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

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?

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Thanks for asking. I feel more grown up now that I know about this theorem. – Richard Kent Aug 13 '11 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, replacing closed sets of positive measure by dense $G_\delta$ on interval). – Gro-Tsen Aug 17 '11 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 '11 at 16:35

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.