Countability of Collection of All Finite Subsets of a Countable Set Let V be a countable set. Ok, first thing to say is that this isn't a question as to whether $S = \{ A \subseteq V \mid A \ \text{finite} \}$ is countable -- there are plenty of other duplicates on SE asking the same question! This question is regarding the legitimacy of a method that I've been thinking about. Wlog let $V$ be infinite, otherwise the result is trivial.

Fix $n \in \mathbb{N}$ and consider $S_n = \{ A \subseteq V \mid |S| = n\}$. ($S_0 = \emptyset$.)


*

*For $n = 1$, this is just the set of singletons of $V$, and so is countable.

*For $n = 2$, this is just the set of pairs of $V$, and so, by the "countable collection of countables is countable" argument, is countable.

*For general $n \in \mathbb{N}$, consider $N \choose n$, which, for fixed $N \in \mathbb{N}$, is a polynomial in $n$. Since we can iterate the "countable collection of countables is countable" argument, we have countability for each $n \in \mathbb{N}$. (That is to say, this follows since the set of all polynomials with integer coefficients of degree $n$ is countable.)
Now, further we know that the set of all polynomials with integer coefficients is countable. Letting $P_n$ be the set of polynomials (with integer coefficients -- assume this whenever talking about polynomials) of degree $n$, we have that the set of all polynomials $P = \cup_{n\in\mathbb{N}} P_n$. Similarly (in our above notation), $S = \cup_{n\in\mathbb{N}} S_n$. From this can we then deduce that $S$ is countable?
My main concern is the following. If $|V| = N \in \mathbb{N}$ (so $V$ is finite), then $|S_n| = {N \choose n}$. Since the union is disjoint, we then have that
$$|S| = \sum_{n=1}^N {N \choose n} = 2^N.$$
But $\aleph_1 = 2^{\aleph_0}$. So if we just let $N \to \infty$ in the above equation, then naively we get $|S| = 2^{\aleph_0} = \aleph_1$, which says that $S$ is uncountable.

Alternatively, we can think of it another way (very similar). Without considering polynomials, we can just say that $S_n$ is countable for each $n \in \mathbb{N}$, and we're summing/unioning over a countable set, and so the result follows from the "countable collection of countable is countable" argument.

Any comments on this would be most appreciated. The 'proof' (if it is a proof) does seem to be fine to me (I'm pretty sure the concern mentioned above is just that if you do the rigorous maths then you'll find that you can't just swap the limits naively). However, in the many SE questions asking if $S$ is countable, everyone (that I've seen) tries to set up some function that is an injection to deduce the result. This is fine, if you can come up with the injection, but my above method seems much more intuitive. (There is one that has a similar method, but they claim that $S_n$ is finite for each $n \in \mathbb{N}$, and that's (a place) where their proof fell down.)
 A: Some remarks:


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*It's unclear to me why the set of pairs is countable because of the fact that a countable union of countable sets is countable. You might want to elaborate there. 

*It's not true that $2^{\aleph_0}=\aleph_1$. That would be the continuum hypothesis which cannot be proved or disproved from the standard axioms of set theory. 

*Cardinal functions have no reason to be continuous. So arguing that something is uncountable just because $N\to\infty$ is a common mistake. 

*In general the statement that countable unions of countable sets are countable depends on the axiom of choice, whereas the statement you are trying to prove does not. There is nothing wrong in using the axiom of choice, but it's worth knowing when you're using it. 

*What's also unclear is that if you know that polynomials with integer coefficients are countable, why aren't you done already? I mean, every finite set defines a unique polynomial (just write the set in increasing order and those are the coefficients). So we're about done. The trick in these proofs is generally to avoid sets like that. 
