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forcing, large cardinals, descriptive set theory, infinite combinatorics, cardinal characteristics, forcing axioms, ultrapowers, measures, reflection, pcf theory, models of set theory, axioms of set theory, independence, axiom of choice, continuum hypothesis, determinacy, Borel equivalence relations, Boolean-valued models, embeddings, orders, relations, transfinite recursion, set theory as a foundation of mathematics, the philosophy of set theory.

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Set Theory and V=L

From http://en.wikipedia.org/wiki/Analytical_hierarchy "If the axiom of constructibility holds then there is a subset of the product of the Baire space with itself which is $\Delta^1_2$ and is the ...
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Arhangel'skii's problem revisited

One of the most well-known problems in set-theoretic topology is Arhangel'skii's question of whether there exists a Lindelöf Hausdorff space with "points $G_\delta$" (meaning, every point is ...
Santi Spadaro's user avatar
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1 answer
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product of power sets

For a set $X$, let $\mathcal P(X)$ denote its power set and let $\mathcal P(X)\otimes\mathcal P(X)$ denote the product $\sigma$-algebra in $X^2$. When $|X|\leq\aleph_0$ then $\mathcal P(X)\otimes\...
Emmanuel Chetcuti's user avatar
10 votes
4 answers
3k views

Naturally occuring groups with cardinality greater than the reals.

In group theory, the single most important piece of information about a group is its cardinality, which is of course either finite, countably infinite, or uncountably infinite. Usually, however, ...
Daniel Miller's user avatar
10 votes
5 answers
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Grothendieck topology for a non-small category

To define a Grothendieck topology of a category, we usually require that the category is small. Question 1: Why do we need to require the category to be small? I thought that the problem was that ...
H. Shindoh's user avatar
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3 answers
841 views

Is every field extension of an ultrafield an ultrafield?

Let $K=\lim(K_{i})$ be an ultrafield (over a non-principal ultrafilter), and let $K\hookrightarrow K'$ be a field extension of $K$. When the field $K'$ is finite over $K$ it is also an ultrafield by ...
user12940's user avatar
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3 answers
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How does categoricity interact with the underlying set theory?

Here's the setup: you have a first-order theory T, in a countable language L for simplicity. Let k be a cardinal and suppose T is k-categorical. This means that, for any two models M,N |= T of ...
Pietro's user avatar
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3 answers
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Reference Request: Independence of the ultrafilter lemma from ZF

I'm looking for references for the following facts concerning the ultrafilter lemma (~ "there exist non-principal ultrafilters"): The ultrafilter lemma is independent of ZF. ZF + the ultrafilter ...
Greg Graviton's user avatar
10 votes
4 answers
978 views

On surjections, idempotence and axiom of choice

The following assertion is trivial in ZFC, or even in much weaker theories. Is it also true in ZF? (I couldn't find it in the Consequences site so far.) If $A$ is an infinite set such that $A$ can ...
Asaf Karagila's user avatar
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Absoluteness of "$\kappa$-homogeneously Suslin" for sets of reals

What is known about the absoluteness, or lack thereof, of the notion of "$\kappa$-homogeneously Suslin" for sets of reals? For example, if $A$ is $\kappa$-homogeneously Suslin and $\lambda > \...
Trevor Wilson's user avatar
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Minimal refinements of open covers of $T_2$-spaces

Let $(X,\tau)$ be a topological space. We say ${\cal U}\subseteq \tau$ is an open cover if $\bigcup {\cal U} = X$, and $X\notin {\cal U}$. ${\cal U}$ is minimal if for all $U_0\in {\cal U}$ we have ...
Dominic van der Zypen's user avatar
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2 answers
2k views

extensions of lebesgue measure

The Hahn-Banach theorem implies that Lebesgue measure can be extended give a "measure" on all subsets of [0,1], but this measure is only guaranteed to be finitely additive. It might magically turn ...
Monroe Eskew's user avatar
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7 answers
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Applications of forcing in model theory

What are the major applications of (set theoretic) forcing in model theory?
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Cofinal monotone maps from $\omega^\omega$ to $\kappa^\kappa$

