I have seen proofs of N×N. The proof by list where you list each element in a pair of elements and then counting them diagonally is the most convincing. I have two thoughts in mind:
- Write it as a composition of functions. Show that both functions are bijections, so the composition is also a bijection.
- Show that N×N×N has the same cardinality as N×N, and N×N has the same cardinality as N.
Generally, I'm having difficulty going from N×N×N to N×N, since it easy to prove the bijection from N×N to N.
Answer
If you already know that N×N is countable, fix a bijection f:N×N→N. Now consider g:N×N×N→N×N defined as: g(n,m,k)=(n,f(m,k)),
or better yet h:N×N×N→N defined as h(n,m,k)=f(n,f(m,k))
and cut out the middle-man.
As my freshman discrete mathematics professor used to tell us, go home and convince yourself this is correct.
No comments:
Post a Comment