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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.

Asaf Karagila
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  • See http://math.stackexchange.com/questions/1231365/monotonous-function-or-monotonic-function – Henricus V. Mar 27 '16 at 00:01
  • It's "monotonic" or "monotone" – carmichael561 Mar 27 '16 at 00:02
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    @carmichael561 Unless the bijection is extremely boring, in which case monotonous is correct. – Plutoro Mar 27 '16 at 00:03
  • Not always. Consider $X=\mathbb{R}$ and $Y=\mathbb{Q}$. You could try to be more specific. – Pedro Sánchez Terraf Mar 27 '16 at 00:06
  • 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.) – Andrés E. Caicedo Mar 27 '16 at 01:10
  • The axiom $\PFA $ implies that the answer to your question is yes for any two $\aleph_1$-dense sets of reals. On the other hand, Sierpinski showed that under $\mathsf {CH} $, there are many non-isomorphic $\aleph_1$-dense sets. In fact, you can find a collection of the same size as the power set of the reals such that none of them embeds into any of the others. – Andrés E. Caicedo Mar 27 '16 at 01:17

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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.

Plutoro
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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.

John Hughes
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