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So I'm doing some practice on set theory, and I am having some trouble proving a lemma.

Basically I want to ask if there is there a bijection between $\mathbb{N} \times \mathbb{R}$ and $\mathbb{R}$

If yes, could someone provide a simple construction of such a bijection?

Any help or insights is deeply appreciated.

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  • $\begingroup$ Observe that $|\Bbb N\times\Bbb R|=|\Bbb R|$. $\endgroup$ – Masacroso Oct 22 '16 at 11:17
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    $\begingroup$ @Masacroso This is precisely what OP is trying to prove. $\endgroup$ – Wojowu Oct 22 '16 at 11:17
  • $\begingroup$ @Wojowu I dont know the background of the question. If he knows some theorems about cardinality he dont need a constructive proof. $\endgroup$ – Masacroso Oct 22 '16 at 11:23
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For just answering the yes/no question, the easiest way is to use the Swiss knife of bijections, the Cantor-Schröder-Bernstein theorem, which just requires us to construct separate injections in each direction $\mathbb R\to\mathbb N\times\mathbb R$ and $\mathbb N\times\mathbb R\to\mathbb R$ -- which is easy:

$$ f(x) = (1,x) $$

$$ g(n,x) = n\cdot \pi + \arctan(x) $$

Because there is an injection either way, Cantor-Schröder-Bernstein concludes that a bijection $\mathbb R\to\mathbb N\times\mathbb R$ must exist.


If you already know $|\mathbb R\times\mathbb R|=|\mathbb R|$, you can get by even quicker by restricting your known injection $\mathbb R\times\mathbb R\to \mathbb R$ to the smaller domain $\mathbb N\times\mathbb R\to\mathbb R$ instead of mucking around with arctangents.

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  • $\begingroup$ ahhh, I forgot about think along that line. Thank you for taking the time to answer my question. This helps $\endgroup$ – some1fromhell Oct 22 '16 at 11:34
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Hint. Consider the bijection $f:\mathbb{Z}\times [0,1)\to \mathbb{R}$ such that $f(n,x)=n+x$.

Is there a bijection between $\mathbb{N}$ and $\mathbb{Z}$?

Is there a bijection between $[0,1)$ and $\mathbb{R}$?

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  • $\begingroup$ @drhab Yes, thanks! $\endgroup$ – Robert Z Oct 22 '16 at 11:14

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