21

No. An early nontrivial constraint on the $\beth$ function comes from Kőnig's Theorem, that for all infinite $\kappa$, $\mathrm{cf}(2^\kappa)>\kappa$. This implies that we cannot have $\beth_\alpha = \aleph_{f(\alpha)}$ for all $\alpha$, when $f(1) = \omega$, nor when $f(\omega+1)$ is a cardinal below $\aleph_\omega$.
Another constraint is Silver's ...

16

The answer is no. Both (I) and (S) are equivalent to AC over ZF. Indeed, for any set $S$ the class of ordinals $\alpha$ such that $\alpha$ injects into $S$ (resp. $S$ surjects onto $\alpha$) is a set, so there is some ordinal $\beta$ outside this set. Assuming (I) (resp. (S)), there is an injection $S\to\beta$ (resp. surjection $\beta\to S$). In the latter ...

12

This is consistent. Kanovei constructed a model $M$ with an infinite Dedekind finite set of reals which is lightface projectively definable. By descending to $L(R),$ we can further assume it satisfies $V=L(R).$
Clearly choice fails in this model. Since it satisfies $V=L(R),$ every set is definable from an ordinal and a real. Of course, any ordinal is ...

11

No. Some, but not all topoi are exponentiable.

11

When working in ZF, one can have more freedom. See
An Easton-like Theorem for Zermelo-Fraenkel Set Theory with the Axiom of Dependent Choice and An Easton-like theorem for Zermelo-Fraenkel Set Theory without Choice.

answered Jan 14 at 17:45

Mohammad Golshani

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9

(Working in ZFC.)
No (re $\mathcal{L}_{\omega_1,\omega}$). Suppose it is. Consider the signature $\Sigma$ with just one binary relation symbol $<$. Let $\Sigma',\eta$ witness SED-ness for $\Sigma$.
Let $\pi:M\to V_\eta$ be elementary,with $\eta$ some sufficiently large limit ordinal, $M$ transitive, $M$ of cardinality $\kappa=2^{\aleph_0}$, with $\mathbb{...

9

please get the pdf here.
note that the text itself starts on page 5.

9

The sets $V_\lambda$ for limit ordinals $\lambda>\omega$ are precisely the models of second-order Zermelo set theory (including foundation scheme, infinity, and choice) plus the cumulative hierarchy axiom CHA, which asserts that every set $x$ is in some $V_\beta$, where this is a set for which there is well-ordered sequence $\langle V_\alpha\mid\alpha\leq\...

answered Jan 17 at 2:33

Joel David Hamkins

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8

In fact, the principle "Every set is analytic" is not consistent with $\mathsf{ZF}$ in the first place. We don't need choice to get a surjection $h$ from Baire space to the set of continuous maps on Baire space. But once we have such an $h$, the "diagonalizing" set $\{x: x\not\in h(x)\}$ can't be analytic.
Let me show how to get such an $...

7

No.
Consider when $2^{\aleph_0}=2^{\aleph_1}=2^{\aleph_2}=\aleph_3$, with $\kappa=\aleph_1$ and $\lambda=\aleph_2$.
If you allow for one of these to be singular, then consider $\kappa=\beth_\omega$ and $\lambda=\kappa^+$. Then $\lambda^{<\lambda}=\lambda^\kappa=2^\kappa\cdot\lambda$, and on the other hand since $\kappa$ is a strong limit cardinal, $\...

answered Dec 29 '21 at 14:43

7

This answer concerns the second question (which asks whether there is a characterization of the theory of models of the form $V_{\alpha}$, where $\alpha$ ranges over limit ordinals), and its elaboration which asks whether there is a characterization of the theory $T$ defined by:
A sentence $\sigma$ is a member of $T$ if and only if, for any models $M\models ...

5

I'm not sure if this gets to the heart of the question, but here are some observations.
For a successor ordinal $\alpha$, let $\kappa_\alpha$ be the least $\kappa>\alpha$ which is $\kappa^{+\alpha}$-supercompact. Then $\kappa_\alpha$ is not $2^{\kappa_\alpha^{+\alpha}}$-supercompact. This is because an embedding $j : V \to M$ with critical point $\...

5

I don’t have a comprehensive answer, but a couple of observations:
$T$ is included in ZFC unconditionally (i.e., not assuming the consistency of worldly cardinals).
Indeed, for any sentence $\sigma$, ZFC proves $\neg\sigma\to\exists\alpha>\omega\text{ limit }V_\alpha\nvDash\sigma$ by Lévy’s reflection principle. Thus, if ZFC proves $\forall\alpha>\...

