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Trying to get started on proving the following (in particular using the Cantor-Bernstein-Schroeder Theorem).

Show that $\vert\mathbb{R}^2\vert=\vert\mathbb{R}\vert$.

Now the book suggests to start by showing $\vert(0,1)\times(0,1)\vert=\vert(0,1)\vert$.

It is easy enough for me to find an injection $f:(0,1)\rightarrow (0,1)\times (0,1)$ as follows. Let $f$ be defined as $f(x)=(x,x)$. But finding an injection $g:(0,1)\times (0,1)\rightarrow (0,1)$ is proving more difficult.

Would the function defined below work as a injection from $(0,1)\times (0,1)\rightarrow (0,1)$?

Let $g$ be defined as $g(0.b_1 b_2 b_3 b_4..., 0.c_1 c_2 c_3 c_4...)=(0.b_1c_1b_2c_2b_3c_3b_4c_4...)$

amrsa
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ClownInTheMoon
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  • $f(x) = (x,x)$ is not a bijection (for example $(1/2,1/3)$ has no preimage). It is only an injection. – C. Dubussy Nov 02 '16 at 14:26
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    @C.Dubussy Sorry, those should all say injection not bijection. I edited the post. – ClownInTheMoon Nov 02 '16 at 14:28
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    There are some numbers that have two different representations, but if you choose always the one without infinite trailing zeros at the right, this is indeed a bijection. – PenasRaul Nov 02 '16 at 14:35
  • https://math.stackexchange.com/questions/1277569/finding-a-bijective-function-from-the-real-plane-to-the-real-line/3243336#3243336 – Akerbeltz Jun 03 '19 at 09:16

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