (which I state here in more generality, to point out which topological assumptions one needs):

Theorem. Let $X,Y$ be topological spaces, where $Y$ is $T_3$, and let $D \subseteq X$ be dense. Let $f: D \to Y$ be continuous and assume that the following property is satisfied for each point $x \in X \setminus D$ (equivalently, each point $x \in X$):

$(*)$ There exists $y_x \in Y$ such that, for each net $(x_j)$ in $D$ that converges to $x$, the net $(f(x_j))$ converges to $y_j$.

Then $f$ has a (obviously, unique) continuous extension to $X$.

(I gave a proof of this in a topology course a year ago, but unfortunately I don't know a reference - maybe somebody else can help out with a reference?)

Application to your situation. For $D = \mathbb{Q}$, let $X$ be the set of all $x \in \mathbb{R}$ which satisfy property $(*)$. Then $f$ can be continuously extended to $X$, and this extension is clearly the largest one.

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  • $\begingroup$ Ah, so it holds in more generality as well. Great answer, thanks! $\endgroup$
    – Nate River
    Dec 1, 2021 at 9:45
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    $\begingroup$ To answer OP's question, we also need to know that if $f$ has a continuous extension to $X\subset\mathbb{R}$, then $f$ has an extension to $\mathbb{R}$ that is continuous on $X$. This is true, but was not completely obvious to me. $\endgroup$ Dec 1, 2021 at 13:14
  • $\begingroup$ @JulianRosen: Good point! When writing the answer, I was a little bit too much focussed on "finding a continuous extension of $f$ to a domain as large as possible", and forgot to take into account that the question is framed somewhat differently. $\endgroup$ Dec 1, 2021 at 17:09
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    $\begingroup$ You probably want to say that $f$ has a continuous extension to $X$! $\endgroup$ Dec 1, 2021 at 18:27
  • $\begingroup$ @DirkWerner: Oops. Yes, sure. Thank you for your correction! $\endgroup$ Dec 1, 2021 at 19:01

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