# Set of continuity points of a real function

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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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. –  Nick Strehlke Sep 26 '11 at 4:56
and how can i prove this nice result? –  Daniel Sep 26 '11 at 5:19
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. –  Nick Strehlke Sep 26 '11 at 6:03
Actually, after checking out item 3) of the below answer, I see that it's the same construction with a better write-up. –  Nick Strehlke Sep 26 '11 at 6:04
For those interested in precise references in published literature for this result (and related results), see groups.google.com/group/sci.math/msg/05dbc0ee4c69898e –  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.

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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...

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Interesting, so the set of continuity points is homemorphic to a complete metric space. –  Asaf Karagila Sep 26 '11 at 6:46
@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... –  Pete L. Clark Sep 26 '11 at 13:05