Consider the Alexandroff duplicate $X\times_{ad} 2$, the space $X\times 2$ where the points of the form $(x,1)$ are isolated and for each open set $U$ in $X$, $(U\times\{0,1\})\setminus (x,1)$ is open. If $X$ is compact then $X\times_{ad} 2$ is compact? How can I see it?
Thank you.
By the Alexander Subbasis Theorem it suffices to take an open cover $\mathcal{U}$ of $X \times_{\text{ad}} 2$ by subbasic open sets (of the kind you've described above).
The subspace $X \times \{ 0 \}$ of the Alexandroff duplicate is homeomorphic to $X$, so it is compact, meaning that we need only finitely many sets $U_1 , \ldots, U_n$ in $\mathcal{U}$ to cover it. Since each $U_i$ is of the form $( U_i^\prime \times \{ 0 \} ) \cup ( ( U_i^\prime \setminus \{ x \} ) \times \{ 1 \} )$ for some open $U_i^\prime \subseteq X$ and $x \in X$ it follows that $U_1 \cup \cdots \cup U_n$ misses only finitely many points of $X \times_{\text{ad}} 2$, so just pick finitely many more sets from $\mathcal{U}$ to cover these.