16
votes

Accepted

### Long chains of amorphous cardinalities

If $A$ is amorphous, then every subset of $A$ is either finite or cofinite in $A$. Since every cardinal below $A$ is determined by a subset of $A$, it follows from this that the cardinals below $A$ ...

15
votes

### Second-order ordinal definability

Since you allow arbitrary sets of ordinals in your second-order definitions, all sets will be in $\text{OD}^2$. The reason is that, for any set $x$, we can code $x$ into a set of ordinals as follows. ...

14
votes

Accepted

### Consistency strength of strongly compact cardinal

Steel [1] showed that if $\square_\kappa$ fails for some singular strong limit $\kappa$, then $\text{AD}$ holds in ${L(\mathbb R)}$. Since Solovay showed a strongly compact cardinal implies the ...

13
votes

Accepted

### Does proper forcing preserve properness under PFA?

In $\sf ZFC$, any forcing of the form $\operatorname{Add}(\kappa,1)$, for any $\kappa$, will destroy the properness of some forcing.
Indeed, much more of that is true.
Theorem. (Yoshinobu) Suppose ...

9
votes

### Can uncountable sets be proved to exist in this variant of ZFC with definability restrictions?

Parameter-free ZFC is equivalent to ZFC. See
Ralf Schindler, Philipp Schlicht, ZFC without parameters.
Therefore, in parameter-free ZFC we can prove all the same theorems as in ZFC, including the ...

9
votes

Accepted

### Can we have an axiom that refers to itself and the prior axioms of the theory it is an axiom of?

Yes, there is a completely robust mechanism for introducing self-reference into mathematics. Any sufficiently robust mathematical language will admit the possibility of self-reference, as shown by the ...

8
votes

Accepted

### Strengthening of a classical set mapping theorem of Lázár

Theorem 1 is due to P. Erdős and E. Specker, On a theorem in the theory of relations and a solution of a problem of Knaster, Colloq. Math. 8 (1961), 19–21 (pdf). This Erdős–Specker paper was cited by ...

8
votes

Accepted

### Would strengthening Foundation and Choice in NBG, make it equi-consistent with MK?

The answer is no.
This follows from a modification of Kameryn's answer at your other question. Namely, KM implies the existence of a transitive model of NBG, and transitive models will always satisfy ...

8
votes

### Why is inner model theory evidence for consistency of large cardinals?

The explanation is philosophical rather than mathematical.
The idea is simply that the inner-model theory provides a rich account of what it would be like for the large cardinal axioms to be true, and ...

8
votes

Accepted

### Is strengthening Foundation in NBG sufficient to make it prove Con(ZFC)?

Joel answered the first question, so here's an answer to the second question.
(Minor point: I would call your schema an induction schema, not a recursion schema. These end up being different in ...

8
votes

### Is strengthening Foundation in NBG sufficient to make it prove Con(ZFC)?

The answer to the first question is no. If ZFC is consistent, then NBG does not prove the second-order $\in$-induction scheme.
To see this, take an $\omega$-nonstandard model of NBG, with only the ...

7
votes

### Most recent results on formulating Kunen's inconsistency theorem in ZF without choice

In "On the consistency of ZF with an elementary embedding from $V_{\lambda+2}$ into $V_{\lambda+2}$" (arXiv 2006.01077, 2020) Farmer Schlutzenberg showed that assuming a theory based on $I0$ ...

7
votes

### Building the real from Dedekind finite sets

Q1: There is no such partition. Let $\langle X_n \rangle$ be a countable partition of $\mathbb{R}.$ We will construct $n,$ an open interval $I,$ and an injection $g: \omega \rightarrow I \cap X_n$ ...

7
votes

Accepted

### Transversal of $\mathbb{N}\times\mathbb{N}$

Counterexample. Let $\mathbb N=\{0,1,2,\dots\}$. Define $f(0,k)=0$, $f(1,k)=1$, and $f(n,k)=k\operatorname{mod}n$ for $n\ge2$. Your conditions are satisfied, and a transversal map $t=t_{(f,c)}$ can't ...

6
votes

Accepted

### Cardinality of maximal diverse families

If $\kappa$ is an infinite cardinal, then the relation $|X\bigtriangleup Y|\lt\kappa$ (where $X\bigtriangleup Y=(X\setminus Y)\cup(Y\setminus X)$) is an equivalence relation on $\mathcal P(\kappa)$. A ...

6
votes

Accepted

### Long chains of Dedekind finite sets

The answer to all of these is yes. Monro constructed a model where for every ordinal $\alpha$ there is a Dedekind-finite set mapped onto $\alpha$.
This alone gives us arbitrarily long chains. Take ...

5
votes

### A problem with a $\Pi_1$ formula of the Lévy hierarchy

This statement has complexity $\Delta_2$, because it is a locally verifiable feature, meaning one that can be decided yes-or-no inside any sufficiently large $V_\theta$. Any $V_\theta$ above $\alpha$ ...

5
votes

### Can uncountable sets be proved to exist in this variant of ZFC with definability restrictions?

The proof below is due to George Boolos, and appears in his article Constructing Cantorian Counterexamples, Journal of Philosophical Logic, Vol. 26, No. 3 (Jun., 1997), pp. 237-23.
Boolos gives an ...

4
votes

Accepted

### Do precipitous ideals "always" come from collapsing?

If $M_1^\#$ exists and is fully iterable, then there is an inner model $M$ in which $\omega_1$ is measurable and an $M$-generic $G$ such that $V_{\omega_1+2}\in M[G].$ Just iterate the first normal ...

4
votes

Accepted

### Convergence of distance

$\newcommand\de\delta$The answer is no. In fact, $d_H(A,C_n(L_n))$ can be however large or even infinite for all $n$.
Indeed, e.g. suppose that $X=[-1,a)$ for some $a\in(0,\infty]$ endowed with the ...

3
votes

Accepted

### Chromatic numbers realised by almost disjoint subsets of $\omega$

This question was answered affirmatively by Theorem 1.1 of Paul Erdős and Saharon Shelah, Separability properties of almost-disjoint families of sets, Israel J. Math. 12 (1972) 207–214 (pdf), ...

3
votes

Accepted

### Simplified method of building an Aronszajn tree

Your argument is basically Kurepa's proof from his thesis Ensembles ordonnées et ramifiés, see page 96 (a footnote has Aronszajn's construction).
As noted in the comments you need to show that what ...

3
votes

### Bernstein's proof of the continuum hypothesis

Ackermann's above-cited review focuses on the second of Bernstein's proposed rules. However, even the first ("axiom of identity") is broken. Gentling tweaking the language, this axiom reads:
...

3
votes

### End-extension in Gödel's constructible universe

This is in the same vein as Monroe Eskew's comments. For $n\geq 1$, given that $\alpha<\beta$ and $L_\alpha\prec_{\Sigma_n}L_\beta$, it is not always possible to produce even a $\Sigma_1$ end-...

2
votes

### Posets obtained from a semigroup by the definition $x \leq y \iff x \cdot y = x$

As Benjamin stated in the comments, semigroups satisfying these identities are called right-regular bands. Let us call the partial orders underlying a right-regular band “associative”. There is no ...

1
vote

Accepted

### End elementary extension in infinitary logic of some $L_\alpha$ producing a $L_\beta$

(Turning some comments into an answer)
The definition of $L(x,\alpha+1)$ was wrong, instead it should have been $$L(x,\alpha+1)\leftrightarrow\bigvee_{n\in\omega}\bigvee_{\varphi}\exists p_1,\ldots,...

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