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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 $(0,1)$ ?

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    $\begingroup$ This is more or less the same as math.stackexchange.com/questions/186427/… $\endgroup$ – Chris Eagle Jul 28 '13 at 16:09
  • $\begingroup$ Try something using $\sin\frac1x$ or its translates perhaps. $\endgroup$ – Karthik C Jul 28 '13 at 16:09
  • $\begingroup$ If you want to explicitly use Chris Eagle's example $f:\mathbb R\to\mathbb R$, you could always say $g(x) = 1+\frac1{\pi}\arctan(\pi f(\tan(\pi(x-\frac12))))$ $\endgroup$ – Omnomnomnom Jul 28 '13 at 16:21
  • $\begingroup$ @Omnomnomnom: Using $g(x)=f(x)$ where it makes sense and $g(x)=0.5$ where it doesn't would be significantly less crazy. $\endgroup$ – Chris Eagle Jul 28 '13 at 16:26
  • $\begingroup$ @ChrisEagle that indeed makes sense and seems much less crazy. $\endgroup$ – Omnomnomnom Jul 28 '13 at 16:30
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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 $(a,b)$

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