Let $X$ be completely regular space, $\beta X$ be Stone-Čech compactification of $X$, and $\upsilon X$ be Hewitt realcompactification of $X$. Then $X\subset \upsilon X\subset \beta X$.
If the remainder $\beta X\setminus X$ is countably compact space, then can $% \beta X\setminus \upsilon X$ be countably compact?
Under what conditions is it countably compact?
Does anyone have an article or book recommendation on this?
There exists a Tychonoff space $X$ such that $\beta X\setminus X$ is countably compact but $\beta X\setminus \upsilon X$ is locally compact, $\sigma$-compact and not countably compact. Such a space can be constructed as follows.
In the Stone-$\check{\mathrm C}$ech remainder $\omega^*=\beta\omega\setminus \omega$ choose a sequence $(U_n)_{n\in\omega}$ of pairwise disjoint nonempty open sets. In each $U_n$ choose a point $x_n\in U_n$ and using the Tychonoff property of $\omega^*$, find a continuous function $f_n:\omega^*\to [0,1]$ such that $f_n(x_n)=1$ and $f_n[\omega^*\setminus U_n]=\{0\}$. Then $K_n=f_n^{-1}(1)$ is a nonempty compact $G_\delta$-set in $\omega^*$ and $(K_n)_{n\in\omega}$ is a discrete sequence of compact $G_\delta$ sets in $\omega^*$ whose union $K=\bigcup_{n\in\omega}K_n$ is a $\sigma$-compact $G_\delta$-set in $\omega^*$ and $K$ is open in its closure $\overline K$ in $\omega^*$. Consider the set $Y=\omega^*\setminus K$ and observe that $Y$ is of type $F_\sigma$ and $G_\delta$ in $\omega^*$.
It is well-known that the space $\omega^*$ has exactly $2^{\mathfrak c}$ closed sets and every infinite closed subset of $\omega^*$ has cardinality $2^{\mathfrak c}$. Using these two facts, it is easy to construct (by transfinite induction of length $2^{\mathfrak c}$) a Bernstein-like set $B\subseteq Y$ such that for every infinite closed subset $F\subseteq Y$ the sets $F\cap B$ and $F\setminus B$ are nonempty. Then the subspace $X=B\cup\omega$ of the space $\beta\omega$ has the desired property.
Indeed, its Stone-$\check{\mathrm C}$ech compactification coincides with $\beta \omega$ (because every bounded continuous function $f:X\to\mathbb R$ uniquely extends to a continuous function on $\beta\omega$). The space $\beta X\setminus X=\beta\omega\setminus X=K\cup(Y\setminus B)$ is countably compact because for every countable infinite discrete subset $C\subseteq K\cup(Y\setminus B)$ either for some $n\in\omega$ the intersection $C\cap K_n$ is infinite and hence has an accumulation point $x\in K_n\subset \beta X\setminus X$ or else for every $n\in\omega$ the intersection $K_n\cap C$ is finite and the set $\overline C\setminus C$ of accumulation points of $C$ is contained in $Y$. Since the closed set $\overline C$ is infinite, it has cardinality $2^{\mathfrak c}$ and hence $(\overline C\setminus C)\setminus B$ is non-empty, which means that the set $C$ has an accumulation point in $Y\setminus B\subseteq \beta X\subseteq X$ and $\beta X\setminus X$ is countably compact.
Next, we show that the Hewitt completion $\upsilon X$ coincides with $\beta\omega\setminus K$. Indeed, for every $x\in K$ we can find $n\in\omega$ such that $x\in K_n$ and hence $K_n$ is a $G_\delta$-subset of $\beta\omega$ that contains $x$ and is disjoint with $X$, witnessing that $x\notin\upsilon X$.
On the other hand, for every $x\in Y$, every $G_\delta$-set $G\subseteq\beta\omega$ that contains $x$ has nonempty intersection with the Bernstein-like set $B\subseteq X$ (because $G$ contains a non-empty and hence infinite closed $G_\delta$-subset, which intersects $B$ by the choice of $B$). This means that $Y\subseteq\upsilon X$ and hence $\upsilon X=\beta\omega\setminus K$.
It remains to observe that the complement $\beta X\setminus\upsilon X=K$ is locally compact, $\sigma$-compact and not countably compact, even not pseudocompact.
