# Why is this proof false?

I know this proof is false, but I don't know why. I need your help.

The false proof says that it is possible to create a bijection between a subset of the rational numbers and the Power set of natural numbers.

We can create orderly the subsets of the natural numbers and create a bijection at the same time to some of the rational numbers:

First pairs:

{1,2},{1,3}.... -> 1/2, 1/3..

{2,3},{2,4}.... -> 2/3, 2/4..

now three:

{1,2,3},{1,2,4}... 12/3, 12/4...

{1,3,4},{1,3,5}....13/4, 13/5...

...

{2,3,4}, {2,3,5}...23/4, 23/5...

four...

And so on...

So this false proof says that the cardinal of the rational numbers are, al least the cardinal of P(N)

How can I explain that is false

Thanks.

-
Your construction does not contain infinite subsets of $\mathbb{N}$. –  Raskolnikov Jan 10 '13 at 7:44
And indeed the set $\mathcal P_f(\mathbb N)$ of the finite subsets of $\mathbb N$ is countable. –  Did Jan 10 '13 at 7:46
Not only have you not defined your function on any infinite subset of $\Bbb{N}$, you haven't defined it on all the finite subsets either. I at least cannot guess from your description how you indeed to deal with sets of size four, let alone larger ones. –  Chris Eagle Jan 10 '13 at 7:53
I think that the main problem is that I don't have the infinite subsets in act...but I don't understand the rules very well. Why in the bijecction between rational and naturals we understand we have contructed all infinite numbers and in this false proof we cannot say e have constructed all even numbers or all quadratic numbers? –  Pedro Jan 10 '13 at 8:00
@did :"the set Pf(N) of the finite subsets of N is countable" –  Pedro Jan 10 '13 at 8:55
show 1 more comment

Ignoring that I don't think that map is going to be injective on finite subsets of the natural numbers. Where do you send $\mathbb N$ or the set of all even numbers?

I will add that if you consider the set $\mathcal A =\{ B \subset \mathbb N : |B| < \infty\}$ that $|\mathcal A |=\mathbb Q$. You can find an injection by taking an enumeration of the primes for instance and mapping $\varphi(B)=\prod_{i \in B}p_i$.

-
Hello Jacob. With the groups of four I have {1,3,5,7}, in the groups of five {1,3,5,7,9} and so on...I'm not havig the even subset in act as a infinite subset. –  Pedro Jan 10 '13 at 7:55
@Pedro I don't understand your question. Where does the set $\{2,4,6,8,10,\dots\} \in \mathcal P(\mathbb N)$ get sent to under your mapping? –  JSchlather Jan 10 '13 at 7:58
That's the problem. Ok, I see –  Pedro Jan 10 '13 at 8:06