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5 votes
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Definable pseudo-standard predicates in Internal Set Theory

Consider the usual language $\mathcal{L}=(\in, \mathrm{st})$ of Nelson's Internal Set Theory, and a unary $\mathcal{L}$-predicate $P$. For an $\mathcal{L}$-formula $\varphi$, let $\varphi^P$ denote ...
Z. A. K.'s user avatar
  • 756
-5 votes
0 answers
250 views

Can Cardinality Theory capture ZFC?

Cardinality Theory "CT" is a theory of sets of cardinals and links between them, only sets of cardinals can be assigned cardinalities. The links are unordered edges linking cardinals, they ...
Zuhair Al-Johar's user avatar
5 votes
1 answer
622 views

Non-atomic probability measures on N

One can intuitively imagine picking a random natural number and ask to what extent the intuition can be axiomatized. Using the axiom of choice, there is a total finitely additive (monotonic) averaging ...
Dmytro Taranovsky's user avatar
6 votes
0 answers
188 views

Is there a characterization of measurables in terms of indiscernibles?

There is a characterization of $\alpha$-Erdős cardinals in terms of sets of indiscernibles of order type $\alpha$. There is also a characterization of Ramsey cardinals in terms of sets of good ...
C7X's user avatar
  • 2,031
10 votes
1 answer
501 views

What is the least $\alpha$ such that $L_\alpha$ contains a non-measurable set

What is the least level of the constructable hierarchy that contains a non-measurable (Lebesgue) subset of $2^\omega$. If it makes a difference assume we are working inside L (V=L). I'm pretty sure it ...
Peter Gerdes's user avatar
  • 3,029
5 votes
1 answer
633 views

Consistency of ZFC with inaccessible cardinals but no measurable cardinals

Let $S$ be a set and k a infinite field. The injection $S \to k\mathrm{Alg}(k^S, k)$ (sending a point to the evaluation in it) is a bijection if and only if $S$ is a non-measurable cardinal (see for ...
Arshak Aivazian's user avatar
2 votes
0 answers
92 views

Geometric interpretation of flags and the role of the rook monoid and Kazhdan–Lusztig theory in $M_n(\mathbb{C})$

Let $G = GL_n(\mathbb{C})$, $B$ be its Borel subgroup, and $P$ a parabolic subgroup. The space $G/B$ corresponds to complete flags in $ \mathbb{C}^n$, and $G/P$ corresponds to partial flags. The ...
Learner's user avatar
  • 141
2 votes
2 answers
172 views

Can the Category of that kind of small sets in $\sf NFU$ be Cartesian closed?

Working in Quine's $\sf NFU$, with urelements being at least as many as sets. Formally the latter is: $|Ur| \geq |Set|$. Where $Ur$ is the set of all urelements and $Set$ is the set of all sets. We ...
Zuhair Al-Johar's user avatar
8 votes
1 answer
228 views

Examples of anti-classical theories in iFOL

An anti-classical axiom $\phi$ is one which is inconsistent with LEM Are there any sources for good examples of anti-classical theories in intuitionstic first-order logic? There are many examples of ...
Jason Carr's user avatar
5 votes
0 answers
138 views

Cone avoidance and $\Pi^0_1$-classes

Suppose $X \subseteq 2^{\omega}$ is nonempty and $\Pi^0_1$ relative to $a$. Assume $c_0 \nleq_T b_0 \oplus a$ and $c_1 \nleq_T b_1 \oplus a$. Must there exist some $y \in X$ such that $c_i \nleq_T ...
Hello World's user avatar
6 votes
2 answers
523 views

Is the logic of the ($\infty$-?)topos of simplicial sets "contradictory up to homotopy"?

