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Questions tagged [independence-results]

This tag is for questions about proving that some statement is independent from a theory, meaning it is neither provable nor refutable from that theory. Common examples are the continuum hypothesis from the axioms of ZFC, and the axiom of choice from the axioms of ZF.

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Richard Laver finishes his seminal paper "On Fraïssé's order type conjecture", with: Finally, the question arises as to how the order types outside of $M$ behave under embeddability. For ...
Agelos's user avatar
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References: Applications of limited information strategies in Menger’s game by Clontz Almost compatible functions and infinite length games by Clontz and Dow Def. 3.7 of [1] $\mathcal{A}(\kappa)$ ...
Jakobian's user avatar
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I originally asked this question on Math StackExchange here, but I have copied it here as I now feel it is more appropriate for this site. There is an explicitly known 549-state Turing machine where, ...
C7X's user avatar
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I was playing with ideas around Gödel’s first incompleteness theorem which, roughly speaking, says that for every ($\omega$-)consistent, recursively axiomatizable formal system $F$ that is ...
Pooya Farshim's user avatar
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The following definition is by Sinclair, G.E. A finitely additive generalization of the Fichtenholz–Lichtenstein theorem. Transactions of the American Mathematical Society. 1974;193:359-74. A function ...
Arkadi Predtetchinski's user avatar
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For a compact Hausdorff space $X$, let $EX$ be the Stonean space corresponding to the Boolean algebra of regular open sets of $X$. Explicitly, if $\text{RO}(X)$ denotes regular open sets of $X$, let $...
Jakobian's user avatar
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In the article A Perfectly Normal, Locally Compact, Noncollectionwise Normal Space Form $\lozenge^\ast$ by Daniels and Gruenhage (I presume "form" is a typo and it should be "from",...
Jakobian's user avatar
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It is known that the Continuum Hypothesis is independent of ZFC. The formulation of the Collatz conjecture looks somehow more simple than that of the Continuum Hypothesis. Is it possible that the ...
Riemann's user avatar
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Although mathematicians usually do not work in constructive mathematics per se, their results often are constructively valid (even if the original proof isn't). An obvious counter-example is the law ...
Christopher King's user avatar
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6 answers
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Let $a, b, c \in \mathbb R$ such that $a \le b \le c$. Let $S$ be some set and $f : [a, b] \cup [b, c] \to S$ be a function. When can we find a function $g : [a, c] \to S$ that meets the following ...
Christopher King's user avatar
5 votes
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250 views

If $\kappa$ is a regular uncountable cardinal, we call a set $S\subseteq\kappa$ fat if for every $\alpha<\kappa$ and every club $C\subseteq\kappa$, there is a closed subset of $S\cap C$ of ...
Hannes Jakob's user avatar
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Say $\kappa$ is small if any set of cardinality $\kappa$ has outer-Lebesgue measure zero. We know that, in the Cohen model of ZFC where CH is false, there is a Borel partition of the unit interval of ...
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By $\textbf{PA}$ I will mean the usual first-order Peano Arithmetic. I will denote an element of $\mathbb{N}$ by $n$, and by $[n]$ I will denote the corresponding term in the language of $\textbf{PA}$:...
jg1896's user avatar
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I am considering the construction in [Peng—Shen—Wu] in which the authors show the consistency of a set $X$ such that there is a surjection from $X^2$ onto the power set of $X$ (henceforth $\mathscr{P}(...
Calliope Ryan-Smith's user avatar
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509 views

