Non Pseudocompact Spaces and C-embedded sets How can I prove that if a Tychonoff space is not Pseudocompact then it contains a countable infinite c-embedded subset?
What we can say for the space which is not countably compact. Of course it will contain an infinite closed and discrete subset. Can we have something more than that. 
 A: Since $X$ is not pseudocompact, there is a continuous function $f:X\to\Bbb R$ whose range is unbounded. Thus, we can choose a set $A=\{x_n:n\in\Bbb N\}\subseteq X$ such that $f(x_{n+1})>f(x_n)+1$ for each $n\in\Bbb N$. For $n\in\Bbb N$ let $$V_n=f^{-1}\left[\left(f(x_n)-\frac12,f(x_n)+\frac12\right)\right]\;;$$ then the family $\mathscr{V}=\{V_n:n\in\Bbb N\}$ is discrete. To see this, let $x\in X$ be arbitrary. Then $f(x)$ has an open nbhd $U$ in $\Bbb R$ that meets at most one of the open intervals $\left(x_n-\frac12,x_n+\frac12\right)$, and $f^{-1}[U]$ is an open nbhd of $x$ in $X$ that meets at most one member of $\mathscr{V}$. $X$ is Tikhonov, so for each $n\in\Bbb N$ there is a continuous function $g_n:X\to[0,1]$ such that $g_n(x_n)=1$, and $g_n(x)=0$ for all $x\in X\setminus V_n$.
Now let $h:A\to\Bbb R$ be arbitrary. Let $\hat h=\sum_{n\in\Bbb N}h(x_n)g_n:X\to\Bbb R$; clearly $\hat h\upharpoonright A=h$, and $\hat h$ is continuous because $\mathscr{V}$ is discrete. Thus, $A$ is $C$-embedded in $X$.
