Independence over ZFC + CH

Question

Acknowledging Woodin's result on $\Sigma^2_1$-absoluteness for forcing-models satisfying the continuum hypothesis (CH), it is natural to ask:

Are there examples of statements $\phi$ in "the usual realm of mathematics" that cannot de decided in ZFC + CH?

Surely modern mathematicians (oblivious of metamathematics) could have naturally come up with problems of complexity beyond $\Sigma^2_1$ that are provably undecidable in ZFC + CH.

A ZFC example to contrast with would be Borel's conjecture.

But what are the ZFC + CH examples if any?

Answer 1

I wanted to add a couple of examples in addition to those mentioned in the comments.

In algebra, Shelah's work on the Whitehead Problem established that the statement ``All Whitehead groups are free'' is independent of ZFC + CH. He originally proved the statement's independence from ZFC, and then developed forcing techniques for getting independence from ZFC + CH.

In set-theoretic topology, I've done a lot of work on building "interesting" models of ZFC+CH using iterated forcing. For example, the following statement is not decided by ZFC + CH:

"a first countable countably compact (=every infinite set has a point of accumulation) space is either compact or contains a subspace homeomorphic to $\omega_1$"

Counterexamples to this can be built from $\diamondsuit$; constructions due to Ostaszewski and Fedorcuk are the most well-known.

Independence from ZFC was established in a flush of results from the late 1980s involving Balogh, Dow, Fremlin, and Nyikos (and others as well). In particular, the statement is a consequence of PFA.

We established that the statement is consistent with ZFC + CH in joint work with Nyikos, so taken with the $\diamondsuit$ results we have the required independence.

The best result we have along these lines (done with Alan Dow) is that ZFC + CH is consistent with "compact countably tight spaces are sequential''.

Here a space $X$ is countably tight if whenever $x$ is in the closure a set $A\subseteq X$ if and only if $x$ is in the closure of a countable subset of $A$. A space $X$ is sequential if a subset $A$ of $X$ is closed if and only if it is sequentially closed. Roughly speaking, the statement "compact countably tight spaces are sequential" says that if the topology of a compact space is determined by its countable subsets, then it is determined by its convergent sequences".

Again, this statement is a consequence of PFA (a famous result of Balogh), and in the other direction $\diamondsuit$ provides us with counterexamples -- in fact, Fedorcuk showed that $\diamondsuit$ entails the existence of a compact countably tight hereditarily separable space of size $2^{\aleph_1}$ with no convergent sequences at all.

The work we did with Dow proves the statement is consistent with (and therefore independent from) ZFC + CH.

Answer 2

Is there a definable non-measurable set?

(Here I mean definable without parameters -- it is cheating if you define it from an ultrafilter, or a well-ordering of $\mathbb R$, without first defining those too.)

The answer to this question is independent of ZFC+CH. The answer is yes if $V=L$, but it becomes no if you add a Cohen real or a random real in certain forcing extensions.

I'm not sure whether or not you consider this question to lie within the realm of "usual mathematics" -- but it is a question that I would imagine a wide variety of mathematicians have asked at some point (not just set theorists or logicians).

EDIT: Over the weekend I realized that I'd made a mistaken claim in this post, namely that adding a Cohen or random real to a model of ZFC+V=L produces a model with no parameter-free-definable non-measurable sets.

(I think I might have made the substitution ''ultrafilter'' $\rightarrow$ ''non-measurable set'' in the back of my mind; adding a Cohen or random real does produce a model without parameter-free-definable ultrafilters.)

For adding a random real, what I claimed is definitely false. If $r$ is random over $L$, then $L[r]$ admits a non-measurable set with a particularly simple definition (only three symbols!), namely $L \cap \mathbb R$. This can be deduced from Example 26.51 in Jech's book (and Jech attributes the example to Solovay).

I don't know whether adding a single Cohen real to $L$ results in a model without parameter-free-definable non-measurable sets.

In his famous paper in which he proved that there is a model of ZF+DC in which every set of reals is measurable, Solovay shows that there is a forcing extension of $L$ where every non-measurable set of reals fails to be definable (in fact, he shows much more: no non-measurable set of reals is definable from a countable sequence of ordinals). His forcing construction uses an inaccessible cardinal, so all I can say for certain is that ``there exists a definable non-measurable set'' is independent of ZFC+CH assuming the consistency of ZFC+"there is an inaccessible cardinal".