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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.
60
votes
7
answers
9k
views
In what respect are univalent foundations "better" than set theory?
It was an ambitious project of Vladimir Voevodsky's to provide new foundations for mathematics with univalent foundations (UF) to eventually replace set theory (ST).
Part of what makes ST so appealing …
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 …
7
votes
1
answer
450
views
Is it consistent that $|[\kappa]^{<\kappa}| > \kappa$?
Let $\kappa>\aleph_0$ be a cardinal, and let $[\kappa]^{<\kappa}$ denote the collection of subsets of $\kappa$ having cardinality strictly less than $\kappa$. Is it consistent that $$|[\kappa]^{<\kapp …
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 …
23
votes
2
answers
3k
views
Does the "three-set-lemma" imply the Axiom of Choice?
Consider the following curious statement:
$(S)$ $\;$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 …
2
votes
1
answer
171
views
Chromatic number and taking duals of hypergraphs
If $H=(V,E)$ is a hypergraph and $\kappa\neq \emptyset$ is a cardinal, then a map $c:V\to \kappa$ is said to be a colouring if for every edge $e\in E$ with $|e|\geq 2$ the restriction $c\restriction_e …
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 …