Bijection between $\{x^2: x \in R \}$ and $\{x: x \in R\}$

Question

Let's consider two sets: $A := \{x^2: x \in R\}$ and $B:= \{x: x \in R\}$.

In my opinion its very intuitive that those two sets have exactly the same cardinality. In other words there has to be a bijection $g$ between $A$ and $B$ but I couldn't find proper form of $g$. I tried to pick $g(x) = x$ or $g(x) = \sqrt x$ but none of them works (second example doesn't work because domain differs).

Could you please help me finding this bijection?

https://math.stackexchange.com/questions/602865/show-that-set-of-real-numbers-is-equipotent-to-the-set-of-positive-real-numbers, https://math.stackexchange.com/questions/654026/real-numbers-and-positive-real-numbers-have-same-cardinality — Martin R, Nov 17 '21 at 09:38
The two sets are not homeomorphic, so your bijection cannot be continuous. Of course, there are many discontinuous functions to try. — GEdgar, Nov 17 '21 at 10:42
In the question, does $\ R\ $ mean $\ \mathbb{R},\ $ the set of real numbers? — Adam Rubinson, Nov 17 '21 at 11:27
The consensus on this site is to use \Bbb R for the reals, and similarly \Bbb Q, \Bbb Z, \Bbb N. — DanielWainfleet, Nov 17 '21 at 11:53
Exercise: Show that a bijection $g:[0,\infty)\to \Bbb R$ cannot be continuous. BTW this will not answer your Q but it will keep you from trying to find a continuous $g$. — DanielWainfleet, Nov 17 '21 at 11:59
What about $g(x)=\frac{(2\lfloor x\rfloor+3)(-1)^{\lfloor x\rfloor}-3}{4}+x-\lfloor x\rfloor$? — Ivan Kaznacheyeu, Nov 18 '21 at 10:27
Correct expression is $g(x)=\frac{(2\lfloor x\rfloor+1)(-1)^{\lfloor x\rfloor}-1}{4}+x-\lfloor x\rfloor$ — Ivan Kaznacheyeu, Nov 19 '21 at 07:34

score 1 · Answer 1 · answered Nov 17 '21 at 11:29

1

Let $A_0=[0,1]$ and $A_n=(n,n+1]$ for $n\in \Bbb Z^+.$ Let $B_0=[0,1]$. For $n\in\Bbb Z^+$ let $B_{2n}=(n,n+1]$ and let $B_{2n-1}=[-n,1-n)$. For each non-negative integer $n$, map $A_n$ bijectively onto $B_n.$

answered Nov 17 '21 at 11:29

DanielWainfleet

59,529

A set-theorist would instead just say "Use the Cantor-Schroeder-Bernstein Theorem." (a.k.a. Cantor-Bernstein, a.k.a. Schroeder-Bernstein). There is a long complicated proof of it and a short simple proof. If $f:A\to B$ and $g:B\to A$ are injective then there exists a bijection $h:A\to B.$ – DanielWainfleet Nov 17 '21 at 11:37

score 0 · Answer 2 · answered Nov 18 '21 at 15:59

$A := \{x: x \in \mathbb{R}\ \text{ and }\ x\geq0\}$ and $B:= \mathbb{R}.$

Now we need to find a bijective map from $A$ to $B$. After fiddling around with some sketches, I found the bijective function $\ g: A\to B\ $ defined by:

$ f(x)= \begin{cases} \frac{n}{2} + (x\mod1), &\ x\in [n,n+1),\quad \text{for each even integer}\ n\geq 0\\ -\left(\frac{n+1}{2}\right) + (x\mod1), &\ x\in [n,n+1),\quad \text{for each odd integer}\ n\geq 0\\ \end{cases} $

I'm not sure this is the clearest way to write that function, so someone feel free to amend this to make it simpler/read better.

It can be written in one expression with floor. See my last comment (I've made mistake in previous comment and lost one interval) — Ivan Kaznacheyeu, Nov 19 '21 at 07:35

Bijection between $\{x^2: x \in R \}$ and $\{x: x \in R\}$

2 Answers2