General Topology – Proving Stone–?ech Compactification is Not Metrizable

compactificationcompactnessgeneral-topologymetric-spaces

Show that the Stone–Čech compactification $\beta \mathbb{Z}$ is not metrizable (here $\mathbb{Z}$ denotes the set of integer numbers in discrete topology).

Definition. Let $X$ be a completely regular space. We say $\beta (X)$ is a *Stone–Čech compactification * of $X$ if it is a compactification of $X$ such that any continuous map $f:X\rightarrow C$ of $X$ into a compact Hausdorff space $C$ extends uniquely to a continuous map $g:\beta (X)\rightarrow C$.

We also have a theorem that states if $X$ is metrizable, then $X$ is first countable. So, if we show that $\beta (\mathbb{Z}) $ is not first countable, then we have our conclusion.

I've looked online for proofs of this fact and can't seem to find any.

Best Answer

In order to show that $\beta(\mathbb Z)$ is not first countable, it will suffice to show that $|\beta(\mathbb Z)|\gt2^{\aleph_0},$ since a Hausdorff space which is separable and first countable has cardinality at most $2^{\aleph_0}$ (each point is the limit of a convergent sequence of points in a countable dense set).

The space $C=\{0,1\}^\mathbb R$ is a separable compact Hausdorff space. Define a countable dense subset $S\subseteq C$ and a surjection $f:\mathbb Z\to S.$ Since $\mathbb Z$ is discrete, $f$ is continuous, and therefore extends to a continuous surjection $g:\beta(\mathbb Z)\to C,$ showing that $|\beta(\mathbb Z)|\ge|C|=2^{2^{\aleph_0}}\gt2^{\aleph_0}.$