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Combinatorial properties of infinite sets. This is a corner-point of set theory and combinatorics.
3
votes
Accepted
Continuous function covers in connected $T_2$-spaces
The answer is strong YES for connected spaces admitting a non-constant continuous function and NO in the opposite case.
If $X$ is a connected topological space that admits a non-constant continuous …
8
votes
2
answers
479
views
Relations between two tower numbers
A tower is a subset $T\subset [\omega]^\omega$ of the family $[\omega]^\omega$ of all infinite subsets of $\omega$ such that $T$ is well-ordered by the relation $\supset^*$ of almost inclusion and has …
5
votes
1
answer
241
views
On filters possessing a countable network
Let $\mathcal F$ be a free filter on $\omega$ and $$\mathcal F^+:=\{E\subset \omega:\forall F\in\mathcal F\;E\cap F\ne\emptyset\}.$$
A family $\mathcal N$ of subsets of $\omega$ is called a network fo …
3
votes
0
answers
183
views
On Khelif's example of a group of countable cofinality having the Bergman property
A group $G$ is defined to have the Bergman property if for any subset $X=X^{-1}$ generating $G$ there exists $n$ such that $X^n=G$.
By a result of Bergman, the permutation group of any set has the B …
4
votes
1
answer
178
views
What is the smallest cardinality of a maximal ultrafamily of infinite subsets of $\omega$?
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 Kuratows …
5
votes
Accepted
Minimal covers in hypergraphs with finite edges
Let $V:=\omega\times\omega$ and $E=\{E_{n,m}:n,m\in\omega\}$ where $$E_{n,m}:=(\{0,\dots,n\}\times\{m\})\cup\{(2n,m+1)\}.$$ It seems that the hypergraph $(V,E)$ has no minimal cover.
A simplification …
0
votes
Accepted
On filters possessing a countable network
Maybe I posed this question too quickly: for the 6 hours that passed since the time of asking this question I have found a (relatively simple) counterexample to my Problem 2.
Example. There exists a …
1
vote
Accepted
What is the smallest cardinality of a maximal ultrafamily of infinite subsets of $\omega$?
To my surprise, I found that this my ``new'' cardinal $\mathfrak{uf}$ is equal to $\mathfrak c$.
Theorem. $\mathfrak{uf}=\mathfrak{c}$.
Proof. Fix any maximal ultrafamily $\mathcal U\subseteq[\omeg …
2
votes
Accepted
Is every Cartesian biaffine plane affine?
The answer to this question is "No". A non-affine biaffine Cartesian plane can be constructed as follows.
First, we fix a suitable terminology. Every function $F:X\to Y$ is identified with its graph …
2
votes
1
answer
101
views
Is every Cartesian biaffine plane affine?
This question concerns the (synthetic) geometry of linear spaces.
Definition 1. A linear space is a pair $(P,\mathcal L)$ consisting of a set $P$ whose elements are called points and a family $\mathca …
6
votes
1
answer
246
views
A combinatorial property of uncountable groups
Let $A,B$ be two uncountable sets in a group $G$ such that for any elements $x,y\in G$ the intersection $xA\cap yB$ is finite. Let $\Phi:G\to 2^G$ be a function assigning to each element $x\in G$ some …
8
votes
1
answer
358
views
A combinatorial property of uncountable groups, II
Problem 1. Is it true that each uncountable group $G$ contains two subsets $A,B\subset G$ such that
1) for any $x,y\in G$ the intersection $xA\cap yB$ is finite and
2) for any function $ …
5
votes
1
answer
154
views
Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgr...
Let $\mathrm{cov}_H(C_2^\omega)$ be the smallest cardinality of a cover of the Boolean group $C_2^\omega=(\mathbb Z/2\mathbb Z)^\omega$ by closed subgroups of infinite index. It can be shown that
$$\m …
1
vote
Accepted
Optimal tiling for a collection of partitions
1) This question (as posed) has a simple negative answer. Just take any countable set $x$ and put $M$ be the family of finite subsets of $x$. Then for the family $\mathcal A$ of all possible partition …
2
votes
Accepted
A combinatorial property of uncountable groups
Unfortunately (for my further plans) this question has negative answer. Just take any two disjoint uncountable sets $A,B$ and consider the free group $G$ over the union $A\cup B$. Let $\Phi:G\to 2^G$ …