# Existence of bijection between $\mathbb{R}$ and $\mathbb{R^2}$

I would like to show the existence of a bijection between $(0,1)$ and $(0,1) \times (0,1)$. Would $f(x)=(x,x)$ work ($x \in (0,1)$)? It seems to be injective, but am I okay in using this even if its domain and range can go outside of $(0,1)$?

Notes:

Define a second function $g: (0,1) \times (0,1) ↦ (0,1)$ which takes two decimal expansions $0.(x_1)(x_2)(x_3)...$ and $0.(y_1)(y_2)(y_3)...$, the first expansion coming from the first component of $(0,1) \times (0,1)$ and the second coming from the second component and maps them to $0.(x_1)(y_1)(x_2)(y_2)(x_3)(y_3)...$. For example, take $0.04265$ and 0.83169. The function g maps the two decimal expansions to 0.0843216659. This seems like it could be an injective function because it takes two numbers and maps them to a unique number. However, I'm not sure how I would formally prove it...

This implies that, by Cantor-Schroder-Bernstein theorem, there exists a bijection between $(0,1)$ and $(0,1) \times (0,1)$.

• This map is not surjective! – Crostul Nov 27 '16 at 18:44
• Fixed, thanks! The more important question is, can I use the function even though its domain/range technically can extend out beyond $(0,1)$? – Timor12 Nov 27 '16 at 18:48
• Yes, it is ok. It is the restriction of the map $\Bbb{R} \to \Bbb{R}^2$ defined by $x \mapsto (x,x)$, but it is a (well-defined) function anyway. – Crostul Nov 27 '16 at 18:51
• @Timor12 And how do you think we should define $g({1\over 3}, {1\over 2})$? Your function needs to be defined on all of $(0, 1)^2$, not just some of it. – Noah Schweber Nov 27 '16 at 19:16
• @Timor12 I meant that you only defined $g$ for the points of the form $(x,x)$. We cannot calculate $g(\frac13,\frac12)$ using your definition. – Vitor Borges Nov 27 '16 at 19:17

Are you sure it's surjective? What gets sent to $({1\over 2}, {1\over 3})$?
EDIT: In the comments below the OP, we've hashed out the following injection from $(0, 1)^2$ to $(0, 1)$ (which, in conjunction with the "diagonal" injection $x\mapsto(x, x)$ given by the OP originally, will establish that $(0, 1)$ and $(0, 1)^2$ have the same cardinality via Cantor-Shroeder-Bernstein): let the standard decimal expansion of a real $r$ be the decimal expansion of $r$ if $r$ has only one decimal expansion, and otherwise the decimal expansion with only finitely many $9$s. So e.g. the standard decimal expansion of $\pi$ is $3.14159...$, and the standard decimal expansion of ${1\over 2}$ is $0.5$ rather than $0.499999...$. Then, if $r, s$ are real numbers in $(0, 1)$ with standard decimal expansions $$r=0.a_1a_2a_3...\quad s=0.b_1b_2b_3,$$ we let $$g(r, s)=0.a_1b_1a_2b_2a_3b_3...$$
We're not done: we still need to show that this is injective. It's usually harder to prove something if you think it's trivial, so: how could $g$ not be injective? Well, for instance, if we had two numbers $r$ and $s$ such that $g(r, s)=0.221999999...$ and two other numbers $r'$ and $s'$ such that $g(r', s')=0.22200000...$, then this would be a problem. So: can you see why this sort of issue can't happen?
• However, the main question is, can I use that function even though its domain/range can go beyond $(0,1)$? – Timor12 Nov 27 '16 at 18:46
• @Timor12 You can define a function on a restricted domain: let $f: (0, 1)\rightarrow (0, 1)^2: x\mapsto (x, x)$. There's no obligation to always have a function be defined on the largest possible domain! – Noah Schweber Nov 27 '16 at 18:47
• @Timor12 Re: your first comment, yes, and you have an $f$ going one way; the difficulty is finding a $g$ to go the other way. – Noah Schweber Nov 27 '16 at 18:48
• @BjörnFriedrich The author's original question was whether a given function provided a bijection between $(0, 1)$ and $(0, 1)^2$; this indeed answers it. – Noah Schweber Nov 27 '16 at 20:28