Given a cardinal $\kappa$ consider the set $\kappa^\kappa$ of all functions from $\kappa$ to $\kappa$, endowed with the partial order $f\le g$ iff $f(\alpha)\le g(\alpha)$ for all $\alpha\in\kappa$. ...
Taras Banakh's user avatar
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Partitioning a set of cardinality $\kappa$ into more than $\kappa$ disjoint subsets

There is an old result due to Mycielski and Sierpiński, and popularized in a Monthly article by Taylor and Wagon (A Paradox Arising from the Elimination of a Paradox; see also this MO answer), that ...
Timothy Chow's user avatar
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How many closed measure zero sets are needed to cover the real line?

This question assumes familiarity with combinatorial cardinal characteristics of the continuum. Let $\mathcal{E}$ be the $\sigma$-ideal generated by closed measure zero subsets of the real line. It ...
Boaz Tsaban's user avatar
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Harrington's unpublished note "The constructible reals can be anything"

Around 1974, Leo Harringto wrote an unpublished note entitled "The constructible reals can be anything", in which he proved that it is consistent that being $\Delta^1_n$ is the same as being ...
Mohammad Golshani's user avatar
9 votes
1 answer
531 views

Existence of infinite groups that are too reluctant to be topological

With ZFC, is there an infinite group $G$ such that there is no non-trivial non-discrete topology on $G$ with the functions $G\times G\to G,~~ (a,b) \mapsto ab$ and $G\to G,~~ a\mapsto a^{-1}$ ...
Minimus Heximus's user avatar
9 votes
1 answer
341 views

Logics detecting their own equivalence notions, take two: $\mathcal{L}_{\omega_2,\omega}$

This question is a follow-up to another question of mine, with different language - see the link below. Say that an infinite regular cardinal $\kappa$ is Fraissean iff the logic $\mathcal{L}_{\kappa,\...
Noah Schweber's user avatar
9 votes
1 answer
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Axiom of class collection

One version of the Axiom of Collection says that any surjection $A\to B$ from a class $A$ to a set $B$ is factored through by some surjection $C\to B$ where $C$ is a set. Note that assuming $B$ is a ...
Mike Shulman's user avatar
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Is this cardinal characteristic trivial? (Number of strategies needed to guarantee at least one win)

(Previously asked at MSE.) Let the determinacy number, $\mathfrak{g}$ (for "game"), be the smallest cardinal such that for every (two-player, perfect-information, length-$\omega$) game on $\...
Noah Schweber's user avatar
9 votes
2 answers
892 views

Can we have an infinite sequence of decreasing cardinality all terms of which have equal sized power sets?

Is the following consistent with $\text{ZF}$? There exists a set $S=\{x_1,x_2,x_3,...\}$ such that: $|x_{i+1}| < |x_i|$ $\forall m,n \in S (|P(m)|=|P(n)|)$ Where cardinality $``||"$ is ...
Zuhair Al-Johar's user avatar
9 votes
1 answer
765 views

Why relative consistency results by forcing arguments are provable in finitistic metatheory

It is claimed in many textbooks that relative consistency results, such as $\text{Con}(\text{ZFC})\rightarrow\text{Con}(\text{ZFC}+2^{\aleph_0}\geq\aleph_2)$, are provable in the finitistic metatheory....
Ruizhi Yang's user avatar
9 votes
1 answer
226 views

Is $\beta\mathbb N$ a unique compactification with the smallest possible permutation group?

For a compactification $c\mathbb N$ of $\mathbb N$ let $\mathcal H(c\mathbb N,\mathbb N)$ be the group of homeomorphisms $h:c\mathbb N\to c\mathbb N$ such that $h(x)=x$ for all $x\in c\mathbb N\...
Taras Banakh's user avatar
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9 votes
1 answer
991 views

Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails?

It is a theorem of Solovay that any stationary subset of a regular cardinal, $\kappa$ can be decomposed into a disjoint union of $\kappa$ many disjoint stationary sets. As far as I know, the proof ...
Norman Lewis Perlmutter's user avatar
9 votes
1 answer
649 views

Can $\mathbb{R}$ be partitioned into dedekind-finite sets?