This question is motivated by my preceding MO-question on (in)consistency of NBG theory of classes. Let $\varphi(x,Y,C)$ be a formula of NBG with free parameters $x,Y,C$ and all quantifiers running ...
Taras Banakh's user avatar
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Thinking on the theory NBG (of von Neumann–Bernays–Gödel) I arrived at the conclusion that it is contradictory using an argument resembling Russell's Paradox. I am sure that I made a mistake in my ...
Taras Banakh's user avatar
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6 votes
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Let us consider the following stronger version of the Axiom Schema of Replacement (let us call it the Axiom Schema of Replacement for Definable Relations): Let $\varphi$ be any formula in the language ...
Taras Banakh's user avatar
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13 votes
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Let $\mathsf{AC}_\mathsf{WO}$: Every well-orderable family of non-empty sets has a choice function. $\mathsf{AC}^\mathsf{WO}$: Every family of non-empty well-orderable sets has a choice function. My ...
Lorenzo's user avatar
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The well-known Axiom Schema of Replacement in ZFC says that for any formula $\varphi$ of the Set Theory with free variables among $w_1,\dots,w_n,A,x,y$ the following holds: $$\forall w_1,\dots,w_n\;\...
Taras Banakh's user avatar
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11 votes
3 answers
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Following Kunen's book, it makes clear that countable transitive models (ctm) exist only for a finite list of axioms of ZFC. So, why can we assume a ctm of the whole ZFC axioms exists and use it as ...
Guest's user avatar
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A topological space $X$ is a $Q$-space if every subset of $X$ is of type $G_\delta$. The smallest cardinality of a metrizable separable space which is not a $Q$-space is denoted by $\mathfrak q_0$ and ...
Taras Banakh's user avatar
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Definition. A topological space $X$ is a $Q$-space if every subset of $X$ is of type $G_\delta$. It is clear that every $Q$-space has countable pseudocharacter (= all singletons are $G_\delta$) and is ...
Taras Banakh's user avatar
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5 votes
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I am interested in a "dynamical" modification of the cardinals $\mathfrak r$ and $\mathfrak r_\sigma$, well-known in the theory of cardinal characteristics of the continuum. For a compact ...
Taras Banakh's user avatar
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7 votes
3 answers
519 views

In $\mathbf{ZF}$, it is possible for a set $A$ to be infinite but not to admit a countable set. In other words, for any $\alpha\in\omega$, there is an injection from $\alpha$ into $A$, but there is no ...
Hannes Jakob's user avatar
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Hi I am new to proofs of consistency and independence with ZFC of some claims. I have read "The uses of set theory" by Judith Roitman, in that article it is mentioned that the Whitehead ...
Gabriel Medina's user avatar
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305 views

Call an antichain (set of pairwise incomparable elements) $A$ of a poset $P$ strong if for every $p,q \in P$ with $p \leq q$ there exists an $a\in A$ which is comparable with both $p$ and $q$. ...
Attila Joó's user avatar
4 votes
2 answers
878 views

My vague intuition is that not only it is common for a simple arithmetic proposition $p$ to be independent of ZFC, but it is common for the statement "$p$ is independent of ZFC" to be ...
Geoffrey Irving's user avatar
54 votes
1 answer
3k views

My question is which player has a winning strategy in the two-player version of the Killing the Hydra game? In their amazing paper, Kirby, Laurie; Paris, Jeff, Accessible independence results for ...
Joel David Hamkins's user avatar
3 votes
1 answer
1k views

The classical Cantor-Schroder-Bernstein Theorem says that there exists a bijective function $X\leftrightarrow Y$ if and only if there exist injective functions $X\hookrightarrow Y$ and $Y\...
Taras Banakh's user avatar
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9 votes
1 answer
593 views

Order invariant graphs and finite incompleteness by Harvey Friedman gives an example of a combinatorial/non-metamathematical $\Pi_1$ sentence that is independent of ZFC. Is there a simpler example of ...
user76284's user avatar
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5 votes
1 answer
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Is Axiom of Choice equivalent to the following statement? Axiom of Ordinal Choice: For any ordinal $\lambda$ and any indexed family of sets $(X_\alpha)_{\alpha\in\lambda}$ there exists a function $f:...
Taras Banakh's user avatar
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5 votes
0 answers
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A family $\mathcal A$ of infinite subsets of $\omega$ is called almost disjoint if for any distinct sets $A,B\in\mathcal A$ the intersection $A\cap B$ is finite. Let $\mathfrak a'$ be the largest ...
Taras Banakh's user avatar
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2 votes
1 answer
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Let $(A_\alpha)_{\alpha\in\mathfrak c}$ be an almost disjoint family of infinite subsets of $\omega$. The almost disjointness of the family means that $A_\alpha\cap A_\beta$ is finite for any ordinals ...
Taras Banakh's user avatar
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4 votes
1 answer
300 views

A family $\mathcal U$ of infinite subsets of $\omega$ is called an ultrafamily if for any sets $U,V\in\mathcal U$ one of the sets $U\setminus V$, $U\cap V$ or $V\setminus U$ is finite. By the ...
Taras Banakh's user avatar
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5 votes
1 answer
625 views

