Covering versus partitioning with the Cantor space

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 $|X|\le {\mathfrak{c}}^{+\omega }$ (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 $|X|\le \mathfrak{c}$, 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.