# Forcing and divisibility

I'm going to bring together a couple of seemingly unrelated questions that I've asked here. This may be silly. Or maybe not?

Imagine that $n$ is some sort of infinitely large integer, and thus so are $n\pm1, n\pm2, n\pm3,\ldots$. In this earlier question, for each (finite!) prime number $p$ and each finite positive integer $k$, we would randomly decide which congruence class modulo $p^k$ to put $n$ into, and consequently which prime powers each member of $\{n, n\pm1, n\pm2, n\pm3,\ldots\}$ is divisible by. As Henry pointed out, this implies that with probability $1$, every member of $\{n, n\pm1, n\pm2, n\pm3,\ldots\}$ has infinitely many prime factors. (I still wonder how I missed that. It follows from the fact that the expected number of prime factors of each such "number" is the sum of the reciprocals of all primes.

Now let's remember a possibly unrelated thing I asked about, concerning forcing. The book by Boolos and Jeffrey that I cited (second edition, before Burgess joined the list of authors) conditions like $\{G(3), \lnot G(5), G(20) \}$ were considered, which give a finite amount of information about membership or non-membership in a set $\{x : G(x)\} \subseteq\{0,1,2,3,\ldots\}$. A "generic" set is defined roughly as a set about which nothing is true except what can be "forced" by such finite "conditions". If one randomly picks $G(x)$ or $\lnot G(x)$ for each $x$ by tossing a coin, then with probability $1$ one does not get a generic set. That is because no finite condition can force a set to have a density (the proof includes some work showing that the property of having a density is actually expressible in the "language of arithmetic", which includes the usual arithmetic relations and operations and also logical connectives and quantifiers). So generic sets are ill-behaved things that are not typical of what you get from ordinary random processes.

So let's go back to our infinite integers and let our conditions be things like $\left\{(2^3 \mid n),\ (7^2 \nmid n+1),\ (13 \mid n-1),\ (3\mid n-1),\ 19 \mid n+2 \right\}$, etc. As before, let a generic "set" (but "set" is no longer literally the right word) be one about which nothing is true but what is forced by some finite condition, and as before we will say that a condition forces the sentence "$\alpha\text{ or }\beta$" iff either it forces $\alpha$ or it forces $\beta$; it forces $\exists x\ \alpha(x)$ iff there is some $x$ for which it forces $\alpha(x)$, and it forces $\lnot\alpha$ iff no more extensive condition forces $\alpha$.

It seems plausible to me that, in contrast to the previous situation, randomly chosen "sets" could be generic. In particular, each of these "numbers" would be forced to have infinitely many prime factors since no finite condition could force it not to. And each would be divisible by a prime $p$ only finitely many times, i.e. by $p^k$ but not by $p^{k+1}$, for some $k$, since, for example, if $2^5\mid m$ then $2^2\nmid m+2$.

Does this thing that seems plausible make sense? Or, more precisely: is it true?

And I fear to ask: Could there be some point to this, or is it just an accidental collision of two of my recent questions to stackexchange?

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The more you write about it, the more it seems that this forcing you are working with is just like symmetric extensions of ZF. – Asaf Karagila Nov 29 '11 at 23:00
It is not clear to me what partial order you are trying to use for your forcing. The key point about forcing with filters is that they ensure consistency of the generic object. What prevents me from making the set $\{ (2 \nmid n), (2 \mid n-1), (2 \mid n+1)\}$, or $\{ 2^m \mid n : m \in \mathbb{N}\}$? – Carl Mummert Nov 30 '11 at 1:59
I don't see what problem yuo have with the first example you give. You seem to be saying $n-1$ and $n+1$ are even and $n$ is odd. That seems perfectly innocent. Your second example, however, is not finite. I had in mind that conditions were simply finite sets of statements of the sort I gave examples of, and they're partially ordered by inclusion. (Or reverse-inclusion, if you like to think of the one that gives more information as "smaller".) – Michael Hardy Nov 30 '11 at 3:27
@CarlMummert : OK, I see: you had "$\nmid$" rather than "$\mid$" there---it looked like "$\mid$" when I saw it earlier. What prevents you from doing that is that we only consider conditions that are consistent; we could take that to be part of the definition of "condition". – Michael Hardy Nov 30 '11 at 21:34