Bijection from two open intervals to one

Question

I was thinking about a bijection between the extended complex plane and the complex plane. Realising that the complex plane could be further reduced to the open unit disc (by mapping every polar coordinate $(r,\phi)$ to $(2\sigma(r)-1, \phi)$ where $\sigma$ is the sigmoid function), the only problem left was to consider a bijection from the open unit disc to an union of the open unit disc with the point at infinity. Simplifying things, I have reduced my problem to finding a bijection between an open interval on the real number line to a closed-open interval, which can further be reduced to this:

Consider open intervals $A$, $B$ and $C$ on the real number line such that $A$ and $B$ don't have any elements in common. Is it possible to define a bijection between $A\cup{B}$ to $C$?

Example case: Does there exist a bijection between $(0,0.5)\cup(0.5,1)$ and $(0,1)$ ?

Yes. It is always possible, even when A and B has some common segment. — ShyamalSayak, Aug 01 '23 at 14:32
Both sets are of size continuum, so there is bijection between them. — Jakobian, Aug 01 '23 at 14:36
@ShyamalSayak could you please prove why it is so by describing an implicit (or if you prefer, explicit) bijection? — Soham Saha, Aug 01 '23 at 14:39
@Jakobian could you please give some kind of implicit bijection for the example case? — Soham Saha, Aug 01 '23 at 14:40
Quick beginner guide for asking a well-received question + please avoid "no clue" questions. Could you please edit your post to include some personal work? — Anne Bauval, Aug 01 '23 at 15:31
@AnneBauval sorry for the mistake. Could you please check if my edit follows the standards? — Soham Saha, Aug 01 '23 at 15:52
Having added you motivation is excellent and might be considered as sufficient. It would have been even better if you did add some attempts even if they failed. But don't bother: I shall find one of the many duplicates showing that $(0,1)$ and $(0,1]$ are equinumerous, since you wrote that is what you want. — Anne Bauval, Aug 01 '23 at 16:01
@AnneBauval, so I should have like written about my thoughts on how to solve it, and then posed my question? Actually it’s my first question on this SE… — Soham Saha, Aug 01 '23 at 16:03
Does this answer your question? Proving intervals are equinumerous to $\mathbb R$ or that? https://math.stackexchange.com/questions/1226659/proving-intervals-are-equinumerous — Anne Bauval, Aug 01 '23 at 16:07
@AnneBauval actually in that question the OP has already considered many things as being already taught while I don’t have that kind of prerequisites. But found this one: https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1 — Soham Saha, Aug 01 '23 at 16:11
Excellent! Have a nice stay on this site :-) https://math.stackexchange.com/search?q=Intervals+equinumerous — Anne Bauval, Aug 01 '23 at 16:13

score 1 · Accepted Answer · answered Aug 01 '23 at 14:49

1

Let $$f=\begin{cases}2x,&x\in\{\frac14,\frac18,\frac16,\dotsb\}\\ x,&\text{otherwise.}\end{cases}$$ Then $f:(0,0.5)\cup(0.5,1)\to(0,1)$ is a bijection.

Note that this function is discontinuous at infinitely many points (all points of the form $1/2^n$). It can be shown that there is no bijection between these sets that only uses finitely many points of discontinuity.

answered Aug 01 '23 at 14:49

Akiva Weinberger

23,062

Is there any way I can mark 2 answers as correct? – Soham Saha Aug 01 '23 at 14:54
I don't think so. – Akiva Weinberger Aug 01 '23 at 14:54
Could you refer a proof for the inexistence of bijection with "finitely many points of discontinuity"? – Soham Saha Aug 01 '23 at 15:01
And that would imply the absence of any continuous bijection, right? – Soham Saha Aug 01 '23 at 15:02
This question goes against our quality standard policy. Instead of posting an answer, you should encourage the person who posted the question to improve it. Moreover, this is certainly a mega-dupe. – Anne Bauval Aug 01 '23 at 15:34
@AnneBauval While I agree that the question is a duplicate, I disagree that it is low-quality. – Akiva Weinberger Aug 01 '23 at 19:22
It was at the time you answered it. I did my best to guide the OP. – Anne Bauval Aug 01 '23 at 23:04
@AkivaWeinberger am I right in thinking that no continuous bijection can exist for my example case since a continuous bijection can’t be established between a connected and a disconnected set? (Inspired from https://www.quora.com/How-do-you-prove-or-disprove-that-there-exists-a-continuous-bijection-from-mathbb-R-2-to-mathBB-R-real-analysis-general-topology-analysis-elementary-set-theory-math) – Soham Saha Aug 02 '23 at 13:06
@SohamSaha There's a continuous bijection from $(0,1)\cup[2,3)$ to $(0,2)$ (subtract $1$ from the second piece). However, the inverse function is not continuous, and there is no continuous bijection from $(0,2)$ to $(0,1)\cup[2,3)$. – Akiva Weinberger Aug 02 '23 at 16:04
(A continuous function with continuous inverse is called a "homeomorphism".) – Akiva Weinberger Aug 02 '23 at 16:04

score 0 · Answer 2 · answered Aug 01 '23 at 14:49

0

Here is an explicit bijection from $(0, 1)$ to $(0, 1/2) \cup (1/2, 1)$: Map 1/2 to 1/4, 1/4 to 1/8, 1/8 to 1/16, etc. For all other values of $x$, map $x$ to $x$.

answered Aug 01 '23 at 14:49

Dan Velleman

2,746

This question goes against our quality standard policy. Instead of posting an answer, you should encourage the person who posted the question to improve it. Moreover, this is certainly a mega-dupe. – Anne Bauval Aug 01 '23 at 15:34

Bijection from two open intervals to one

2 Answers2