# Tychonoff space with unique compactification and 3 disjoint non-compact closed subsets

Prolog : The only compactification of a non-compact normal space $S$ is the one-point (Alexandroff) compactification IFF whenever $A,B$ are disjoint closed subsets of $S$, at least one of $A,B$ is compact.(Example. $\omega_1$ with the $\epsilon$-order topology.) In Engelking,General Topology, it is shown that $[(\omega+1) \times (\omega_1+1)]\backslash \{(\omega,\omega_1)\}$ is not normal, which generalizes verbatim, replacing $\omega$ and $\omega_1$ with,respectively, infinite cardinals $a,b$ where $cf(b)>a.$ I showed that the only compactification of $T=[(\omega_1+1)\times (\omega_2)]\backslash \{(\omega_1,\omega_2)\}$ is the Alexandroff. Now if $A,B,C$ are pairwise-disjoint closed subsets of $T$ then at least one of them is compact, although $\{\omega_1\}\times \omega_2$ and $\omega_1 \times \{\omega_2\}$ are disjoint,closed, and not compact. QUESTION: Is there a non-compact Tychonoff space $X$ whose only compactification is the Alexandroff, such that $X$ has $3$ or more pairwise disjoint closed non-compact subsets?

Let $Y=(\omega+1)\times(\omega_1+1)\times(\omega_2+1)$, and let $X=Y\setminus\{\langle\omega,\omega_1,\omega_2\rangle\}$. The sets
\begin{align*} &\{\omega\}\times\{\omega_1\}\times\omega_2\;,\\ &\{\omega\}\times\omega_1\times\{\omega_2\}\;,\text{ and}\\ &\omega\times\{\omega_1\}\times\{\omega_2\} \end{align*}
are pairwise disjoint, closed, non-compact subsets of $X$. The same kind of argument that shows that the one-point compactification of the Tikhonov plank is its Čech-Stone compactification works for $X$ as well.
• @user254665: Good question. The answer is no: see Exercise $\mathbf{3.12.16}$ in Engelking. – Brian M. Scott Feb 3 '16 at 19:27