I have a question about subsets $$ A \subseteq \mathbb R $$ for which there exists a function $$f : \mathbb R \to \mathbb R$$ such that the set of continuity points of $f$ is $A$. Can I characterize this kind of sets? In a topological,measurable or in some way? For example, does there exist a function continuous on $\mathbb Q$ and discontinuous on the irrationals?

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    $\begingroup$ A subset $A$ of $\mathbb R$ is the set of continuity of a function $f:\mathbb R\to\mathbb R$ if and only if $A$ is a $G_{\delta}$ set. See, e.g., Wikipedia. In particular, no function has $\mathbb Q$ for its set of continuity. $\endgroup$ – Nick Strehlke Sep 26 '11 at 4:56
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    $\begingroup$ and how can i prove this nice result? $\endgroup$ – Daniel Sep 26 '11 at 5:19
  • $\begingroup$ Presumably you've seen the very clear, very complete answer below. A less fleshed out version of that argument appears on the Wikipedia page to which I linked. If you are interested to see another explicit construction of a function whose set of continuity is a given $G_{\delta}$, check out this Wikipedia article. $\endgroup$ – Nick Strehlke Sep 26 '11 at 6:03
  • $\begingroup$ Actually, after checking out item 3) of the below answer, I see that it's the same construction with a better write-up. $\endgroup$ – Nick Strehlke Sep 26 '11 at 6:04
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    $\begingroup$ For those interested in precise references in published literature for this result (and related results), see groups.google.com/group/sci.math/msg/05dbc0ee4c69898e $\endgroup$ – Dave L. Renfro Sep 26 '11 at 16:49

At Pete L. Clark's suggestion, I'm making a comment of mine into an answer.

For those interested in precise references in the published literature for this result (and related results), see


I may repost that older post here, perhaps with additions that I didn't know about then (if I happen find any additional references in my stuff at home), but it'll take at least a day or two before I can correctly format all of that older post for posting here.


1) Let $X$ be a metric space, and let $f: X \rightarrow \mathbb{R}$ be any function. For any $\epsilon > 0$, let us say that a point $x \in X$ has property $C(f,\epsilon)$ if there exists $\delta > 0$ such that $d(x,x'), d(x,x'') < \delta \implies |f(x')-f(x'')| < \epsilon$. If $x \in X$ has property $C(f,\epsilon)$, then so does every point in a sufficiently small $\delta$-ball about $x$, so the locus of all points satisfying property $C(f,\epsilon)$ is an open subset. Moreover, $f$ is continuous at $x$ iff $x$ has property $C(f,\frac{1}{n})$ for all $n \in \mathbb{Z}^+$. This shows that the locus of continuity of f -- i.e., the set of $x$ in $X$ such that $f$ is continuous at $x$ -- is a countable intersection of open sets, or in the lingo of this subject, a $G_{\delta}$-set.

2) If $x \in X$ is an isolated point -- i.e., if $\{x\}$ is open in $X$; or equivalently, if for some $\delta > 0$ the $\delta$-ball around $x$ consists only of $x$ itself -- then every function $f: X \rightarrow \mathbb{R}$ is continuous at $x$. This places a further restriction on the locus of continuity: it must contain the subset of all isolated points. For instance, if $X$ is discrete then the locus of continuity of any $f: X \rightarrow \mathbb{R}$ is all of $X$, so certainly not every $G_{\delta}$-set is a locus of continuity!

3) Conversely, let $Y \subset X$ be a $G_{\delta}$-set which contains all isolated points of $X$. Then $Y$ is a locus of continuity: there exists a function $f: X \rightarrow \mathbb{R}$ which is continuous at $x$ iff $x \in Y$. A short, elegant proof of this is given in this 1999 note of S.S. Kim.

Note that since $\mathbb{R}$ has no isolated points, here the result of 3) reads that every $G_{\delta}$-subset of $\mathbb{R}$ is a locus of continuity. But one might as well record the general case -- it's no more trouble...

  • $\begingroup$ Interesting, so the set of continuity points is homemorphic to a complete metric space. $\endgroup$ – Asaf Karagila Sep 26 '11 at 6:46
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    $\begingroup$ @Asaf: Yes, if $X$ itself is complete (as it is if $X = \mathbb{R}$, of course). (For others: he is referring to the Mazurkiewicz Theorem: en.wikipedia.org/wiki/G%CE%B4_set#Properties.) It would be more interesting if there were some specific application; whether this is the case I don't know offhand... $\endgroup$ – Pete L. Clark Sep 26 '11 at 13:05
  • $\begingroup$ I have not read your link. But I'd like to mention that if $Y$ is a $G_{\delta}$ subset of $X$ and contains all the isolated points of $X$ then there is a function $f:X\to \mathbb R$ which is continuous only at points of $Y$ and for which $Y=f^{-1}\{0\}.$ $\endgroup$ – DanielWainfleet Mar 9 '17 at 3:40
  • $\begingroup$ Hi Pete, this is an excellent answer! It's so good, in fact, that I think the formatting should be improved a bit to help future visitor's to the site. What do you think? $\endgroup$ – goblin Jun 9 '18 at 2:23

The answer to your last question is No.

The Wikipedia article on Thomae's function gives an elegant proof of why this is impossible. It includes links to terms it uses. I'll present what it says here but I suggest you view it on Wikipedia:

A natural follow-up question one might ask is if there is a function which is continuous on the rational numbers and discontinuous on the irrational numbers.

This turns out to be impossible; the set of discontinuities of any function must be an Fσ set. If such a function existed, then the irrationals would be an Fσ set and hence, as they don't contain an interval, would also be a meager set.

It would follow that the real numbers, being a union of the irrationals and the rationals (which is evidently meager), would also be a meager set. This would contradict the Baire category theorem.


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