In a sense this is a followup to my earlier question Does the (1-)topos structure on simplicial sets have any homotopy-theoretic significance?. In the topos of simplicial sets, the subobject ...
მამუკა ჯიბლაძე's user avatar
4 votes
0 answers
100 views

Explicit superexponential growth for Presburger Arithmetic

Fischer and Rabin proved a superexponential bound $2^{2^{cn}}$ for the worst-case length of a proof of a proposition of length $n$ in Presburger arithmetic. The result is in Michael J. Fischer and ...
Mikhail Katz's user avatar
  • 16.6k
15 votes
3 answers
3k views

Finite verification for theorems due to Busy Beaver numbers

I recently learned about the Busy Beaver function, and a formulation of it that essentially tells us if a turing machine of $n$ states takes over $BB(n)$ steps, it will never halt. One consequence I ...
Ivan Galakhov's user avatar
4 votes
1 answer
239 views

True or false? Every left or right cancellative, duo semigroup is cancellative

A semigroup $S$ is duo if $aS = Sa$ for all $a \in S$, where $aS := \{ax: x \in S\}$ and similarly for $Sa$; for instance, every commutative semigroup is duo, and so is every group. On the other hand, ...
Salvo Tringali's user avatar
8 votes
2 answers
596 views

If a semigroup embeds into a group, then is it a subdirect product of groups?

The title has it all: Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups? If yes, then $S$ is a subdirect product of subdirectly irreducible groups,...
Salvo Tringali's user avatar
5 votes
1 answer
268 views

What oracles make finding isomorphism (of finite structures) easy?

Below, all structures are finite, in a finite language, with underlying set an initial segment of the natural numbers. This has been edited to fix errors pointed out by Emil Jerabek in his answer ...
Noah Schweber's user avatar
7 votes
2 answers
490 views

Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?

By a classical result of Birkhoff (that is, Theorem 2 in [G. Birkhoff, Subdirect unions in universal algebra, Bull. AMS, 1944]) and the trivial fact that the class of semigroups is closed under the ...
Salvo Tringali's user avatar
6 votes
3 answers
554 views

Conjecture about commutative semigroups

Conjecture: given any commutative semigroup $S$ of order $n \ge 4$, there exist $a, b \in S$ with $a \ne b$, an integer $m \ge \lfloor (n-1)/2 \rfloor$, and two $m$-element subsets $X = \{x_1, \ldots, ...
Fabius Wiesner's user avatar
6 votes
1 answer
162 views

Can there exist a set of all transitive sets in a model of NF or NFU?

Is it consistent with $\sf NF$ or $\sf NFU$ to have a set of all transitive sets? Formally: $\exists t \forall x (x \in t \leftrightarrow x \text { is transitive})$ Where "$x$ is transitive" ...
Zuhair Al-Johar's user avatar
1 vote
1 answer
154 views

Looking for Fitch-style Natural Deduction system that allows for open formula

I find most Natural Deduction proof systems only allow for close formulas, which are not convenient for FOLs without a constant. Most Sequent Calculus systems instead allow for open formulas, but it ...
William's user avatar
  • 127
18 votes
1 answer
555 views

When can we add choice to a model of ZF

For countable transitive models of ZF, is existence of a ZFC extension with the same height a first order property? In other words, is there a statement $τ$ (in the language of set theory) such that ...
Dmytro Taranovsky's user avatar
11 votes
1 answer
416 views

Examples of natural algebraic irreflexive relations

To motivate the question, consider the theory of rings. Define $x \parallel y$ to mean $\exists w \exists z .((x - y) z = w (x - y) = 1)$, or in words, "$x - y$ is a unit". Then $\parallel$ ...
Zhen Lin's user avatar
  • 15.9k
2 votes
0 answers
189 views

Semantic equivalence between mathematical proofs

Sometimes, we recognize two proofs of the same claim to be the "same" proof. In some cases, this sameness is obvious -- for example, the proofs that $\sqrt{2}$ and $\sqrt{3}$ are irrational ...
safsom's user avatar
  • 225
1 vote
0 answers
96 views

Determine equivalences in the generated collection of subgroups and quotients

Let $A$ be an abelian group, and $B_1, B_2, \dots, B_m$ be subgroups of $A$. Define the family of subgroups $\mathcal{D}_0 = \{ \{0\}, A, B_1, B_2, \dots, B_m \}$. Let $\mathcal{C}_1$ be the ...
Zhang Yuhan's user avatar
8 votes
1 answer
322 views

Does every cancellative duo semigroup embed into a group?

