The Suslinian number and other cardinal invariants of continua

TL;DR

This paper introduces the Suslinian number as a new invariant of continua, relating it to classical properties like weight and dimension, and explores its implications under different set-theoretic assumptions.

Contribution

It defines the Suslinian number for continua, establishes bounds on weight and structure, and connects these to the Suslin Hypothesis and set-theoretic principles.

Findings

Each compact space has weight at most the successor of its Suslinian number.

Under the Suslin Hypothesis, all Suslinian continua are metrizable.

Existence of non-metrizable Suslinian continua is equivalent to the negation of the Suslin Hypothesis.

Abstract

By the {\em Suslinian number} $\Sln(X)$ of a continuum $X$ we understand the smallest cardinal number $\kappa$ such that $X$ contains no disjoint family $\C$ of non-degenerate subcontinua of size $|\C|\ge\kappa$. For a compact space $X$, $\Sln(X)$ is the smallest Suslinian number of a continuum which contains a homeomorphic copy of $X$. Our principal result asserts that each compact space $X$ has weight $\le\Sln(X)^+$ and is the limit of an inverse well-ordered spectrum of length $\le \Sln(X)^+$, consisting of compacta with weight $\le\Sln(X)$ and monotone bonding maps. Moreover, $w(X)\le\Sln(X)$ if no $\Sln(X)^+$-Suslin tree exists. This implies that under the Suslin Hypothesis all Suslinian continua are metrizable, which answers a question of \cite{DNTTT1}. On the other hand, the negation of the Suslin Hypothesis is equivalent to the existence of a hereditarily separable non-metrizable Suslinian continuum. If $X$ is a continuum with $\Sln(X)<2^{\aleph_0}$, then $X$ is 1-dimensional, has rim-weight $\le\Sln(X)$ and weight $w(X)\ge\Sln(X)$. Our main tool is the inequality $w(X)\le\Sln(X)\cdot w(f(X))$ holding for any light map $f:X\to Y$.