Bijective map of omega tuples

[EV33]

Homework Statement

Find a bijective map : χ^ωxχ^ω$\rightarrow$χ^ω

Homework Equations

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

χ^ω is the set of all omega tuples of elements of χ.

A bijective function is both injective and surjective.

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 χ^ω?

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.

 

[mighty2000]

Your interpretation of the solution is correct. The function f is a bijection because it maps every element in χω to a unique element in χω, and every element in χω has a preimage in χω. Your proof for injectivity and surjectivity is also correct. However, you can also prove surjectivity by showing that for every element in χω, there exists an element in χω that maps to it under f. Overall, your solution appears to be correct.