Question

Prove or disprove: If \(A=\\{X \subseteq \mathbb{N}: X\) is finite \(\\},\) then \(|A|=\aleph_{0}\).

Short answer

The statement is correct. The set \(A\) of all finite subsets of \(\mathbb{N}\) is countably infinite, i.e., \(|A|=\aleph_0\).

Step-by-step solution

Definition of finite subset of \(\mathbb{N}\)

Understand that any set \(X\) belongs to set \(A\) if it is a finite subset of \(\mathbb{N}\). Therefore, \(A\) includes all possible subsets of natural numbers with a finite number of elements.

Countability of finite subsets

To consider combinations intending to count subsets, you need to take each natural number \(n\) and all subsets of \(\mathbb{N}\) of size \(n\). Each of these is finite and we have a countable number of such \(n\), hence each collection is countable.

Using the principle of countable union of countable sets

Unite all countable collections obtained in Step 2. The result will be the set of all finite subsets of \(\mathbb{N}\), which is the set \(A\). By using the principle of countable union of countable sets, we can confirm that \(A\) is countably infinite, i.e., \(|A|=\aleph_0\). So, the statement holds true.