Exercise 4.6 - clarification needed

[abhinav-upadhyay]

The definition of is not clear from the question. I could think of two possible interpretations: is a set consisting of any n natural numbers, or another interpretation is that consists of numbers from 1 to n.

In the latter interpretation, it is pretty straightforward to prove that the union of all such will essentially be N

which has an injective mapping from

In the former case I am confused if it can be shown that there will be a injection from to the union because I can always select arbitrary which always excludes certain numbers from (say none of the has 1) then the union will essentially be . In this case it is easy to show a surjection from to but will it also be an injection because of the their infinite cardinality?

[j2kun]

I should have instead wrote the question as:

Suppose for each natural number n we chose a different, countably infinite set $A_n$. I.e., each $A_n$ has a bijection with the natural numbers, but each $A_n$ is itself different. Prove the union is countable.

As an example, you could have the set of all pairs $A_n = {(n, i) : i \in \mathbb{N}}$. Perhaps I can make it simpler and just use that concrete choice in the next version of the book...

[abhinav-upadhyay]

Thank you, that helps :-)

[j2kun]

Sorry, I was mistaken. The previous exercise already uses that concrete construction. I suppose the point of this exercise was to show that N x N is not special in this regard. I.e., you can take countably infinite sets that don't look like N and give the necessary bijection to N x N