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Combinatorial properties of infinite sets. This is a corner-point of set theory and combinatorics.
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 $ …
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 …
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 …
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 …
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
$$\m …
5
votes
Accepted
Minimal covers in hypergraphs with finite edges
Let $V:=\omega\times\omega$ and $E=\{E_{n,m}:n,m\in\omega\}$ where $$E_{n,m}:=(\{0,\dots,n\}\times\{m\})\cup\{(2n,m+1)\}.$$ It seems that the hypergraph $(V,E)$ has no minimal cover.
A simplification …
5
votes
1
answer
356
views
Calculate the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of th...
Let $(P,\le)$ be a poset. For a point $x\in P$ let
$${\downarrow}x=\{p\in P:p\le x\}\quad\text{and}\quad{\uparrow}x=\{p\in P:x\le p\}$$be the lower and upper sets of the point $x$, and for a subset $ …
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 …
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 …
3
votes
Accepted
A combinatorial property of uncountable groups, II
Problems 1 and 2 both have affirmative answers (implying that the finitary ballean of any uncountable group is normal).
Two cases are possible:
I. There exists a countable subgroup $A\subset G$ and …
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 …
3
votes
Accepted
Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgr...
Lyubomyr Zdomskyy proved that in the Laver model $\mathrm{cov}_H(2^\omega)=\omega_1<\mathfrak b=\mathfrak c$.
His argument used the following known Laver property of the Laver model $V'$: for every
fu …
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 $\Phi:G\to 2^G$ …
2
votes
Calculate the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of th...
At the moment we have the following information on the cofinalities of the poset $\mathfrak P$ (see Theorem 7.1 in this preprint).
Theorem.
1) ${\downarrow}\!{\uparrow}\!{\downarrow}(\mathfra …
2
votes
0
answers
81
views
A convex version of the small uncountable cardinal $\mathfrak b$
Let us recall that $\mathfrak b$ is the smallest cardinality of a subset of $\omega^\omega$, which cannot be covered by countably many compact subsets of $\omega^\omega$.
The definition of $\mathfrak …