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?



If $X$ is compact, is $[X]^2$ compact, too?
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.


