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I was wondering if there exists a monotonic bijection from a dense set $X$ and a dense subset of $X$, $Y$, where $X$ and $Y$ are not equal. For example a monotonic bijection from the reals to the reals minus the integers. Thank you in advance.

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  • $\begingroup$ See math.stackexchange.com/questions/1231365/… $\endgroup$ – Henricus V. Mar 27 '16 at 0:01
  • $\begingroup$ It's "monotonic" or "monotone" $\endgroup$ – carmichael561 Mar 27 '16 at 0:02
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    $\begingroup$ @carmichael561 Unless the bijection is extremely boring, in which case monotonous is correct. $\endgroup$ – Alex S Mar 27 '16 at 0:03
  • $\begingroup$ Not always. Consider $X=\mathbb{R}$ and $Y=\mathbb{Q}$. You could try to be more specific. $\endgroup$ – Pedro Sánchez Terraf Mar 27 '16 at 0:06
  • $\begingroup$ If $X $ and $Y $ are both countable dense subsets of the reals, the answer is yes. This is a classic result due to Cantor. The answer is in general no otherwise, but additional set-theoretic assumptions may give you some information both on positive and negative cases. For example, $\mathsf {PFA} $, the *proper forcing axiom*, implies that the continuum hypothesis fails and there is exactly one intermediate size ($\aleph_1$) between the size of $\mathbb N $ and that of $\mathbb R $. A set of reals is $\aleph_1$-dense if it meets each interval in exactly $\aleph_1$ points. (Cont.) $\endgroup$ – Andrés E. Caicedo Mar 27 '16 at 1:10
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This is not a complete answer, but if $X=\mathbb R$, and $Y$ is any dense proper subset of $\mathbb R$, the answer is no. For if there were a monotonic function $f:X\to Y$, it would have to be discontinuous (a continuous function maps connected sets to connected sets. $Y$ is not connected). Suppose W.L.O.G. that $f$ is increasing. Discontinuities of monotonic functions are always jump discontinuities, so if the discontinuity occurs at $x=a$ and $\sup_{x<a} f(x)=c$ and $\inf_{x>a} f(x)=d$, then $c<d$. Thus, $f(\mathbb R)$ contains no point of the set $(c,d)$ except possibly $f(a)$. Therefore, the image of $f$ is not dense.

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On the set $X$ of rationals whose denominator (in lowest terms) is divisible by 2 at most once, consider the map $x \mapsto 2x$. This maps onto the subset $Y$ consisting of those rationals whose denominators contain no factors of 2.

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