Friday, 20 May 2016

discrete mathematics - Prove that the set of mathbbNtimesmathbbNtimesmathbbN is countable.




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:




  1. Write it as a composition of functions. Show that both functions are bijections, so the composition is also a bijection.

  2. 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×NN. Now consider g:N×N×NN×N defined as: g(n,m,k)=(n,f(m,k)),

or better yet h:N×N×NN 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.


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