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I am trying to show $\mathbb{R}^\mathbb{N}$ ~ $\mathbb{R}$ using the Cantor Bernstein method. Here is my proof so far:

Let $f: \mathbb{R}\to\mathbb{R}^\mathbb{N}$ be defined as for each $r_n\in\mathbb{R}$ where $n\in\mathbb{N}$, $f(r_n)=(r_n,0,0,...)$. Thus, we have an injection from $\mathbb{R}$ to $\mathbb{R}^\mathbb{N}$. (For me, $0$ is in the natural numbers.)

Now, for the other direction, let $g:\mathbb{R}^\mathbb{N}\to\mathbb{R}$ be defined as follows. Let $(r_0,r_1,...,r_n,...)\in\mathbb{R}^\mathbb{N}$ where $r_n\in\mathbb{R}$ and let $r_n=Z_n.a_{0,n}a_{1,n}a_{2,n}...$ be the unique corresponding decimal expansion where $n\in\mathbb{N}$, $Z\in\mathbb{Z}$, $0\leq a_{n,n}\leq 9$, and the sequence does not end in infinitely many 9's.

Consider the list:

$r_0=r_n=Z_0.a_{0,0}a_{1,0}a_{2,0}...$

$r_1=r_n=Z_1.a_{0,1}a_{1,1}a_{2,1}...$

$r_2=r_n=Z_2.a_{0,2}a_{1,2}a_{2,2}...$

Then, let $r=0.a_{0,0}a_{1,0}a_{0,1}a_{0,2}a_{1,1}a_{2,0}...\in\mathbb{R}$.

So, $g((r_0,r_1,...,r_n,...))= r$.

Then, we have an injection from $\mathbb{R}^\mathbb{N}$ to $\mathbb{R}$. By Cantor Bernstein Theorem, $\mathbb{R}^\mathbb{N}$ ~ $\mathbb{R}$.

Does this proof work or how can I modify/improve it?

Andrés E. Caicedo
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Sam
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  • Hi, what you mean by $r_n$? – sti9111 Dec 05 '17 at 03:37
  • @sti9111 Okay, I defined it. – Sam Dec 05 '17 at 03:38
  • If you look around, you can probably find a more general proof of the fact, that if $X$ is an infinite set and $n$ is a natural number, then $X^n \simeq X$, and then adapt the proof to this situation. – Duncan Ramage Dec 05 '17 at 03:41
  • @DuncanRamage Okay, but does what I have here not work then? – Sam Dec 05 '17 at 03:42
  • A nit: you need to biject $\Bbb R$ with $(0,1)$ because you have not dealt with numbers before the decimal point. – Ross Millikan Dec 05 '17 at 04:02
  • @RossMillikan I am not sure what you mean? Would it be best then to do an injection from $(0,1)^\mathbb{N}$ to $\mathbb{R}$ instead since we know $\mathbb{R}$~$(0,1)$? – Sam Dec 05 '17 at 04:04
  • Just that your work only deals with numbers to the right of the decimal point. There are lots of numbers in $\Bbb R$ that have digits to the left of the point. We know $\Bbb R$ is equinumerous with $(0,1)$ so that is not a problem. You can just chain your favorite bijection to get to $(0,1)$, then go to $(0,1)^{\Bbb N}$, then use the inverse bijection to get $\Bbb {R^N}$ – Ross Millikan Dec 05 '17 at 04:49
  • @RossMillikan Okay, I think I see the issue but I am not clear on how to fix this. Why do we need the bijection in this case if we just chose to use $(0,1)$ instead? – user376239 Dec 05 '17 at 16:30

1 Answers1

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Maybe, for the first direction you want to express the following, given $r\in\mathbb{R},$ this have the form $a_0,a_1a_2\dots$ where every $a_i$ is a digit, then we can define $f(r)=(a_0,a_1,\dots,a_n,\dots)$

the second part is o.k

sti9111
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