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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 …

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 …

The answer to this question is negative. Given a linearly ordered family $\mathfrak C$ of finitary partitions of $\omega$, write $\mathfrak C=\bigcup_{n=1}^\infty\mathfrak C_n$ where $\mathfrak C_n=\{ …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 $ …

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 …

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 …

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 …

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 $ …