Can we have a map from $(0,1)$ to $(0,1)$ such that the image of every open interval in $(0,1)$ is all of $(0,1)$ ?
Asked
Active
Viewed 340 times
8
Martin Sleziak
- 53,687
-
3This is more or less the same as http://math.stackexchange.com/questions/186427/function-whose-image-of-every-open-interval-is-infty-infty/186434#186434 – Chris Eagle Jul 28 '13 at 16:09
-
Try something using $\sin\frac1x$ or its translates perhaps. – Karthik C Jul 28 '13 at 16:09
-
If you want to explicitly use Chris Eagle's example $f:\mathbb R\to\mathbb R$, you could always say $g(x) = 1+\frac1{\pi}\arctan(\pi f(\tan(\pi(x-\frac12))))$ – Ben Grossmann Jul 28 '13 at 16:21
-
@Omnomnomnom: Using $g(x)=f(x)$ where it makes sense and $g(x)=0.5$ where it doesn't would be significantly less crazy. – Chris Eagle Jul 28 '13 at 16:26
-
@ChrisEagle that indeed makes sense and seems much less crazy. – Ben Grossmann Jul 28 '13 at 16:30
-
A suitable modification of Example 2.25 in Gelbaum and Olmsted's Counterexamples in Analysis will work. – David Mitra Jul 28 '13 at 16:38
1 Answers
7
Yes. My favorite, though it takes a little cleaning in the corners is this one. Let $x\in (0,1)$ and express it in base $3$. If there are an infinite number of $2$'s in the expansion, set $f(x)=x$ and ignore it. Otherwise, cut off all the leading digits through the last $2$ and read the resulting number in binary to get $f(x)$. Given any $y \in (0,1)$ expressed in binary and an interval $(a,b)$ we can find $c \in (a,b)$ expressed in ternary ending in $2$ that we can append $y$ to and stay in $(a,b)$
Ross Millikan
- 374,822