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Whitehead problem is a rather well known problem:

Suppose that $G$ is an abelian group and $\mathrm{Ext}^1(G,\Bbb Z)=0$, is $G$ free?

It wasn't long before it was proved that if $G$ is countable (and thus countably-generated) then the answer is positive. However the question was open for uncountable groups, and it took more than a decade until Shelah came up with the interesting answer: In ZFC the problem is undecidable.


I was wondering what other open problems have somewhat of a Whitehead-like nature, namely they were proved for objects with a countable nature (e.g. separable topological spaces, countably generated), but the question remains open for the general case.

One example which comes to mind is Naimark's problem which was solved for separable $C^\ast$-algebras, and was partially solved in the general case in the sense that it is consistent to have a negative answer.

(As with CW big-lists, please post one problem per answer.)

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What happens if you assume that the cardinality of $G$ is bounded? For example, suppose the cardinality of $G$ is at most that of $\mathbb R$, and the conditions hold - is $G$ free? (I don't know any logic or set theory, so this could be a naive question.) –  Peter Samuelson Jun 14 '13 at 17:32
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@Peter: Shelah's independence result applies even in cardinality $\aleph_1$. –  Andreas Blass Jun 14 '13 at 19:00
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One that - to the best of my knowledge - is still a huge open problem is about non-standard models of $PA$ and Scott systems. A Scott system $S$ is a set of reals such that $(\omega, S)\models WKL_0$; that is, $S$ is closed under Turing reducibility and has paths through all of its trees. Now, given a non-standard model of arithmetic $M\models PA$, we say a real $X$ is coded in $M$ if $$ \exists a\in M\forall b\in M\cap\omega(M\models b\vert a\iff b\in X). $$ (It's a slight abuse of notation to write "$M\cap \omega$" for the standard part of $M$, but this is common.) The set of reals coded in $M$ is called the standard system of $M$ and is denoted by "$SSy(M)$."

A beautiful short argument shows that $SSy(M)$ is always a Scott set: infinite binary trees in $SSy(M)$ come from binary trees of nonstandard height in $M$, by an overspill argument, which must have paths of internally-finite but externally-infinite length; and these long paths corresond to infinite paths in $SSy(M)$. Conversely, it is easy to prove that every countable Scott set is the standard system of some model of $PA$.

The major open question is whether the stronger statement, "Every Scott set is the standard system of a model of $PA$," is true.

(Actually, we can do a bit better than "countable": by a union of chains argument, every Scott set of size $\le\aleph_1$ is a standard system, so in the presence of $CH$ the problem is solved. However, I still think this fits the question, because what's really going on is that countable=easy and $\aleph_1$ is "close enough" to countable to still be tractable.)

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