Showing that no point of $X^*$ is a point at which $\beta X$ is first countable.

Question

I am having a tough time with this problem. It is from Munkree's Topology book. I am unsure where my proof is heading. Can someone please help me prove the problem? Thank you for your time and help.

Recall that for a completely regular topological space $X, \beta X$ denotes the Stone-Cech compactification of $X.$ Assume that the space X is normal, and show that no point of $X^*$ is a point at which $\beta X$ is first countable.

$\def\R{{\mathbb R}} \def\N{{\mathbb N}} $ $\textbf{Solution:}$ Let $\beta X$ be first countable at $\beta\in X^*$. Then, there exists a countable collection $\{U_n\}_n$ of open sets in $\beta X$ which is a local base at $\beta$. Because $\beta X$ is $T_1$, $$\bigcap_{n=1}^\infty U_n =\{\beta\}.$$ Therefore, $\beta$ is a $G_\delta$-point. Then, there exists $f\in C^*(X)$ such that the zero set of $f^\beta$, $Z(f^\beta) = \{\beta\}$ and $f^\beta \colon \beta X \to \R$ such that $f^\beta \vert_X=f.$ Thus, $Z(f^\beta) \cap X =\emptyset$ and therefore, $Z(f) =\emptyset.$ Then $f$ takes values arbitrarily near $0$ implying $\frac{1}{f}$ is unbounded so $X$ is not pseudocompact. So, there exists a $C$-embedded copy $\N$ in $X$ such that $\displaystyle{\lim_{n\to\infty}\frac{1}{f^{(n)}} = \infty}$ and $\N$ must be closed in $X$. However, $\N$ is not closed in $\beta X$. Therefore, $$\emptyset \ne Cl_{\beta X} (\N)\setminus \N \subseteq BX\setminus X.$$ Moreover, $\beta\N\cong Cl_{\beta X}(\N)$ so $\beta \N \setminus \N \subseteq \beta X\setminus X.$ Again, $\displaystyle{\lim_{n\to\infty} f(n) = 0}$ implies $f^\beta(Cl_{\beta X} \N\setminus \N) = 0$ then $Cl_{\beta X}(\N) \setminus \N \subseteq Z(f^\beta)$ as $$\left\vert Cl_{\beta X} (\N) \setminus \N\right\vert = Z^{-C}.$$

Answer 1

If $\beta X$ is first countable at $\beta$, there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ converging to $\beta$ in $\beta X$. The sets $H=\{x_{2n}:n\in\Bbb N\}$ and $K=\{x_{2n+1}:n\in\Bbb N\}$ are disjoint closed subsets of $X$, and $X$ is normal, so there is a continuous $f:X\to[0,1]$ such that $f[H]=\{0\}$ and $f[K]=\{1\}$. What does this say about $f^\beta(\beta)$?