33

An interpretation of $(\mathbb R,+,\cdot)$ in $(\mathbb C,+,\cdot)$ in particular provides an interpretation of $\DeclareMathOperator\Th{Th}\Th(\mathbb R,+,\cdot)$ in $\Th(\mathbb C,+,\cdot)$. To see that the latter cannot exist in ZF:
The completeness of the theory $\def\rcf{\mathrm{RCF}}\rcf$ of real-closed fields is an arithmetical ($\Pi_2$) statement, ...

22

(1) $\mathbb{C}$ is stable, (2) $\mathbb{R}$ is unstable, and (3) stability is preserved under interpretations. Of course the usual development of stability uses a lot of choice, but (2) and (3) are elementary. (1) follows from QE for algebraically closed fields. One of the reasons why model theorists care about the whole alphabet soup of classification-...

19

Bertrand Russel, On denoting
Mind, October 1905, pages 479-493.
When we say: "George IV wished to know whether so-and-so", or when we
say "So-and-so is surprising" or "So-and-so is true", etc., the
"so-and-so" must be a proposition. Suppose now that "so-and-so"
contains a denoting phrase. We may either ...

answered Jul 10 at 15:11

Carlo Beenakker

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9

Inner model theorists use the word "canonical" to explain the problem in intuitive terms, it is indeed a vague problem, though it is as precise as anything in the region of superstrong cardinals.
Inner model theorists understand the importance of the problem, the way it sits within our understanding of models of set theory, which is why they ...

8

For your first question, the definition of $e\Vdash x\in y$ and $e\Vdash x=y$ seems circular, but $\mathsf{CZF}$ provides a way to avoid the circularity, called inductive definition.
Definition. An inductive definition is a class $\Phi\subseteq \mathcal{P}(V)\times V$. For each inductive definition $\Phi$, define
$\Gamma_\Phi(X)=\{a \mid \exists Y\subset X :...

7

No, the random graph cannot interpret the random binary relation.
I’ll just answer the title question with the goal of illustrating a technique; I haven't considered $k$-ary structures. The approach is to use the property of least supports to set up a counting argument. Some equivalences are discussed at Least supports and weak elimination of imaginaries. ...

7

For the first question (distinct regular cardinals $>\aleph_1$): Force ZFC + MA + $2^{\aleph_0}=\aleph_3$ over $L$ in the usual way (see Jech, Theorem 16.13; note the forcing is ccc and it forces MA + $2^{\aleph_0}=\aleph_3$, which is all we need here). Then in $L[G]$, $\aleph_2$ and $\aleph_3$ are both $\mathcal{R}_{\mathbb{N}}$-detectable.
$\aleph_2$: ...

7

Back in the mid 1980s I remember convincing myself (alas, in unpublished work) that there is an ill-founded model $M$ of ZFC that has no end extension to another model of ZFC. Such a model $M$ by default is powerful and technically answers the question.
My unpublished work above used techniques employed in the following related results from the same time ...

6

You are asking about three choice principles, two of which have practically no research around them. Mainly due to the lack of tools we have for dealing with them.
The most you can find is the following paper,
Howard, Paul; Tachtsis, Eleftherios, No decreasing sequence of cardinals, Arch. Math. Logic 55, No. 3-4, 415-429 (2016). ZBL1339.03038.
Where the ...

6

Amorphicity implies strong minimality and $\omega$-categoricity, which together imply $\kappa$-amorphicity for any $\kappa$.
Assume that $T$ is amorphic. To see that $T$ is $\omega$-categorical, we proceed by induction. It is clear that the type space $S_1(T)$ must be finite, otherwise we could form a bi-infinite set $\bigvee$-definable over a finite set of ...

6

For a counterexample, let $G$ be the complete graph on the ordinal $\omega_1$, let the $W$'s be the countable ordinals, and let $\kappa=\aleph_0$.

6

The answer is yes. In fact, you can make the difference enormous.
First, let me construct a certain set of ordinals.
Lemma. There is a countable set $X$ of ordinals such that:
Every OTM program that halts with $X$ as input, halts strictly before $\sup X$.
If a program does not halt with $X$ as input, then there is no countable end-extension $Y\supset X$ for ...

answered Jul 13 at 7:27

Joel David Hamkins

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6

Harry de Swart's PhD from the University of Nijmegen (the Netherlands) was (explicitly) about this kind of topic. He establishes the completeness of IPC using search trees, in an intuitionistic meta theory. I do not know how search trees relate to your question.
H.C.M. de Swart: Intuitionistic logic in intuitionistic metamathematics, Dissertation, 1976, ...

5

A fairly complete answer to this question appears in Woodin's "In search of Ultimate $L$" at the beginning of the section on Martin-Steel extender sequences. Woodin omits the proofs, so I'll fill in some of the details.
If $E$ is a short extender, then $L[E] = L[U]$ where $U$ is the normal measure of $E$. To see this, let $j : V\to M$ be the ...

