Constructing the power set of natural numbers

Question

Consider the set of Natural numbers, $\mathbb{N}$, and a particular natural number, $n$.

Consider $A_n$ to be the set of all subsets of $\mathbb{N}$ whose size is $\leq n$.

Now as we take $n$ to infinity, does this $A_n$ become the power set of $\mathbb{N}$?

The answer should be NO since $A_n$ is countable for all $n$, whereas the power set of $\mathbb{N}$ is not countable. We also know that a countable union of all countable sets is countable, but the power set of natural numbers is not countable.

So is there some glitch in the above argument?

Answer 1

I suppose by "take $n$ to infinity" you mean the set $$\bigcup_{n\in\mathbb{N}}A_n.$$ Indeed, this set is not the power set of $\mathbb{N}$. You can either (as you have done in your question) argue that this is a countable set where the power set is uncountable or you can see directly that the above union does not contain any infinite subsets of $\mathbb{N}$ while the power set certainly does.

Answer 2

If you are thinking along the lines of forming $$A_{\infty} = \bigcup_{n=0}^{\infty}A_n$$ then note that this is not the power set of $\mathbb N$. Instead, this is the collection of all finite subsets of $\mathbb N$, which is not the same set. For example, it doesn't include $\{2, 4, 6, 8, \ldots\}$.