Unanswered Questions
1,316 questions with no upvoted or accepted answers
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How many algebraic closures can a field have?
Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is ...
- 3,419
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Concerning proofs from the axiom of choice that ℝ³ admits surprising geometrical decompositions: Can we prove there is no Borel decomposition?
This question follows up on a comment I made on Joseph O'Rourke's
recent question, one of several questions here on mathoverflow
concerning surprising geometric partitions of space using the axiom
of ...
- 13.5k
33
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Defining $\mathbb{Z}$ in $\mathbb{Q}$
It was proved by Poonen that $\mathbb{Z}$ is definable in the structure $(\mathbb{Q}, +, \cdot, 0, 1)$ using $\forall \exists$ formula. Koenigsmann has shown that $\mathbb{Z}$ is in fact definable by ...
CommunityBot
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Peano Arithmetic and the Field of Rationals
In 1949 Julia Robinson showed the undecidability of the first order theory of the field of rationals by demonstrating that the set of natural numbers $\Bbb{N}$ is first order definable in $(\Bbb{Q}, +,...
- 18.3k
29
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Did Grothendieck overestimate topoi?
I was reading the Russian translation of Recoltes et Semailles and in the footnote where Grothendieck lists his 12 contributions (including schemes) we find the following lines:
Из этих тем наиболее ...
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28
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Is Feferman's unlimited category theory dead?
In 2013 Solomon Feferman in Foundations of unlimited category theory: what remains to be done (The Review of Symbolic Logic, 6 (2013) pp 6-15, link) laid out three desirable axioms for "...
- 1,466
28
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Can one divide by the cardinal of an amorphous set?
This question arose in a discussion with Peter Doyle.
It is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \times \{1,\ldots,k\} \simeq Y \times \{1,\ldots,k\}$ then $X \...
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27
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Supercompact and Reinhardt cardinals without choice
A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:
Definition. A cardinal $\kappa$ is supercompact if for all ordinals $\...
- 1,153
26
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Where do uncountable models collapse to?
Suppose $T$ is a complete first-order theory (in an finite, or at worst countable, language). Given any model $\mathcal{M}\models T$ of cardinality $\kappa$, we can ask whether $\mathcal{M}$ can be ...
- 20.1k
24
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What's the smallest $\lambda$-calculus term not known to have a normal form?
For Turing Machines, the question of halting behavior of small TMs has been well studied in the context of the Busy Beaver function, which maps n to the longest output or running time of any halting n ...
- 1,774
23
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CH and automorphisms of ultrapowers of $\mathbb{Z}$ and $\mathbb{R}$
Notation and motivation. Given an algebraic structure $\mathbb{M}$ of cardinality at most the continuum and with countably many operations, and a nonprincipal ultrafilter $\cal{U}$ on a countably ...
- 18.3k
21
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443
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Ado's theorem and the reduction to positive characteristic
The synopsis: proofs of Ado theorem in positive characteristic are simple, and in characteristic $0$ are difficult. Can one infer the characteristic $0$ case from the positive characteristic case?
The ...
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21
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"Compactness for computability" - does it ever happen?
Throughout, "computable structure" means "first-order structure in a computable language with domain $\omega$ whose atomic diagram is computable."
Say that a computable structure $...
- 20.1k
21
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Density of first-order definable sets in a directed union of finite groups
This is a generalization of the following question by John Wiltshire-Gordon.
Consider an inductive family of finite groups:
$$
G_0 \hookrightarrow G_1 \hookrightarrow \ldots \hookrightarrow G_i \...
CommunityBot
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20
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Non-rigid ultrapowers in $\mathsf{ZFC}$?
Originally asked and bountied at MSE:
Question: Can $\mathsf{ZFC}$ prove that for every countably infinite structure $\mathcal{A}$ in a countable language there is an ultrafilter $\mathcal{U}$ on $\...
- 20.1k