Constructing a model of set-thory via forcing requires dealing with some complications, e.g., either "names" for elements of the model, or boolean-valued models, or sheaves, or something along those lines.

I am looking for an application of forcing, where the concept appears in close to a "pure" form. This is for pedagogical reasons. The audience will be familiar with the basic concepts of first-order logic.

I was wondering if there are examples in recursion theory. It's easy to define a subset of $\mathbb{N}$ that is generic with respect to the countable family of r.e. sets of conditions (or alternately, arithmetic conditions). This would require minimal background from the audience. Are there any interesting results based on this approach? It wouldn't matter if there are "non-forcing" proofs of the same results, so long as the "forcing" proofs are short and sweet.

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    $\begingroup$ There are many examples. Sacks original paper on forcing with perfect trees is mostly about recursion-theoretic applications. See also MR0611176(82h:03038). Odifreddi, Piergiorgio. Trees and degrees. Cabal Seminar 77–79 (Proc. Caltech-UCLA Logic Sem., 1977–79), pp. 235–271, Lecture Notes in Math., 839, Springer, Berlin,1981. $\endgroup$ – Andrés E. Caicedo Jan 22 '18 at 18:45
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    $\begingroup$ For more sophisticated applications, see the manuscript by Slaman and Woodin on definability in degree structures. $\endgroup$ – Andrés E. Caicedo Jan 22 '18 at 18:46
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    $\begingroup$ A silly example is that you can violate $\mathsf{CH} $ by adding Cohen reals. Since each real is only Turing above countably many other reals, if $\mathsf{CH}$ fails, there are Turing incomparable degrees. By Shoenfield absoluteness, it follows that Turing incomparable degrees exist (in the ground model, regardless of $\mathsf{CH}$). This is obviously a silly proof, but the same argument applies to infinite time Turing reducibility a la Hamkins-Lewis. $\endgroup$ – Andrés E. Caicedo Jan 22 '18 at 20:19
  • $\begingroup$ @AndrésE.Caicedo Re: that silly example, see here and the related disclaimer. $\endgroup$ – Noah Schweber Jan 22 '18 at 21:08
  • $\begingroup$ @Noah Funny. I came up with this "proof" years ago. It is obviously not serious. I saw it again in the Hamkins-Lewis paper, and now in your link. Probably it also shows up in a thousand other places. $\endgroup$ – Andrés E. Caicedo Jan 22 '18 at 21:12

There are many examples. Very quickly they get quite complicated - e.g. I think the Low Basis Theorem might not be pedagogically appropriate - but there are indeed examples.

In particular, the easiest (in my opinion) proof that there are incomparable Turing degrees is a forcing argument. Specifically, one shows that for any pair of finite binary strings $\sigma,\tau$ and any $e\in\mathbb{N}$, there are extensions $\sigma'\supseteq\sigma$ and $\tau'\supseteq\tau$ such that there are no infinite binary sequences $f\supseteq\sigma', g\supseteq\tau'$ with $\Phi_e^f=g$. This lets us inductively build a pair of infinite binary sequences which are Turing incomparable, by diagonalizing against possible reductions (the phrase "finite extensions" is often used here). The Friedberg-Muchnik theorem improves on this by making the degrees c.e. by moving from "pure" forcing to the more complicated setting of priority arguments; see this old question.

Now, you brought up the issue of "generic" versus "generic for some specific collection of dense sets" (e.g. the r.e. ones). There is a rich hierarchy of "levels of genericity" in computability theory, and this is also the case for non-Cohen forcing notions (unfortunately the word "generic" in computability theory usually means Cohen-generic). In particular, examining the argument above we see the following:

  • The dense sets we want to meet are those of the form "No infinite extension of the left string (resp. right) computes any infinite extension of the right string (resp. left) via $\Phi_e$."

  • On the face of it, each of these is quite complicated - namely, $\Pi^1_1$. There are still only countably many such dense sets, so we can trivially meet them all.

