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?



A question about realcompact spaces
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.


