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The title says it all: if $f\colon \mathbb{R} \to \mathbb{R}$ is any real function, there exists a dense subset $D$ of $\mathbb{R}$ such that $f|_D$ is continuous.

Or so I'm told, but this leaves me stumped. Apart from the rather trivial fact that one can find a dense $D$ such that the graph of $f|_D$ has no isolated points (by a variant of Cantor-Bendixson), I don't know how to start. Is this a well-known fact?

flag	asked Aug 12 2011 at 23:28 Gro-Tsen 928●3●13

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Thanks for asking. I feel more grown up now that I know about this theorem. – Richard Kent Aug 13 2011 at 0:58

@Cole → I don't think that's what you meant (maybe I misunderstood): $f$ itself might be continuous nowhere: Blumberg's theorem only promises continuity of the restriction $f|_D$. And even concerning continuity of $f|_D$, it's not true that one can strengthen the conclusion to $D$ being comeager (=residual): I believe there exists a real function which takes every real value on any $G_\delta$ which is dense in a non-trivial interval (along the lines of [Sierpiński](matwbn.icm.edu.pl/ksiazki/fm/fm1/fm1113.pdf), replacing closed sets of positive measure by dense $G_\delta$ on interval). – Gro-Tsen Aug 17 2011 at 15:20

Gro-Tsen, what I said is not what I thought I said, since I misunderstood the statement of Blumberg's theorem. I'll delete my comment so that it doesn't confuse anyone else. – Cole Leahy Aug 17 2011 at 16:35

Samuele · Accepted Answer · 2011-08-13 00:13:02Z UTC

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It is a theorem due to Blumberg (New Properties of All Real Functions - Trans. AMS (1922)) and a topological space $X$ such that every real valued function admits a dense set on which it is continuous is sometimes called a Blumberg space.

Moreover, in Bredford & Goffman, Metric Spaces in which Blumberg's Theorem Holds, Proc. AMS (1960) you can find the proof that a metric space is Blumberg iff it's a Baire space.

answered Aug 13 2011 at 0:13

Samuele
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Blumberg's paper is online: ams.org/journals/tran/1922-024-02/… – Richard Kent Aug 13 2011 at 0:18

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Thanks for the reference! For completeness, there's a self-contained and rather nice proof (or at least nicer than Blumberg's original one) of the statement, plus an extensive discussion, in the more general setting, in chapter 8 of Goffman, Nishiura, and Waterman's book Homeomorphisms in Analysis, available online at ams.org/publications/online-books/surv54-index – Gro-Tsen Aug 13 2011 at 23:09

Every real function has a dense set on which its restriction is continuous

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