Does the "continuous locus" of a function have any nice properties? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:58:04Z http://mathoverflow.net/feeds/question/165 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/165/does-the-continuous-locus-of-a-function-have-any-nice-properties Does the "continuous locus" of a function have any nice properties? Anton Geraschenko 2009-10-07T02:21:14Z 2010-08-28T07:58:52Z <p>Suppose f:<b>R</b>&rarr;<b>R</b> is a function. Let S={x&isin;<b>R</b>|f is continuous at x}. Does S have any nice properties?</p> <p>Here are some observations about what S could be:</p> <ul> <li>S can be any closed set. For a closed set S, let g be a continuous function whose vanishing locus is S (for example, you could take g(x) to be the distance of x from S if S is non-empty). Then define f(x)=g(x) if x&isin;<b>Q</b> and f(x)=0 otherwise. Then the continuous locus of f is exactly S.</li> <li>S can be an open interval. For an open interval S, define f(x)=0 if x&isin;S or x&isin;<b>Q</b> and f(x)=1 otherwise. Then the continuous locus of f is exactly S.</li> <li>S can be the complement of any countable set. Let T={t<sub>1</sub>,t<sub>2</sub>,t<sub>3</sub>,...} be a countable set, and let &sum;a<sub>i</sub> be some absolutely convergent series all of whose terms is non-zero (like a<sub>i</sub>=1/2<sup>i</sup>). Define<br> f(x) = &sum;<sub>i such that t<sub>i</sub> &lt; x</sub> a_i.<br> Then the continuous locus of f is exactly the complement of T.</li> </ul> <p>Here are some questions I'd like to know the answers to:</p> <ul> <li>Can S be any open set?</li> <li>Can S be non-measurable? (if f(x)=0 if x&isin;S and f(x)=1 otherwise, what will the continuous locus be?)</li> </ul> http://mathoverflow.net/questions/165/does-the-continuous-locus-of-a-function-have-any-nice-properties/167#167 Answer by Eric Wofsey for Does the "continuous locus" of a function have any nice properties? Eric Wofsey 2009-10-07T03:02:54Z 2009-10-07T19:42:21Z <p>It's a standard result that the continuous locus is always G-delta. For each r>0, let U(r) be the set of points x such that some neighborhood of x maps into some ball of radius r. Then each U(r) is open, and the continuous locus is their intersection. Conversely, given a G-delta set, I'm pretty sure it's not hard to construct a function with that continuous locus, though I don't remember how off the top of my head.</p> http://mathoverflow.net/questions/165/does-the-continuous-locus-of-a-function-have-any-nice-properties/174#174 Answer by Alon Amit for Does the "continuous locus" of a function have any nice properties? Alon Amit 2009-10-07T18:03:29Z 2010-08-28T07:58:52Z <p>Yes, here's a quick proof that any given $G_\delta$ (in $\mathbb{R}$) can be realized as the set of continuity points of some real-valued function.</p> <p>Let $G$ be a given $G_\delta$ set in $\mathbb{R}$, meaning $G = \cap_{i=1}^\infty G_i$, each $G_i$ an open set. Define a function $f:\mathbb{R} \to \mathbb{R}$ as follows: $f(x)=0$ if $x$ is in $G$. If x is not in $G$, there is some $k$ such that $x$ is not in $G_k$; let $k$ be minimal with that property. Define $f(x)=1/k$ if $x$ is rational and $f(x)=-1/k$ if $x$ is irrational. </p> <p>If I'm not very much mistaken, $G$ is precisely the set of continuity points of this $f$. I'm happy to leave this as an exercise for now :-) Let me know if you're not sure how to do it, or - worse - if I'm just wrong about the construction.</p> http://mathoverflow.net/questions/165/does-the-continuous-locus-of-a-function-have-any-nice-properties/1690#1690 Answer by watchmath for Does the "continuous locus" of a function have any nice properties? watchmath 2009-10-21T17:07:55Z 2009-10-21T17:07:55Z <p>I hope nobody would mind if I try to do the exercise.</p> <p>Clearly f is continuous on G. Suppose f is continuous on x and f(x)=1/k. Take epsilon=1/k. Let U be any neighborhood of x. U\cap G_1\cap .. \cap G_{k-1} contains an irrational number y. Hence |f(x)-f(y)|=2/k > epsilon. (if f(x)=-1/k, take y to be a rational number)</p>