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first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.
5
votes
1
answer
150
views
Intersection cardinalities in MAD families
Let $\newcommand{\o}{[\omega]^\omega}\o$ denote the collection of infinite subsets of the set of nonnegative integers $\omega$. We say ${\cal A}\subseteq \o$ is almost disjoint if $A\cap B$ is finite …
4
votes
1
answer
151
views
Minimal dominating sets in thin hypergraphs
Let $H=(V,E)$ be a hypergraph. We say that $H$ is thin if for every $v\in V$ the set $E_v=\{e\in E:v\in e\}$ is finite.
A subset $D\subseteq V$ is dominating if
$\bigcup \{e\in E:e\cap D \neq \emptyse …
20
votes
1
answer
534
views
Almost orthogonal maps $f:\omega \to \{-1,1\}$
Let $\omega$ denote the set of non-negative integers. For sets $A,B$, let $B^A$ denote the set of maps $f:A\to B$. For $f,g\in\{-1,1\}^\omega$ we say that $f,g$ are almost orthogonal if there is $C_0\ …
3
votes
1
answer
196
views
Mutually equal Hamming distance of members of ${\cal P}(\mathbb{N})$
This is inspired by an older, as of yet unanswered question.
If $X$ is a set and $A,B\subseteq X$, we let the Hamming distance of $A, B$ be defined as $d_H:=\text{card}\big((A\setminus B)\cup (B\setmi …
5
votes
1
answer
166
views
Cardinality of separating families on an infinite cardinal $\kappa$
Let $\kappa$ be an infinite cardinal. We say ${\cal S}\subseteq {\cal P}(\kappa)$ is separating if whenever $a\neq b\in \kappa$ there is $T\in {\cal S}$ such that $|T\cap\{a,b\}| = 1$. Let $\newcomman …
5
votes
1
answer
212
views
Image-catching families in $\omega$
Let $[\omega]^\omega$ be the collection of infinite subsets of the set of nonnegative integers $\omega$, and let $\newcommand{\I}{\cal{I}}\I=$ $\{S\in[\omega]^\omega: (\omega\setminus S)\in[\omega]^\o …
2
votes
1
answer
167
views
Smallest ${\mathbf B}$-function $f:\omega\to( \omega\setminus\{0\})$
Motivation. Every hypergraph $(\omega, E)$ where $E$ is countable and consists of infinite sets has property $\newcommand{\B}{\mathbf{B}}\B$. On the other hand, if the members of $E$ are allowed to be …
3
votes
1
answer
143
views
Large almost disjoint family on $\mathbb{N}$ with property $\mathbf{B}$
Let
$\newcommand{\oo}{[\omega]^\omega}\oo$ denote the collection of all infinite subsets of the set of nonnegative integers $\omega$. We say that $\newcommand{\ss}{{\cal S}}\S\subseteq \oo$ haspropert …
2
votes
0
answers
46
views
Chromatic number of the dual hypergraph [duplicate]
Let $H = (V,E)$ be a hypergraph. For $v\in V$ we set $E_v = \{e\in E: v\in E\}$. The dual of $H$ is defined by $H^* =(E, V^*)$ is, where $V^* = \{E_v:v\in V\}$.
If $\kappa>0$ is a cardinal, a map $c:V …
10
votes
2
answers
234
views
Additive, multiplicative, and Dedekind infiniteness in ${\sf (ZF)}$
We call a set $X$
Dedekind-infinite if there is an injective map $f:X\to X$ that is not surjective,
addititvely infinite if $X \neq\emptyset$ and there is an injective map $f:\big((X\times\{1\})\cup( …
5
votes
1
answer
166
views
The equivalence of Dedekind-infinite and dually Dedekind-infinite as a weak form of (AC)
A set $X$ is Dedekind-infinite if there is an injective map $f: X\to X$ that is not surjective.
A set $X$ is dually Dedekind-infinite if there is a surjective map $f: X\to X$ that is not injective.
In …
12
votes
1
answer
678
views
Graphs $G$ with $G \cong \text{Aut}(G)$
Let $G=(V,E)$ be a simple, undirected graph. By $\newcommand{\Aut}{\text{Aut}}\Aut(G)$ we denote the collection of graph isomorphisms $\varphi:G\to G$. We let $$E(\Aut(G)) =\big\{\{\varphi, \psi\}:\va …
4
votes
Accepted
Proof of the axiom of choice for finite sets in ZF
It holds vacuously for $A = \emptyset$: then $\bigcup A = \emptyset$ and the empty function $\emptyset: \emptyset \to \emptyset$ trivially fulfills $f(a) \in A$ for all $a\in A$ -- because it is impos …
10
votes
1
answer
262
views
Does every linear cover contain a minimal cover?
This is a follow-up question to an older question.
Let $X\neq \emptyset$ be a set. We say that ${\cal C}\subseteq {\cal P}(X)$ is a cover if $\bigcup {\cal C} = X$, and we call ${\cal C}$ linear if $| …
5
votes
1
answer
158
views
(Weakly) minimal subcovers of linear covers
Motivation. The starting point of this question is the trivial observation that if we cover $\mathbb{N}$ with $$\big\{\{0,\ldots n\}: n\in \mathbb{N}\big\},$$ then this cover doesn't have a minimal su …