Set of continuity points of a real function

I have a question about subsets AR
for which there exists a function f:RR 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 Q and discontinuous on the irrationals?

Answer

  1. Let X be a metric space, and let f:XR be any function. For any ϵ>0, let us say that a point xX has property C(f,ϵ) if there exists δ>0 such that d(x,x),d(x,x. 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...

Attribution
Source : Link , Question Author : Daniel , Answer Author : Pete L. Clark

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