Covering versus partitioning with the Cantor space

TL;DR

This paper investigates conditions under which topological spaces can be partitioned into or covered by copies of the Cantor space, establishing cardinality bounds and constructing counterexamples.

Contribution

It proves when covering with copies of the Cantor space implies partitioning, providing sharp bounds and characterizing spaces with no isolated points.

Findings

Partitioning and covering are equivalent under certain cardinality conditions.

Constructed spaces show the bounds are sharp and the conditions are necessary.

A completely metrizable space can be partitioned into Cantor spaces iff it has no isolated points.

Abstract

What topological spaces can be partitioned into copies of the Cantor space $2^\omega$? An obvious necessary condition is that a space can be partitioned into copies of $2^\omega$ only if it can be covered with copies of $2^\omega$. We prove three theorems concerning when this necessary condition is also sufficient. If $X$ is a metrizable space and $|X| \leq \mathfrak{c}^{+\omega}$ (the least limit cardinal $>\!\mathfrak{c}$), then $X$ can be partitioned into copies of $2^\omega$ if and only if $X$ can be covered with copies of $2^\omega$. To show this cardinality bound is sharp, we construct a metrizable space of size $\mathfrak{c}^{+(\omega+1)}$ that can be covered with copies of $2^\omega$, but not partitioned into copies of $2^\omega$. Similarly, if $X$ is first countable and $|X| \leq \mathfrak{c}$, then $X$ can be partitioned into copies of $2^\omega$ if and only if $X$ can be covered with copies of $2^\omega$. On the other hand, there is a first countable space of size $\mathfrak{c}^+$ that can be covered with copies of $2^\omega$, but not partitioned into copies of $2^\omega$. Finally, we show that a completely metrizable space can be partitioned into copies of $2^\omega$ if and only if it can be covered with copies of $2^\omega$ if and only if it has no isolated points.