Assuming $ZF$ itself is consistent, it is consistent that there are sets $D$ which are infinite but cannot be placed in bijection with any of their proper subsets; such sets are called "strictly ...
Noah Schweber's user avatar
9 votes
2 answers
426 views

Can local $0^\#$ exists in L?

Assume $0^\#$ exists and there is an inaccessible cardinal. Are there two transitive sets $M,N$ s.t. $M\in N,M\vDash ZF+V=L[0^\#],N\vDash ZF+V=L$?
Reflecting_Ordinal's user avatar
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0 answers
345 views

The global dimension of fields

In the absence of the Axiom of Choice, it is not necessarily true that all vector spaces over a field have bases. What are the possible global dimensions of fields in a model of ZF in which AoC ...
Mariano Suárez-Álvarez's user avatar
9 votes
1 answer
1k views

Definable set in ZF that cannot be proved to be Borel

Is there a predicate $P(x)$ such that $\mathrm{ZF} \vdash \exists! x. P(x)$, and $\mathrm{ZF} \vdash \forall x. P(x) \to (x \subseteq \mathbb R)$, but $\mathrm{ZF} \nvdash \forall x. P(x) \to \mathsf{...
Trebor's user avatar
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9 votes
1 answer
1k views

Large cardinals and constructible universe

We know that if $V=L$ holds, then $|\cal{P}(\omega)|=|\cal{P}(\omega)\cap \textrm{L}|=\aleph_1$ whereas, in the presence of a measurable cardinal (in fact, even Ramsey) $|\cal{P}(\omega)\cap \textrm{L}...
kvagk's user avatar
  • 207
9 votes
4 answers
2k views

What kinds of operations are well-defined when working with sets, classes, conglomerates, and yet higher order collections?

There are many foundations to set theory, ZFC, NBG, SEAR, to name a few, and while they differ in how sets, classes, and higher-order collections are represented as mathematical objects, they all ...
Benjamin Braun's user avatar
9 votes
0 answers
256 views

A bi-modal logic related to determinacy

The short version of my question is as follows. There is a natural (I hope!) way to associate a bimodal theory to a game (two-player, perfect-information, length-$\omega$, on $\omega$); are there "...
Noah Schweber's user avatar
9 votes
1 answer
557 views

Is it possible to prove that any two points of a convex complete metric space are connected by some metric segment without the axiom of choice?

We say that a point $m$ is between points $p$ and $q$ of a metric space $(M, d)$ if $d(p, q) = d(p, m) + d(m, q)$ and $p ≠ m ≠ q$. A metric space $M$ is said to be metrically convex if given any two ...
Juan Atacama's user avatar
9 votes
1 answer
1k views

Can the continuum be a singular cardinal?

Martin's Axiom implies that $2^{\aleph_0}$ is a regular cardinal. But can $2^{\aleph_0}$ be a singular cardinal? By Konig's Lemma, it can never be $\aleph_{\omega}$ since cf($2^{\aleph_0}$)>$\...
Rachid Atmai's user avatar
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9 votes
1 answer
526 views

Pontryagin dual of the surreal numbers?

Has any work been done on the Pontryagin dual of the surreal numbers (suitably topologized)? I have not been able to find anything and am not sure if this is still unknown. Alternatively, has this ...
Mike Battaglia's user avatar
9 votes
1 answer
479 views

Is there any forcing free proof for hard independence results?

We are forced to use forcing for almost all "hard" independence results such as: $Con(ZFC)\longrightarrow Con (ZFC+\neg CH) $. The question simply is: Primary Question: Is there any "forcing free" ...
user avatar
9 votes
1 answer
495 views

How much choice does a linear or well-order on cardinals imply?

It is well-known that if the natural (partial) order on the class of cardinal numbers is a linear order, then it is in fact a well-order and the axiom of choice holds. I was, however, interested in ...
Wojowu's user avatar
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9 votes
4 answers
3k views

Incompleteness and nonstandard models of arithmetic

The following are a collection of doubts, some of which shall have concrete answers while others may have not. Any kind of help will be welcome. Reading Peter Smith's "Gödel Without (Too Many) Tears",...
Marc Alcobé García's user avatar
9 votes
1 answer
782 views

Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$?

