I can prove that given a function $f:X \rightarrow Y$, where $X,Y$ are metric spaces, the set $A \subseteq X$ of points on which $f$ is continuous, is $G_{\delta}$.

(Take $U_n = \bigcup_{y \in Im(f)} f^{-1}(B_{\frac{1}{n}}(y))$ and $V= \bigcap_{n \in \mathbb{N}}U_n$, and $V$ is $G_\delta$).

My question is: Is the converse direction true? Is it true that, given a $G_\delta$ set $A \subseteq X$, there exists a metric space $Y$, and a function $f:X \rightarrow Y$, such that, the set of points on which $f$ is continuous is $A$?

Thank you! Shir

  • 2
    For $Y=\Bbb R$, see Pete L. Clark's answer here. – David Mitra Jan 9 '14 at 15:25
  • 1
    See in my 20 December 2006 sci.math post, especially my comments in the section titled PROOFS OF 2' (note the prime ' on two 2). – Dave L. Renfro Jan 9 '14 at 16:14
  • I will have a look. Thank you! – topsi Jan 12 '14 at 10:25
up vote 5 down vote accepted

Yes. See Theorem 2.1 on page 2 of

http://artsci.kyushu-u.ac.jp/~ssaito/eng/maths/Gdelta.pdf

In fact, every dense $G_\delta\subset\mathbb R$ is also the set of continuity of a derivative of a differentiable function!

  • ok I will have a look. Thank you! – topsi Jan 12 '14 at 10:24
  • Not every $G_\delta\subset\mathbb R$ is the set of continuity points of some function $f^\prime$, since $\emptyset$ is $G_\delta$, but no $f^\prime$ is continuous nowhere – 2016 Jun 16 '14 at 15:56
  • Right. Corrected now! – Yiorgos S. Smyrlis Jun 16 '14 at 19:01
  • @2016: Is there any exception other than the empty set? – gary Dec 6 '17 at 22:53
  • 1
    @gary Every DENSE $G_\delta$ is a set of continuity of of an $f'$. – Yiorgos S. Smyrlis Dec 7 '17 at 5:59

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