# Prove that the set $A=\{S:S\subseteq \mathbb N,|S|\le 1\}$ is countable

Prove that the set $$A=\{S:S\subseteq \mathbb N,|S|\le 1\}$$ is countable.

Hello everyone. I understand that this set is countable because it is equinumerous to $$\mathbb{N}$$. But I am not sure how to write a formal proof for this.

• That is a finite set???? – Prince M May 29 '20 at 0:10
• Let $n \in \mathbb{N}$, then $S_n = \{n\} \in A$, thus $A$ is not a finite set because |$A$| $\geq$ |$S$| – Prince M May 29 '20 at 0:11
• The first thing I would do would be describe $S$ better. – gen-ℤ ready to perish May 29 '20 at 0:39

Since $$S\subset\Bbb N$$ and $$|S|\le1$$, you get that either $$S=\emptyset$$ or $$S=\{k\},k\in\Bbb N$$.
Thus $$A=\{\emptyset,\{k\}\;:\;k\in\Bbb N\}$$ which is clearly countable (and NOT FINITE!!), since its elements are in bijection with elements of $$\Bbb N$$.
No, it is not a finite set! $$A$$ is the collection of all sets that either have $$0$$ elements, or only have $$1$$ element. So $$A$$ consists of the emptyset $$\varnothing$$ and all singletons $$\{n\}$$, $$n\in\mathbb{N}$$, i.e. $$A=\{\varnothing\}\cup\{\{n\}: n\in\mathbb{N}\}$$ You can count $$A$$, the first element is $$\varnothing$$, the second is $$\{1\}$$ the third is $$\{2\}$$ and so on. I believe you can write this enumeration explicitly by yourself!