Step 1

Given that, A is collection of sets such that \(\displaystyle{x}\in{A}\) if and only if \(\displaystyle{X}\subset{N}\) and \(\displaystyle{\left|{X}\right|}={n}\) for some \(\displaystyle{n}\in{N}\)

Here, it means that A is collection of finite subset of N

Now need to show \(\displaystyle{\left|{A}\right|}={\left|{N}\right|}.\)

It is sufficient to show A is countable.

Step 2

Use induction method to show A is countable.

a: The number of subsets of N with cardinality 1 is finite. It is trivially true.

b: Assume that, the number of subsets of N with cardinality K is finite.

Step 3

c: Need to show the number of subsets of N with cardinality \(\displaystyle{K}+{1}\) is finite.

The number of subsets of N with cardinality \(\displaystyle{K}+{1}\) is exactly rwo times of number of subsets of cardinality K as each subset has two choices either \(\displaystyle{\left({K}+{1}\right)}^{{{t}{h}}}\) term belongs to the set or not.

Thus, the number of subsets of N with cardinality \(\displaystyle{K}+{1}\) is finite.

Hence, for all \(\displaystyle{n}\in{N}\) the number of subsets of N with cardinality n is finite.

Therefore, \(\displaystyle{A}={\bigcup_{{{n}={1}}}^{{\infty}}}{A}_{{{n}}}\)

Step 4

It is known that, countable union of countable set is countable.

Here, \(\displaystyle{A}_{{{n}}}\) is countable for each \(\displaystyle{n}\in{N}\)

So, A is countable.

Hence, \(\displaystyle{\left|{A}\right|}={\left|{N}\right|}\) is proved.

Given that, A is collection of sets such that \(\displaystyle{x}\in{A}\) if and only if \(\displaystyle{X}\subset{N}\) and \(\displaystyle{\left|{X}\right|}={n}\) for some \(\displaystyle{n}\in{N}\)

Here, it means that A is collection of finite subset of N

Now need to show \(\displaystyle{\left|{A}\right|}={\left|{N}\right|}.\)

It is sufficient to show A is countable.

Step 2

Use induction method to show A is countable.

a: The number of subsets of N with cardinality 1 is finite. It is trivially true.

b: Assume that, the number of subsets of N with cardinality K is finite.

Step 3

c: Need to show the number of subsets of N with cardinality \(\displaystyle{K}+{1}\) is finite.

The number of subsets of N with cardinality \(\displaystyle{K}+{1}\) is exactly rwo times of number of subsets of cardinality K as each subset has two choices either \(\displaystyle{\left({K}+{1}\right)}^{{{t}{h}}}\) term belongs to the set or not.

Thus, the number of subsets of N with cardinality \(\displaystyle{K}+{1}\) is finite.

Hence, for all \(\displaystyle{n}\in{N}\) the number of subsets of N with cardinality n is finite.

Therefore, \(\displaystyle{A}={\bigcup_{{{n}={1}}}^{{\infty}}}{A}_{{{n}}}\)

Step 4

It is known that, countable union of countable set is countable.

Here, \(\displaystyle{A}_{{{n}}}\) is countable for each \(\displaystyle{n}\in{N}\)

So, A is countable.

Hence, \(\displaystyle{\left|{A}\right|}={\left|{N}\right|}\) is proved.