By a dynamical system I understand a pair $(K,G)$ consisting a compact Hausdorff space and a subgroup $G$ of the homeomorphism group of $K$. We say that a dynamical system $(K,G)$ $\bullet$ is ...
Taras Banakh's user avatar
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11 votes
1 answer
791 views

A function $f:\omega\to\omega$ is called $\bullet$ 2-to-1 if $|f^{-1}(y)|\le 2$ for any $y\in\omega$; $\bullet$ almost injective if the set $\{y\in \omega:|f^{-1}(y)|>1\}$ is finite. Let us ...
Taras Banakh's user avatar
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3 votes
0 answers
155 views

Looking at this MO-problem, my collegue Igor Protasov suggested to ask on Mathoverflow his old question on $T$-ultrafilters hoping that somebody on MO can solve it. First I recall the necessary ...
Taras Banakh's user avatar
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11 votes
1 answer
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In this post I will discuss some cardinal characteristic of the continuum, related to partitions of $\omega$ and would like to know if it is equal to some known cardinal characteristic. By a partition ...
Taras Banakh's user avatar
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3 votes
1 answer
313 views

Given two set $A,B$ we write $A\subset^* B$ if the complement $A\setminus B$ is infinite. A Hausdorff gap is a transfinite family $\langle A_\alpha,B_\alpha\rangle_{\alpha\in\omega_1}$ of infinite ...
Taras Banakh's user avatar
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6 votes
0 answers
429 views

This question concerns proper forcings of size $\aleph_1$. In the context of $\rm ZFC+\neg CH$, I couldn't find any counter example to the following property. Suppose $\mathbb P$ is a proper forcing ...
Rahman. M's user avatar
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6 votes
1 answer
750 views

This question concerns the possibility of the bi-interpretation synonymy of the structure $\langle H_{\omega_1},\in\rangle$, consisting of the hereditarily countable sets, and the structure $\langle ...
Joel David Hamkins's user avatar
5 votes
1 answer
614 views

The question is pretty much the title. I'm wondering if anything is known about the smallest size $\kappa$ of a non-measurable subset of the real numbers (regarding the Lebesgue measure). Since we ...
Hannes Jakob's user avatar
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1 vote
1 answer
188 views

If consistent, is existence of a proper class of extendible cardinals provably equivalent to a $Σ^V_5$ statement? Recall that in ZFC, a cardinal $κ$ is extendible iff for every $λ>κ$ there is an ...
Dmytro Taranovsky's user avatar
5 votes
1 answer
255 views

As in the title, I'm looking for examples of $\Sigma^1_4$ (preferably complete) sentences which are independent from ZFC in both ways, namely given a model $V$ we can extend it to $V'$ where such a ...
Jarek Swaczyna's user avatar
4 votes
1 answer
446 views

Let $\mathcal P$ be a family of nonempty subsets of a topological space $X$. A subset $D\subset X$ is called $\mathcal P$-generic if for any $P\in\mathcal P$ the intersection $P\cap D$ is not empty. A ...
Taras Banakh's user avatar
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9 votes
1 answer
547 views

In regular ZF, AC, WO, and Zorn's Lemma are equivalent, but every proof I know (of the implication AC -> WO and AC -> Zorn) uses the axiom of choice on the powerset of X (where X is the Set which is ...
Hannes Jakob's user avatar
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4 votes
1 answer
267 views

A regular topological space $X$ is called $\bullet$ analytic if $X$ is a continuous image of a Polish space; $\bullet$ $K$-analytic if $X$ is the image of a Polish space $P$ under an upper ...
Taras Banakh's user avatar
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2 votes
0 answers
143 views

A subset $A$ of the real line is called a Q-set if any subset of of $A$ is of type $F_\sigma$ in $A$. Let $\mathfrak q_0$ be the smallest cardinality of a subset $X\subset\mathbb R$ which is not a Q-...
Taras Banakh's user avatar
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7 votes
0 answers
199 views

Definition 1. A ultrafilter $\mathcal U$ on $\omega$ is called discrete (resp. nowhere dense) if for any injective map $f:\omega\to \mathbb R$ there is a set $U\in\mathcal U$ whose image $f(U)$ is a ...
Taras Banakh's user avatar
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8 votes
0 answers
266 views

On Thuesday I was in Kyiv and discussed with Igor Protasov the system of MathOverflow and its power in answering mathematical problems. After this discussion Igor Protasov suggested to ask on MO the ...
Taras Banakh's user avatar
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