Possible Duplicate:
Construct a monotone function which has countably many discontinuities
How would I construct a monotone function on $[0,1]$ that is discontinuous at each rational point.
Asked
Active
Viewed 224 times
2
-
One rational at a time? – davidlowryduda Oct 31 '11 at 04:35
-
2I think titi is looking for a monotone function $f$ such that the set of points of continuity of $f$ is $[0,1] \setminus\mathbb{Q}$ – leo Oct 31 '11 at 04:40
-
@mixedmath: I don't understand what you mean by one rational at a time. – titi Oct 31 '11 at 04:42
-
@leo: That's exactly what I'm looking for. – titi Oct 31 '11 at 04:43
-
Take $r:\mathbb{N}\to\mathbb{Q}$ an enumeration of the rational numbers in $[0,1]$. I think the function $$f(x)=\sum_{r(k)\lt x}2^{-k}$$ is a example. This function was discussed here before but I can't find the post. – leo Oct 31 '11 at 04:45
-
2Just clarifying, I mean $$f(x)=\sum_{k:r(k)\lt x} 2^{-k}$$ – leo Oct 31 '11 at 04:52
-
3Finally found, actually exact duplicate: http://math.stackexchange.com/q/69317/8271 – leo Oct 31 '11 at 05:07
-
@leo: Thanks for the link. I'd appreciate it though if you or anyone could explain to me why the function is not right continuous. – titi Oct 31 '11 at 05:20
-
ok, consider $x$ rational in $[0,1]$. Then $x+1/n\to x$ and note that $$f(x+1/n)\to \sum_{k:r(k)\leq x}2^{-k}\neq f(x)$$ – leo Oct 31 '11 at 16:59
-
@leo: Thanks.... – titi Oct 31 '11 at 23:00