Prompted by the comments to a recent answer by YCor to a related question (here), I'd like to ask the following: Q. Does every cancellative duo semigroup embed into a group? A (multiplicatively ...
Salvo Tringali's user avatar
8 votes
1 answer
385 views

Is "every infinite set of strictly subnumerous sets is supernumerous to its union" equivalent to AC?

Is the following sentence equivalent to $\sf AC$ over the rest of axioms of $\sf ZF$? For each infinite set $X$: if for all $y \in X$ we have $|y| < |X|$, then $| \bigcup X|\leq |X| $? Note: ...
Zuhair Al-Johar's user avatar
7 votes
2 answers
399 views

Numerical choice and reverse mathematics

Consider the following fragment of numerical choice in the language of second-order arithmetic: for any arithmetical $\varphi$, we have: $$ (\forall n\in \mathbb{N})(\exists m\in \mathbb{N})(\forall X\...
Sam Sanders's user avatar
  • 4,359
11 votes
3 answers
781 views

Is every recursively axiomatizable and consistent theory interpretable in the true arithmetic (TA)?

I am looking for a scholarly text that discusses this issue in detail.
Juan Atacama's user avatar
20 votes
5 answers
1k views

Uniqueness results that follow from CH

Recently, Joel David Hamkins presented a historical thought experiment that shows that CH could have been adopted as an axiom if we had been using the hyperreal field $\mathbb{R}^*$ instead of $\...
6 votes
1 answer
228 views

Can we computably escape infinitely many functions (allowing partiality)?

Let $(p_i)_{i\in\omega}$ be a uniformly computable sequence of partial functions (i.e. the partial function $q(i,x)=p_i(x)$ is computable) such that infinitely many $p_i$s are total. Must there be a ...
Noah Schweber's user avatar
1 vote
0 answers
109 views

Name For Effective Cantor-Bendixsonish Derivitive

When dealing with a tree (substring closed subset of $\omega^{< \omega})$ a useful operation will frequently be to remove any nodes with finite ordinal rank (i.e., all nodes whose extensions on the ...
Peter Gerdes's user avatar
  • 3,029
-2 votes
1 answer
218 views

If existence of a pre-isomorphism implies existence of an isomorphism, would AC follow?

Let a surjection $f: M \to N$ be called a pre-isomorphism on membership, if and only if: $\begin{align} \forall x \in M \,\forall y \in M \,\exists x' \in M \exists y' \in M : \ & f(x')=f(x) \...
Zuhair Al-Johar's user avatar
15 votes
3 answers
2k views

Exponentials of truth values

I noticed that the exponentiation identity $$\exp(r + s) = \exp(r) \cdot \exp(s)~,$$ which is of course completely standard for real or complex numbers also holds in a Boolean setting. That is, when I ...
blk's user avatar
  • 349
13 votes
1 answer
291 views

Descriptive complexity of analytic continuation

Consider the set of complex power series $$ f(z)=\sum_{n=0}^\infty a_nz^n $$ that have radius of convergence $1$ and can be analytically continued to the neighborhood of some point on the unit circle. ...
183orbco3's user avatar
  • 623
4 votes
0 answers
214 views

Algebraic logic in the style of algebraic geometry

I am writing a thesis on algebraic logic, I wonder if there is any recent research on an idea mentioned in Yuri Manin's book on algebraic geometry and in another Russian textbook on differential ...
YKY's user avatar
  • 558
7 votes
1 answer
556 views

Does this ZFC+V=L like theory, have a limit on large cardinal properties?

Let $\sf T$ be a theory that has as axioms every axiom of $\sf ZFC$, and every theorem of $\sf ZFC + [V=L]$ that is neither provable nor disprovable by $\sf ZFC$, whose addition or addition of its ...
Zuhair Al-Johar's user avatar
14 votes
1 answer
642 views

Example of a forcing notion with finite-predecessor condition that does not add reals

This question seems very basic but I cannot seem to find any literature on it. Let $\mathbb{P}$ be a forcing notion. If $p$ is a condition of $\mathbb{P}$, define the predecessor set of $p$ to be $$\{...
Zoorado's user avatar
  • 1,328
2 votes
0 answers
117 views

Can we have the set world obeying Quine's New Foundations with its well-founded realm obeying $\sf ZFC$?

