If $p_n$ is the $n^{th}$ prime, is it ever appropriate to speak of $p_{\aleph_0}$? If $p_n$ is the $n^{th}$ prime, is it ever appropriate to speak of $p_{\aleph_0}$?
I'm no math student.  Your pardon if this is just some clearly obvious and easy answer, I'm just not seeing it.
When talking about a function such as $p_n$, where $p_n$ is the $n^{th}$ prime, and $n \in \mathbb{N}$, is it ever appropriate to talk of $p_{\aleph_0}$, since $\left|\mathbb{N}\right| = \aleph_0$?
Further is there any proof that $p_n \in \mathbb{N}$ for all $n \in \mathbb{N}$?  The point isn't so much in using primes, they are merely convenient.
My basic issue points to a fact discussed in this question, A "number" with an infinite number of digits is a natural number?, which states, "By definition, a natural number has a finite number of digits..."
While it may not be as easily seen with base-10 numbers, I could easily construct a numbering system based on the primes themselves, such that each digit is mulitipled by $p_{digit}$, so that $p_n$ is always 1 followed by (n-1) zeros.  This numbering system, just to help see it, demonstrates that while $\mathbb{N}$ is an infinite set comprised of numbers with a finite number of digits, $\mathbb{P}$, constructed over the same range, grows one digit in length each time and so has one number of each count of digits, from 1 to $\aleph_0$, and thus would contain many numbers with an infinite number of digits.  But, I believe the numbering system is irrelevant, and just illustrates the point.  The real point is that over the range of $\mathbb{N}$, it seems like $p_n$ itself must become infinite in length, and thus no longer be in $\mathbb{N}$ (which, of course, makes no intuitive sense, that a prime not be an integer).
What it seems like, thinking about it, I relate to the percentage of primes.  If one looks at the entire number line, there are effectively 0% primes across the whole of the number line, out to infinity.  Out at infinity, you have to go an appropriate "infinite distance" to find the next suitable prime, on average, which leaves you with a non-finite number of digits on your next $p_n$.  But, this is arrived at through an infinite set of finite-lengthed $n \in \mathbb{N}$.
And, here's what I'm trying to wrap my head around.  This fact seems to indicate that you would run out of primes before you would natural numbers, except you can keep picking primes.  But, according to the illustration given above, $p_{\aleph_0} \notin \mathbb{N}$, or so it seems.. 
It seems I'm running into somekind of logical falacy here.  Is it just wrong to think of $p_{\aleph_0}$, which of needs must be of infinite length, and hence is not natural, or something else?  What am I missing?
 A: Ok, a few quick things. First of all, it still isn't true that the "number system" that you've described contains anything with an infinite number of digits; in fact, I recommend that you read through the post you cite again to better understand this.
To paraphrase though (assuming I understand what you meant), you are looking at number system consisting of objects of the form $1\underbrace{000...0000}_{n}$ i.e. a 1 followed by some number of zeros. However, just because these strings get longer doesn't mean that there will ever be infinitely long strings! If you start with a string of length 50, then adding a zero gets a string of length 51, never one of infinite length.
Furthermore, you are correct that the density of primes---the notion of density must be defined carefully!---drops off to zero despite the fact that there are infinitely many primes. But this isn't really surprising. Consider numbers (in base 10 representation) of the form $10^k$. These get further and further apart, but there are always more of them out there. The same is true of primes.
Anyhow, finally, the question you ask about whether or not $p_n \in \mathbb{N}$ is in some sense backwards; primes are first and foremost elements of $\mathbb{N}$, but they are simply elements of this set that have special properties. So it doesn't make sense to speak of a proof that $p_n \in \mathbb{N}$ because they are defined to be in this set to begin with.
As for $p_{\aleph_0}$, well, that is perhaps a more interesting question.  Maybe a simpler (and related) question: Is $\aleph_0 \in \mathbb{N}$? If you can answer this, then you should be able to answer the other question.
