# Stone-Cech compactification of completely regular noncompact space is not metrizable.

I have problem with the statement in Munkres, topology (section28, 9)

If $$X$$ is completely regular and noncompact space Then, $$\beta (X)$$ (Stone-Cech compactification of X) is not metrizable.

I solved this statement with condition normality of X. But I am still failed wih condition completely regularity. Could you give me some help??

• FYR – Arctic Char Oct 22 '19 at 3:23
• The assumption $\beta(X)$ metrisable implies $X$ is normal, so we can assume $X$ is normal "for free". – Henno Brandsma Oct 22 '19 at 4:19
• Ah I didn't catch it. Thanks a lot!! – HooMun Oct 22 '19 at 7:44

Suppose $$\beta(X)$$ is metrisable. The assumption that $$X$$ is completely regular is necessary for $$\beta(X)$$ to exist. And $$X$$ is non-compact implies that there must be some $$p \in \beta(X)\setminus X$$.

The supposed metrisability of $$\beta(X)$$ implies two things: there is a sequence $$(x_n)_n$$ from $$X$$ such that $$x_n \to p$$ (as $$X$$ is dense in $$\beta(X)$$) and $$X$$ is normal (as $$X$$ is normal from being metrisable too, as a subspace of $$\beta(X)$$.

It follows from the convergence of the sequence (and the fact that we're in a Hausdorff space) that $$A=\{x_2n: n \in \Bbb N\}$$ and $$B=\{x_{2n+1}: n \in \Bbb N\}$$ are closed disjoint subsets of $$X$$, so by normality (and Urysohn) we have a continuous $$f: X \to [0,1]$$ with $$f[A]=\{0\}$$ and $$f[B]=\{1\}$$. By the fact that we're in $$\beta(X)$$ we can extend $$f$$ to a continuous $$\beta(f): \beta(X) \to [0,1]$$, but then we have a contradiction as

$$\beta(f)(p) \in \beta(f)[\overline{A}] \subseteq \overline{\beta(f)[A]} = \overline{f[A]}=\{0\}$$

but also

$$\beta(f)(p) \in \beta(f)[\overline{B}] \subseteq \overline{\beta(f)[B]} = \overline{f[B]}=\{1\}$$

So $$\beta(X)$$ is not metrisable.

If $$X$$ is not normal, then $$X$$ itself is not metrizable (all metrizable spaces are normal), so neither is $$\beta X$$, since it contains $$X$$ as a subspace.