Given that, A is collection of sets such that x in A if and only if \(X \subset N and |X|=n\) for
some n in N.
Here, it means that A is collection of finite subset of N.
Now need to show \(|A|=|N|\).
It is sufficient to show A is countable.
Use induction method to show A is countable.
The number of subsets of N with cardinality 1 is finite.
It is trivially true.
Assume that, the number of subsets of N with cardinality K is finite.
Need to show the number of subsets of N with cardinality \(K +1\) is finite.
The number of subsets of N with cardinality K +1 is exactly two times of number of
subsets of cardinality K as each subset has two choices either \((K +1)^{th}\) term belongs to the set or not.
Thus, the number of subsets of N with cardinality \(K +1\) is finite.
Hence, for all n in N the number of subsets of N with cardinality n is finite.
So, A, is equal to collection of subsets of N with cardinality n .
Therefore, \(A=\bigcup_{n-1}^{\infty}A_{n}\).
Tt is known that, countable union of countable set is countable.
Here, A_n is countable for each n in N .
So, A is countable
Hence, \(|A|=|N|\) is proved.