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Lavrentieff proved a Theorem which implies that every real valued continuous function defined on a dense subset $D\subseteq \mathbb R$ admits a continuous extension to some $G_\delta $ subset of $\mathbb R$. See Theorem (4.3.20) in "General Topology" by Engelking, or this Mathematics Stack Exchange post.

If the dense subset $D$ is already a $G_\delta $, such as the set of irrational numbers, Lavrentieff's Theorem of course says nothing. So let us be a bit more audacious:

Given a real valued continuous function $f$ defined on $\mathbb R\setminus \mathbb Q$, is there an open subset $U$ of $\mathbb R$, containing all irrational numbers, and a continuous extension of $f$ to U?

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Enumerate the rationals as $\{q_n\}$ and define $f(x) = \sum_{n : q_n < x} 2^{-n}$. Then $f$ is continuous on $\mathbb{R} \setminus \mathbb{Q}$ but cannot be extended continuously to any proper superset of $\mathbb{R} \setminus \mathbb{Q}$.

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  • $\begingroup$ Wonderful !!!!! $\endgroup$ – Ruy Dec 12 '18 at 13:16
  • $\begingroup$ I am writing a paper in which your nice example will fit marvelously, so I plan to cite this post, but I wonder if perhaps you have other sources I should also cite? $\endgroup$ – Ruy Dec 12 '18 at 16:02
  • $\begingroup$ I don't remember where I first learned about this function, but I feel like it's "well known". Gelbaum's Counterexamples in Analysis might have it. I can check my copy later today. $\endgroup$ – Nate Eldredge Dec 12 '18 at 17:24
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    $\begingroup$ This is Remark 4.31 in Rudin’s book Principles of Mathematical Analysis. A generalization of this is presented to show that the set of discontinuities of a monotonic function can be any countable set. $\endgroup$ – Ramiro de la Vega Dec 19 '18 at 12:25
  • $\begingroup$ @RamirodelaVega, thanks for the reference. $\endgroup$ – Ruy Jan 24 at 4:02

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