Assume that $V\neq HOD$ and let $\kappa = \min \{\alpha\in On \mid \mathcal{P}(\alpha) \not\subseteq HOD\}$. Clearly, $\kappa$ is a cardinal. Question: Is it consistent that $\kappa = \aleph_\...
Yair Hayut's user avatar
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9 votes
1 answer
698 views

Source for NBG+Equipollence conservative over ZFC?

The "conservative" class theory, NBG, proves no new theorems about sets (with respect to ZFC). The choice function used here is set choice, and it's not too hard to prove (if M is a ctm for ZFC, then ...
Richard Rast's user avatar
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9 votes
1 answer
336 views

Cardinals without choice: interpolation (reference wanted)

Is there a published reference for this ZF theorem? Let $m,n\in\mathbb{N}$. If $a_1,\dots,a_m$ and $b_1,\dots,b_n$ are cardinals such that $a_i\le b_j$ for all $i$ and $j$, then there is a cardinal $...
user avatar
9 votes
0 answers
367 views

A $\mathsf{ZF}$ example of two Baire spaces whose product is not Baire?

Motivated by this question I'm looking for a pair of Baire topological spaces whose product is not Baire and whose construction does not need the axiom of choice. The example of such spaces I'm ...
Alessandro Codenotti's user avatar
9 votes
1 answer
552 views

"Towers" on singular cardinals with countable cofinality

Let $\lambda$ be a singular cardinal of countable cofinality. Is there necessarily a sequence $\{A_\alpha\mid\alpha<\lambda^+\}$ of countable subsets of $\lambda$, such that $\alpha<\beta$ if ...
Asaf Karagila's user avatar
  • 39.8k
9 votes
1 answer
489 views

Intuition behind Pincus' "injectively bounded statements"

In David Pincus, Zermelo-Fraenkel Consistency Results by Fraenkel-Mostowski Methods, The Journal of Symbolic Logic, Vol. 37, No. 4 (Dec., 1972), pp. 721-743 Pincus introduces the notion of ...
David Roberts's user avatar
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9 votes
0 answers
388 views

On the role of $\diamondsuit$

The well-known axiom $\diamondsuit$ states that there is a sequence $\langle A_\alpha:\alpha<\omega_1\rangle$ (a $\diamondsuit$-sequence) of countable sets with the property that for any $A\...
Todd Eisworth's user avatar
9 votes
0 answers
530 views

How many steps does it take to "Tarski-Vaughtify" second-order logic?

Given a regular logic $\mathcal{L}$, let $\preccurlyeq_\mathcal{L}$ be the usual elementary submodelhood relation for $\mathcal{L}$. There is also a separate submodelhood relation coming from the ...
Noah Schweber's user avatar
9 votes
2 answers
1k views

What sort of structure can amorphous sets support?

Assuming the Axiom of Choice, every cardinal is either finite (i.e., an element of $\omega$) or Dedekind-infinite (i.e., in bijection with a proper subset of itself). This dichotomy is not true in ZF, ...
Noah Schweber's user avatar
9 votes
1 answer
496 views

Can two versions of $\omega_1^{CK}(\mathsf{Ord})$ ever coincide?

The goal of this question is to fill in the gap in this old answer of mine. For a transitive set $M$, thought of as an $\{\in\}$-structure, we define the following ordinals (this is not the notation ...
Noah Schweber's user avatar
9 votes
0 answers
174 views

A model where the uniformity of the meager ideal is strictly below the almost disjointness number

I'm looking for a model satisfying the inequality described in the title. Recall that the uniformity of the meager ideal, denoted $\operatorname{non}(\mathcal M)$ (or $\operatorname{non}(\mathcal B)$) ...
Corey Bacal Switzer's user avatar
9 votes
1 answer
713 views

How much choice is necessary to prove this statement?

Consider the following statement (in $\mathsf{ZF}+\text{AC}_\omega (\mathbb{R})$): There exists $(\varphi_\alpha)_{\alpha\in\omega_1}$ with $\varphi_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ ...
Lorenzo's user avatar
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