If $X$ is compact, is $[X]^2$ compact, too? Let $(X,\tau)$ be a Hausdorff space. Let $[X]^2 = \big\{\{x,y\}: x,y\in X \land x\neq y\big\}$. For $U,V\in \tau$ with $U\cap V = \emptyset$ we set $[U,V] = \big\{\{x,y\} \in [X]^2: x\in U\land y\in V\big\}$.
We endow $[X]^2$ with the topology $[\tau]^2$, which is generated by $\{[U,V]: U,V\in \tau\land U\cap V =\emptyset\}$.
If $X$ is compact, is $[X]^2$ compact, too?
 A: Not necessarily, because in some cases of $X$ Hausdorff, compact, infinite, $[X]^2$ is isomorphic to an open subset of $X\times X$ that is not closed (therefore not compact).
Here's an example. Let $I$ be the real interval $[0,1]$. Then it is not hard to prove that $[I]^2 \cong \{(x,y) \in I^2: x < y\}$. The latter set is open in $[0,1]^2$, therefore not closed (since $[0,1]^2$ is connected). So it is not compact, and therefore $I^2$ is not compact either.
A: If $[X]^2$ is compact then $X$ is finite.
Let $\Delta \subseteq X^2$ be the diagonal. It is easy to show that the map $X^2 \setminus \Delta \to [X]^2$ given by $(x,y) \mapsto \{x,y\}$ is continuous, closed and has compact fibers (since it is two-to-one). So if $[X]^2$ is compact then $X^2 \setminus \Delta$ is also compact and hence closed in $X^2$. Thus $\Delta$ is open in $X^2$ and therefore $X$ is discrete. So $[X]^2$ is discrete and hence finite.
A: Consider $X$ the disjoint union of an arbitrary compact Hausdorff space $Y$ and a single point. Then for any $U, V \subseteq X$, we can add that point to both $U$ and $V$, so $(U \cup p) \times (V \cup p)$ is open. Because $[Y]^2$ with the usual product topology is usually not compact (as Dominic points out), we may find a set of pairs $U$, $V$ such that $U \times V$ covers $[Y]^2$ but no finite subset does. Then because the $U$ and $V$ individually cover $Y$, $(U \cup p) \times (V \cup p)$  cover $[X]^2$, but no finite subset does.