Is this theory consistent? Language: first order language of set theory, Extra-logical axioms: 1. Extensionality: as in $\sf NF$. 2. Stratified Comprehension: as in $\sf NF$. Define: a set is said ...
Zuhair Al-Johar's user avatar
20 votes
1 answer
694 views

Is the theory of ordinals in Cantor normal form with just addition decidable?

This seems like it should be a pretty well-studied question but I can't seem to find an easy answer: Is the theory $(\varepsilon_0, +, \omega^{\ \cdot}, 0, 1)$ decidable? From Is the theory of $(\...
cody's user avatar
  • 1,452
-5 votes
1 answer
233 views

First research papers in mathematical logic [closed]

Hello I'm a software engineer who just wants to start research in mathematics soon. I'm interested in the foundations and hence I'm picking mathematical logic. As I have never touched undergraduate-...
john_delta's user avatar
19 votes
6 answers
2k views

Book recommendation introduction to model theory

Next semester I will be teaching model theory to master students. The course is designed to be "soft", with no ambition of getting to the very hardcore stuff. Currently, this is the syllabus....
Ivan Di Liberti's user avatar
7 votes
1 answer
716 views

What is the flaw in Cooper's argument?

Lately I have been studying in the subject of degree theory, specifically definability results related to $\mathcal{D}$. A famous conjecture in the field due to Slaman and Woodin is that the only ...
H.C Manu's user avatar
  • 893
11 votes
0 answers
430 views

Is $(\mathbb{R}, +)$ still injective as long as $(\mathbb{Q},+)$ is?

It is known that the existence of nontrivial injective abelian groups is independent of choice in ZF (or, rather, ZFA). In particular, $\mathbb{Q}$ is not provably injective, much less $\mathbb{R}$, ...
Garrett Figueroa's user avatar
-4 votes
1 answer
140 views

About the definitions of well-foundedness in this extension of NFU that interprets ZFC?

Lets see how the world of sets could look like from the perspective of $\sf NFU$. So, here we work within the first order language of set theory, with the following extra-logical axioms: 1. Quine atom:...
Zuhair Al-Johar's user avatar
2 votes
0 answers
232 views

Is the poset in the following construction stationary $\aleph_{\alpha + 2}$-linked?

Definition: A poset P is $\mathbf{stationary}$ $\kappa^{+}$-$\mathbf{linked}$ if for every sequence of conditions $(p_{\gamma} | \gamma < \kappa^{+})$, there is a regerssive function $f: \kappa^{+} ...
George Marangelis's user avatar
2 votes
0 answers
102 views

Direct construction of an arithmetically high degree below $0^{(\omega)}$

The existence of a high arithmetic degree (meaning the degrees induced by the notion of relative arithmetic definability) below $0^\omega$ can be established by using Harrington/Simpson's ...
Peter Gerdes's user avatar
  • 3,029
5 votes
1 answer
157 views

Intersection cardinalities in MAD families

Let $\newcommand{\o}{[\omega]^\omega}\o$ denote the collection of infinite subsets of the set of nonnegative integers $\omega$. We say ${\cal A}\subseteq \o$ is almost disjoint if $A\cap B$ is finite ...
Dominic van der Zypen's user avatar
7 votes
0 answers
262 views

A version of determinacy for all sets

Under ZF + AD, some games are undetermined because of lack of choice. In fact, the axiom of choice is equivalent to determinacy of games of length 2. However, we can ask whether the lack of choice ...
Dmytro Taranovsky's user avatar
82 votes
3 answers
20k views

Czelakowski's claimed proof of the Twin Prime Conjecture

It seems like the article "The Twin Primes Conjecture is True in the Standard Model of Peano Arithmetic: Applications of Rasiowa–Sikorski Lemma in Arithmetic (I)" by Janusz Czelakowski ...
Glycerius's user avatar
  • 1,083
20 votes
1 answer
557 views

Almost orthogonal maps $f:\omega \to \{-1,1\}$

Let $\omega$ denote the set of non-negative integers. For sets $A,B$, let $B^A$ denote the set of maps $f:A\to B$. For $f,g\in\{-1,1\}^\omega$ we say that $f,g$ are almost orthogonal if there is $C_0\...
Dominic van der Zypen's user avatar