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Definitions:

Let $E$ be a subset of $X$. By an extension of a function $f: E \to \mathbb R$, I mean a function $\bar f: X \to \mathbb R$ such that $f = \bar f$ on $E$.

Question: For every continuous function $f: \mathbb Q \to \mathbb R$, does there exist an extension $\bar f: \mathbb R \to \mathbb R$ that is continuous at each $q \in \mathbb Q$?

In other words, can every continuous function on the rationals be extended to a function on the reals that is continuous at every rational point?

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  • $\begingroup$ Just pick a rational function with an irrational zero in the denominator and no rational zero in the denominator, like $f(x) = 1/(x^2-2)$. $\endgroup$
    – KConrad
    Jun 5, 2021 at 6:05
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    $\begingroup$ This would still be continuous on each $q \in \mathbb Q$ though. $\endgroup$
    – Nate River
    Jun 5, 2021 at 6:05
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    $\begingroup$ Sorry, which part was unclear? I will edit it. $\endgroup$
    – Nate River
    Jun 5, 2021 at 6:11
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    $\begingroup$ Your update has an error: you don't want to declare the extension to the reals to be continuous there, since that would make it automatically continuous on the subset $\mathbf Q$. I still think it would be nice to know what your motivation is (or just random curiosity?). $\endgroup$
    – KConrad
    Jun 5, 2021 at 6:25
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    $\begingroup$ Oh, so this is a “toy case” of the following problem: if a function is continuous on a full measure set, does it agree a.e. with a function that is continuous a.e.? The motivation for this problem being... idle curiosity... $\endgroup$
    – Nate River
    Jun 5, 2021 at 6:27

1 Answer 1

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You may extend by upper limit.

Details.

Denote $g(y)=\limsup_{x\to y, x\in \mathbb{Q}} f(x)$ for all real $y$. So, possibly $g$ takes the value $+\infty$ or $-\infty$ at some points. But we have $g(x)=f(x)$ for rational $x$, and for each rational $x$ and each $\varepsilon>0$ there exists $\delta>0$ such that $|g(y)-f(x)|<\varepsilon$ whenever $|y-x|<\delta$, in particular $g(y)\ne \pm \infty$ if $|y-x|<\delta$. Thus if we replace all infinite values of $g$ to 0, $g$ remains continuous at rational points.

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  • $\begingroup$ Wow this just works huh.. I suppose this same proof would work if we replaced $\mathbb Q$ with any dense subset $S$ of $\mathbb R$? $\endgroup$
    – Nate River
    Jun 5, 2021 at 6:30
  • $\begingroup$ @NateRiver yes of course, and also with $\mathbf{R}$ (at the origin) replaced with any topological space. $\endgroup$
    – YCor
    Jun 5, 2021 at 6:33
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    $\begingroup$ @FedorPetrov I said arbitrary dense subset of arbitrary topological space. For arbitrary subset some additional argument seems needed. $\endgroup$
    – YCor
    Jun 5, 2021 at 6:38
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    $\begingroup$ ah, I am afraid that in the non-normal spaces the general counterexample to Tietze extension theorem is still a counterexampe, like this math.stackexchange.com/a/143072/166817 $\endgroup$ Jun 5, 2021 at 6:45
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    $\begingroup$ And the target space can be assumed to be an arbitrary locally compact space (replacing limsup with an arbitrary cluster point, and an arbitrary point when none). But this is not optimal, since it then follows that it works when the target space is an arbitrary closed subset of an arbitrary product of locally compact spaces. $\endgroup$
    – YCor
    Jun 5, 2021 at 8:32

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