# Bijective map of omega tuples

1. Oct 20, 2011

### EV33

1. The problem statement, all variables and given/known data

Find a bijective map : χωω$\rightarrow$χω

2. Relevant equations
An omega tuple is a function x:N$\rightarrow$χ, where χ is a set.

χω is the set of all omega tuples of elements of χ.

A bijective function is both injective and surjective.

3. The attempt at a solution

I know that a solution is

f[(x1,x2,...),(y1,y2,...)]=(x1,y1,...)

I am trying to interpret the answer...

First of all, (x1,x2,x3,...) and (y1,y2,y3,...)represent arbitrary omega tuples. Since they are arbitrary they represent all omega tuples and thus both represent χω?

Second thing is I must recognize that this function is a bijection. So I must show that this function is both injective and surjective.

So I look at this function and it appears to be the union of (x1,x2,x3,...) and (y1,y2,y3,...).

If we let (x1,y1,x2,y2,...)=(x1',y1',x2',y2',...) then x1=x1',y1=y1' and so on. This would imply that (x1,x2,x3...)=(x1',x2',x3',...), and thus this function is injective.

If we have an arbitrary omega tuple (x1,y1,x2,y2,...) then we can find (x1,x2,x3,...) and (y1,y2,y3,...) such that (x1,x2,x3,...)X(y1,y2,y3,...)=(x1,y1,x2,y2,...). Thus the function is surjective.

Thus the function is bijective.

Are there any flaws here?

Thank you.