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.

  • $\begingroup$ Observe that $|\Bbb N\times\Bbb R|=|\Bbb R|$. $\endgroup$ – Masacroso Oct 22 '16 at 11:17
  • 3
    $\begingroup$ @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

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.

  • $\begingroup$ 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

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 $\mathbb{R}$?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.