5

As clarified in the comments, the existence of a regular S-space is independent of ZFC, but a "real" $T_2$ example can be constructed by taking a well-ordering of a set of reals in order type $\omega_1$ and refining the Euclidean topology by declaring initial segments open. This example is $T_2$ since it refines a $T_2$ topology and is still ...

5

Yes, assuming ZF is consistent it is, because it follows from ZFC + V=HOD. This is because the ordertype of the class of (ordinal-)cardinals is just that of the ordinals. That is, if $X$ is definable over $V_\alpha$ from ordinal parameter $\beta<\alpha$, then $X$ is definable over $V_{\aleph_\alpha}$ from ordinal parameter $\aleph_\beta$, since $\xi\...

4

David Chodounský and Osvaldo Guzmán showed in arXiv:1703.02082 that There are no P-points in Silver extensions. They prove that
after adding a Silver real no ultrafilter from the ground model can be extended to a P-point, and this remains to be the case in any further extension which has the Sacks property.
In particular, any free ultrafilter from the ...

4

Working on a related topic for a new paper with Jonathan Schilhan I've stumbled into a disproof of your ridiculous conjecture.
Definition. The class $W$ is the closure of $\{\{a\}\mid a\in V\}$ under well-ordered unions. Sets in $W$ are called "almost well-orderable sets".
The axiom $V=W$ is consistent with $\lnot\sf AC$, for example in Gitik's ...

answered Dec 23 '21 at 19:28

4

One can collapse $2^\lambda$ to have cardinality $\lambda^+$ without adding $\lambda$-sequences even if $\lambda$ is singular. Therefore one could start with $2^{\aleph_\alpha} = \aleph_{\alpha+2}$ for all $\alpha < \omega_1$ by Foreman-Woodin and then force $2^{\aleph_{\omega_1}} = \aleph_{\omega_1+1}$ without changing $P(\aleph_{\omega_1})$ and hence ...

4

One can get fairly close bounds by observing that if $\kappa$ is $\lambda$-supercompact, then by using a supercompactness measure of least Mitchell rank, there is a $\lambda$-supercompactness embedding $j:V\to M$ for which $\kappa$ is not $\lambda$-supercompact in $M$. But $\kappa$ will be $\theta$-supercompact in $M$ whenever $2^{\theta^{<\kappa}}\leq\...

answered Jan 17 at 16:09

Joel David Hamkins

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3

Your mistake is in misunderstanding $\mathrm{Card}^M=\mathrm{Card}^N$. Indeed, it is easy to take it at face value and simply say that the classes are exactly the same.
However, much like the case with Borel sets, when we move to a different model, what we really mean is the reinterpretation of the Borel code, not the actual set itself.
The way to make sense ...

3

Theorem. For any cardinals $\alpha,\beta\ge2$ there is an $\alpha$-uniform hypergraph $H$ with $\chi(H)=\beta$ and $\chi(H^\partial)=2$.
Proof. Let $V=\bigcup_{\xi\in\beta}V_\xi$ where the sets $V_\xi$ are pairwise disjoint and $|V_\xi|\gt\alpha\beta$. Let $E=\{e\in[V]^\alpha:|\{\xi\in\beta:e\cap V_\xi\ne\varnothing\}|\ge2\}$.
Plainly $H=(V,E)$ is an $\alpha$...

3

This doesn't answer your question, but gives some information that
might be useful.
Main Claim: ZFC + "There is a proper class of inaccessible limits of measurable cardinals" + "There is some measurable cardinal $\kappa$ and an
ordinal definable normal measure on $\kappa$" disproves the statement
in question.
Corollary: If ZFC + VP is ...

3

A set $X \subseteq \mathbb R$ is of ``strong measure zero'' if the following holds:
For every sequence $(\varepsilon_n:n\in \mathbb N)$ of positive numbers there is a sequence $(I_n:n\in \mathbb N)$ of intervals such that $I_n$ has length $\varepsilon_n$, and $X$ is covered by the union of these intervals.
The smz (strong measure zero) sets are a proper ...

2

Here's a somewhat trivial answer.
Note that $V_\alpha$, for an infinite $\alpha$, have a particularly nice set of properties which follow from the fact that $|V_\alpha\times V_\alpha|=|V_\alpha|$.
Now, easily, if $\sf AC$ fails, we can find arbitrarily high such cardinals. Simply take $|V_\alpha|+\aleph(V_\alpha)$, where $\aleph(x)$ is the Hartogs number of $...

2

It is shown in
Tachtsis, E.
On the existence of permutations of infinite sets without fixed points in set theory without choice.
Acta Math. Hungar. 157 (2019), no. 2, 281-300.
that ZF+(every infinite set supports a permutation with no fixed points)
does not imply AC. It is easy to see in ZF that a finite set supports a cyclic permutation, so that should be ...

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