# $C^n$ And Forcing: Reading a Recent Paper By Kunen

While reading a recent paper by Kunen arxiv.org/abs/0912.3733, which deals with PFA and the existence of certain differentiable functions, (defined on all of $\mathbb{R}$) which map certain $\aleph_1$-dense subsets of $\mathbb{R}$ onto other $\aleph_1$-dense subsets of $\mathbb{R}$. The technical details aside, he was able to show:

Theorem 1.6 Assume PFA, and let $D,E,\subset \mathbb{R}$ be $\aleph_1$-dense. Then there exist exists an order preserving bijection $f:\mathbb{R} \rightarrow \mathbb{R}$, and $D^\ast \subseteq D$ such that $D^\ast$ is $\aleph_1$-dense, $f(D^\ast)=E$, and

1. For all $x \in \mathbb{R}$, $f'(x)$ exists and $0\leq f'(x)\leq 2$
2. $f'(d) = 0$ for all $d\in D^\ast$

It occurred to me that with a few modifications his method/forcing notion might be used to add other differentiable, or Lipschitz functions to some ground model. It follows that these new functions would in turn produce new $C^1$ functions, and so on. And in the end could result in new systems of differential equations, with absolutely strange behavior.

So my questions are the following:

• What is known about "messing with" the class of $C^n$ functions, via forcing?
• Are there other examples of more exotic forcing notions which add smoother functions?

Edit: took out weakly bit..

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Perhaps you could clarify what kind of strange behavior you are seeking? In terms of ordinary behavior, there are a few easy observations:

• Any forcing notion that adds a new real number will add many new $C^\infty$ functions, such as new lines and polynomials.

• If a forcing notion does not add a new real number, then it cannot add any new continuous functions on $\mathbb{R}$, since every such function is determined by a countable list of data: the rational approximations to its value on the rationals.

• The question of whether a given function is $C^n$ or not cannot be affected by forcing, by the Shoenfield absoluteness theorem.

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The kind of strange behavior I was thinking about would be something that would make Hilbert's 16th unresolvable. But now that I look at Shoenfields theorem, this is certainly impossible. Thank you for your answer. As a side request would you happen to know of any forcing notions which allow for more control over the functions you are adding? Kunens notion exhibits a level of control over what is being added, and is what made me post the question. –  Michael Blackmon Jan 29 '11 at 14:12

Michael, why do you say that Kunen's result is about weakly differentiable functions?
This seems to talk about functions that are plainly differentiable on all of $\mathbb R$.
Or do you mean they are differentiable on $D^*$?

Adding either $C^1$-functions or Lipschitz functions by forcing seems to be nontrivial.
I have been interested in the question of how many continuous, Lipschitz, or $C^n$-functions are needed so that the functions together with their inverses (as relations) cover all of $\mathbb R^2$. In the Sacks model there is a family of $\aleph_1$ $C^1$-functions so that the functions and their inverses cover all of $\mathbb R^2$ (this is due to Steprans). The same is true for Lipschitz functions.

You can use ccc forcing to add a "small" family of continuous functions that covers $\mathbb R^2$ in the above sense. However, the argument fails for Lipschitz and for $C^1$-functions.
Kubis and Vejnar showed that one can force a countable family of Lipschitz functions on the Cantor set such that the functions and their inverses cover an uncountable square.
An absoluteness argument now shows that already in the ground model some uncountable square is covered by countably many Lipschitz functions.
So in some sense, the Kubis-Vejnar forcing does not give you anything new.

On the negative side, $\mathbb R^2$ cannot be covered by less than continuum many functions that are twice differentiable. This seems to indicate that adding twice differentiable functions with interesting properties by forcing is difficult.

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honestly I didn't know what else to call it when I was typing it (because $f'$ cannot be continuous in this particular case.) I guess I have such a preference for continuously differentiable functions, that over time the terms differentiable and continuously differentiable have erroneously become synonymous to me. Thank you, for your answer, and I will hunt down the two papers you mention. The prospect of using forcing to effect objects of analysis fascinates me. –  Michael Blackmon Jan 29 '11 at 18:11
"Continuous Ramsey theory on Polish spaces and covering the plane by functions" on my homepage has the references concerning covering $\mathbb R^2$ by functions. The Kubis-Vejnar paper is on Wieslaw Kubis' homepage. –  Stefan Geschke Jan 29 '11 at 19:34
Another good reference for the things that I mentioned abouve is the book "The Covering Property Axiom CPA" by Ciesielski and Pawlikowski. –  Stefan Geschke Jan 29 '11 at 19:41
Your covering problems are very interesting. But one quibble: you say that adding $C^1$ functions by forcing is nontrivial, but any forcing notion adding a real adds a new piece-wise linear continuous function, whose integral will be $C^1$ but not $C^2$. –  Joel David Hamkins Jan 29 '11 at 20:38
I was going to suggest the Ciesielski-Pawlikowski book, but I see Stefan did already. It has a nice developed mathematical theory, and as with PFA it reaches a point where a lot can be proved without needing to reference forcing. –  Andres Caicedo Jan 29 '11 at 20:47