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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??

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

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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.

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