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.


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 $|\beta(\mathbb Z)|\ge|C|=2^{2^{\aleph_0}}\gt2^{\aleph_0}.$

  • $\begingroup$ Is your notation for $X$ acting as the $C$ in the definition of Stone-Cech compactification? $\endgroup$ – Michael Aug 15 '16 at 3:02
  • $\begingroup$ So, you've shown the space is not separable. Therefore, it is not first countable? Is that a theorem we have? $\endgroup$ – Michael Aug 15 '16 at 3:07
  • $\begingroup$ Okay, I follow. Yes, that works. $\endgroup$ – Michael Aug 15 '16 at 3:20
  • $\begingroup$ Answer edited to point out that the same argument shows that $\beta(\mathbb Z)$ is not first countable. $\endgroup$ – bof Aug 15 '16 at 6:17

Suppose $\beta \mathbb Z$ is metrizable. Then, since $\mathbb Z$ is not compact, there is an $b\in \beta \mathbb Z\setminus \mathbb Z$ and since $\mathbb Z$ is dense in $\beta \mathbb Z$, there is a sequence of integers $X=\left \{ x_i \right \}_{i\in \mathbb N}$ that converges to $b$.

Now take any two disjoint subsequences $X_1$ and $X_2$ of $X$ and note that they are closed in $\mathbb Z$. As $\mathbb Z$ inherits the metric on $\beta \mathbb Z$, it is a normal space. Therefore, there is a continuous $f:\mathbb Z\to [0,1]$ s.t.$f(X_1)=0$ and $f(X_2)=1$, which extends continuously to a $g:\beta \mathbb Z\to [0,1]$.

But $X_1$ and $X_2$ both converge to $b$ and as $g$ is continuous we must have $g(b)=g(\lim X_1)=\lim g(X_1)=\lim f(X_1)=0$ and also $g(b)=g(\lim X_2)=\lim f(X_2)=1$, which is a contradiction.

  • $\begingroup$ We can just use $\{0,1\}$ instead of $[0,1]$ and not speak about normality/Urysohn's lemma: any function $\mathbb{Z}→\{0,1\}$ is continuous since $\mathbb{Z}$ is taken as a discrete space. $\endgroup$ – Idéophage Jun 20 '20 at 22:56

A compact metric space is separable.

Take an uncountable family of subsets of $\mathbb Z$, any two of which have finite intersection. Then their closures intersected with the "remainder" $\beta \mathbb Z \setminus \mathbb Z$ give you an uncountable family of pairwise disjoint open sets in the compact space $\beta \mathbb Z \setminus \mathbb Z$. So $\beta \mathbb Z \setminus \mathbb Z$ is not metrizable, and therefore of course neither is $\beta \mathbb Z$.


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