# Is there a bijection between $\mathbb{N} \times \mathbb{R}$ and $\mathbb{R}$

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.

• Observe that $|\Bbb N\times\Bbb R|=|\Bbb R|$. – Masacroso Oct 22 '16 at 11:17
• @Masacroso This is precisely what OP is trying to prove. – Wojowu Oct 22 '16 at 11:17
• @Wojowu I dont know the background of the question. If he knows some theorems about cardinality he dont need a constructive proof. – 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.

• ahhh, I forgot about think along that line. Thank you for taking the time to answer my question. This helps – 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 $\mathbb{R}$?

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