Talk:Type Space is Compact

Is there a reference on here for the first line of the proof?:
 * It will suffice to show that every open cover of $S_{n}^{\mathcal{M}}(A)$ by the basic open sets $[\phi]$ of the topology  has a finite subcover.

Qedetc 00:06, 7 June 2011 (CDT)


 * Well yes, the definition of compact will do, surely? (puzzled by question) --prime mover 00:39, 7 June 2011 (CDT)


 * ... actually no, you're just showing that a subcover of a basis is finite, from what I can tell. Not sure if this is the same thing, it might just amount to being locally compact. --prime mover 00:41, 7 June 2011 (CDT)

The general theorem I'm hoping to appeal to would be something like
 * If $B$ is a basis for the topology on $(X,\vartheta)$, then if every open cover of $X$ by open sets from $B$ has a finite subcover, $(X,\vartheta)$ is compact.

The proof should be something like noting that if $U$ is a (usual) open cover of $X$, then (by the way a basis generates a topology), each open set in $U$ can be written as a union of sets in $B$. So $U$ yields a cover using elements from $B$. Then you use the hypothesis to get finitely many sets $B_1, \dots, B_n$ which cover $X$, and you select finitely many sets $U_1, \dots, U_n$ from $U$ such that $B_i \subseteq U_i$.

Qedetc 01:20, 7 June 2011 (CDT)