5

Writing an answer because it was suggested/advised. I will assume constructibility for simplicity. Also, I will assume writeable to mean as written in second last paragraph of question. Going through various examples:
(1) All elements of $A$ are countable and all elements of $B$ are uncountable.
Take $A=\{x \in \omega_1+\omega \,|\, x \geq \omega_1\}$. Now ...

5

If $M$ is supertransitive and satisfies $\sf ZFC$, then $\omega\in M$, and more importantly, $V_\omega\in M$.
Now by recursion, if $\alpha$ is an ordinal in $M$, then $V_\alpha\in M$ as well.
Therefore $M$ must agree on the $V_\alpha$ hierarchy, and therefore it must have the form $V_\kappa$ for a worldly cardinal $\kappa$, or $M=V$.
So the smallest ...

4

Yes! There are other set theories explored along this way generally speaking.
(1) Quine's Mathematical Logic $\sf ML$ adopted a similar approach on top of his $\sf NF$, and easily one can get a similar treatment on top of $\sf NFU$. See: Quine, Willard Van Orman (1951), Mathematical logic (Revised ed.), Cambridge, Mass.: Harvard University Press, ISBN 0-674-...

3

I believe there are some simpler direct possibilties here? Let $A=\omega +1$. Then $(W_\varnothing =) W_A = \beta$ for some countable ordinal $\beta$. Let $B\subseteq \omega$ be a code of a wellorder of type $\beta$. Then $W_B>W_A$. If you want (either of) $A$ or $B$ to have uncountable ordinals, let $A'=\omega \cup\{\omega_1+1\}$ and $B'=\{\omega_1 \}\...

3

I claim that if the language is countable, or merely well-orderable, then the witness property by itself prevents amorphous domains, and even infinite Dedekind-finite domains. Strong minimality has nothing to do with it.
Theorem. Any satisfiable complete theory $T$ in a well-ordered language with the witness property is never true on an amorphous domain, nor ...

answered Jul 9 at 19:29

Joel David Hamkins

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3

A "convention" floating around in (co)homology theory is that given a space $X$ the quotient with the empty set is $X/\emptyset = X_+ = X \sqcup *.$ This seems rather arbitrary if one defines $X/A$ to be given by coequalizing the inclusion $A \subseteq X$ and a constant map $\smash{A \rightarrow * \underset{a}{\rightarrow} X}$ to some point $a \in ...

3

This is another update of my answer, in which I also give a quick summary about solvability obstructions as suggested in Will Sawin's answer.
The smallest equation with easy-to-find solution is $0=0$ with $H=0$. The smallest equations with no solutions are $1=0$ and $-1=0$ with $H=1$. The smallest equations with at least one variable and no solutions are $x^...

2

No, except in trivial cases.$\newcommand{\ck}{\omega_1^{\mathrm{ck}}}
\newcommand{\nmereo}{n_{\mathrm{mereo}}}
\DeclareMathOperator{MT}{MT}
\DeclareMathOperator{rk}{rk}
\DeclareMathOperator{rks}{rks}$ If $\nmereo$ is the cardinality of the set of isomorphism classes of reducts of countable models of ZF to their inclusion relation, then $\nmereo\neq 2.$
A ...

2

In order to obtain a valid and consistent extension all we need is to define how we calculate one fractional operation for example $H_{\frac{3}{2}}(x,y)$ which is something between addition and multiplication. So let us try to extend one work that says that we can use arithmetic-geometric mean and obtain such consistent extension. First thing first.
Notice ...

2

$$A(x,y)=\begin{cases} x+y+1 & \text{ if } 0 \leq x \lt 1 \\
A(x-1,y+1) &\text{ if } 0 \leq y \lt 1, x \geq 1 \\
A(x-1,A(x,y-1)) &\text{ otherwise } \end{cases}
$$
Very nice and smooth and gives no headache in any particular sense.
If you look at the definition the only missing element is $A(p,q), 0 \leq p<1$ ...

2

It's going to depend on exactly what you mean by Heyting algebra semantics, but there is a proof in Troelstra and Van Dalen, Constructivism in Mathematics, Vol 2, for what they call $\Theta$-models, where $\Theta$ is a Heyting algebra. The completeness theorem appears as Theorem 6.12.

1

The following is not precisely what I was looking for in my question (one might very well call what I did "cheating"), but since it did give me some new insight into the problem that led me to ask the question, I am posting it anyway in hope that it will also be helpful to others.
For simplicity suppose that all our Laver trees are increasing, ...

1

In a general sense, yes. What follows is perhaps not the answer you are looking for, but seems to me to be relevant nonetheless.
Recall that the following are equivalent:
$\Bbb P$ is a weakly homogeneous forcing,
If $V[G]=V[H]$ for two $V$-generic filters, then there is an automorphism of $B(\Bbb P)$ such that $\pi``G=H$.
Since you assume that $M[x]=M[y]$, ...

1

Firstly, the question was already answered before my comment. This is how I thought about it (perhaps this slow thinking might be helpful for other non-experts, though I am not quite certain about this).
(1) In general, first we can observe that, given an ordinal parameter $\alpha$, if an ordinal $x$ is eventually writeable then every ordinal less than $x$ ...

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