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All subsets of Z that have an even number of members. For example, the set{1,2,3,4}would be in this set, but{1,2,3}would not be in the set.

I thought that it would be something like {A ⊆ ℤ | |A| / 2 ∈ ℤ} in set-builder and {infinity?} in set-roster. I believe it to be infinity because you could have 100, 1000, 10000, etc, elements and they would have an even amount of elements. Such as, A = {1, 2, 80, -7, -5, 43}. The cardinality of A is 6 but I could put in two more arbitrary elements and it would be at 8 and therefore still an even amount of members.

Can anybody please explain how I am thinking incorrectly? I think the answer is infinity but I do not know how to express infinity in set-roster notation.

Thanks!

  • Not really a Stack Overflow question -- but why do you think that every definable set can be adequately described in "set roster" notation? Set roster notation is quite limited. – John Coleman Sep 17 '15 at 01:35
  • I guess I don't. I just assumed. Are you saying this can not be expressed in set roster? –  Sep 17 '15 at 02:07
  • Not in any useful way. The set of finite subsets of even cardinality is of course countable and can be enumerated in fairly simple ways. You can fix such an enumeration, list the first few terms, and put an ellipsis -- but the result will not be very readable and will strictly speaking be ambiguous since any finite sequence can be extended in infinitely many ways. – John Coleman Sep 17 '15 at 02:17

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