  • ... But we might also want to see how simple we can go! There are two ways that a computation can "fail:" it can make an error, or it can fail to give any answer at all. This lets us redefine the dense sets we care about: "EITHER there is already some $n$ such that $\Phi_e^{left}(n)$ halts, $right(n)$ is defined, and $\Phi_e^{left}(n)\not=right(n)$, OR there is some $n$ such that no extension of the left string makes $\Phi_e^?(n)$ halt." (And similarly, swapping "left" and "right.")

  • This definition is easily checked to be $\Sigma^0_2$. So we have:

If $x, y\in 2^{\omega}$ are mutually $\Sigma^0_2$-Cohen generic - that is, if the set $\{(\sigma,\tau)\in 2^{<\omega}\times 2^{<\omega}: \sigma\prec x, \tau\prec y\}$ is a filter meeting every $\Sigma^0_2$ dense subset of $2^{<\omega}\times 2^{<\omega}$ - then $x$ and $y$ are Turing incomparable.

Here's another example. I'm not sure this fits the bill - the proof is quite complicated if one hasn't gotten comfortable with forcing in computability theory already - but the forcing itself is simple, the question is easy to state (although the answer is a bit complicated), and the whole situation is quite interesting.

Let's say we want to build a fast-growing function $\omega\rightarrow\omega$. There is a natural forcing which leaps to mind (there are many variations on this idea):

  • Conditions are pairs $(p, f)$ with $p$ a map from some finite subset of $\omega$ to $\omega$, and $f:\omega\rightarrow\omega$ is total with $p\subseteq f$.

  • The ordering is given by $(p, f)\le (q, g)$ (remember, "$\le$" means "is stronger than") if:

    • $p\supseteq q$,

    • $f(n)\ge g(n)$ for all $n$, and

    • for each $m$ in $dom(p)\setminus dom(q)$, we have $p(m)\ge g(m)$.

Basically, what's going on here is the following. We're trying to build a function $\omega\rightarrow\omega$. In a condition $(p, f)$, the $p$-part is the amount of the function we've built so far, and the $f$ part is a "promise" that from now on the function we're building will grow at least as fast as $f$.

This is called Hechler forcing. It's not the only forcing that adds a fast-growing function, of course, but it is in my opinion the most natural (and it has some nice set-theoretic universality properties - e.g. Truss showed that if $V[d]$ is a forcing extension of the ground model $V$ where $d:\omega\rightarrow\omega$ dominates every function in $V$, and $c\in\omega^\omega$ is Cohen generic over $V[d]$, then $d+c$ is Hechler generic over $V$).

Now, an important class of questions in computability theory are of the form:

Suppose I can compute a real with a certain property. What are the specific reals which I am guaranteed to be able to compute?

For example, it's not hard (thinking about the argument above) to show that two sufficiently mutually Cohen generic reals don't compute anything in common besides the computable sets (they form a "minimal pair"), so there are no specific reals which "being sufficiently Cohen generic" lets me compute.

One particularly natural question of this form, phrased in the contrapositive, is:

Suppose $x$ is computable from any sufficiently fast-growing function - that is, there is some $f:\omega\rightarrow\omega$ such that any $g$ dominating $f$ computes $x$. What do we know about $x$?

It turns out that the set of such reals has a very snappy characterization:

Theorem. A real $x$ is computable from any sufficiently fast growing function iff it is hyperarithmetic.

The hyperarithmetic sets are essentially those which can be computed by taking repeated Turing jumps - possibly infinitely many jumps, but only "computably many." (There are many other characterizations as well.) Precisely defining this is tricky, but it's not too hard to $(i)$ show that "the $\omega$th jump of $0$" has an obvious definition and $(ii)$ argue that stuff gets really weird if we try to take the $\alpha$th jump of $0$ for a really really complicated countable ordinal $\alpha$ (namely, any which doesn't have a computable copy).

The proof of the theorem above - due to Solovay if I recall correctly - is quite nice. Showing that every hyperarithmetic real is computable from a sufficiently fast-growing function is a proof by transfinite induction, using a generalized notion of the Busy Beaver functions. The converse is the interesting direction, and involves a clever analysis of $\Vdash_{Hechler}$. If you're interested, Peter Gerdes' thesis is a good place to start reading if I recall correctly.


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