Two equivalent notions of compactness? Let $X$ be a topological space $Sh(X)$ sheaves of $X$. Then  it is stated that compactness of $X$ can be expressed  equivalently as


*

*Every cover of $1$ by subobjects has a finite subcover.


*The global sections functor $\Gamma:Sh(X) \rightarrow Set$ maps directed covers of $1$ to covers of the $1$-point set.

I see how the usual notion of $X$ as a compact space is equivalent to 1. But not 2. Elaboration would be appreciated.
 A: Any subobject of $1$ in $Sh(X)$ has the form $F_U(V)=\begin{cases}\{*\}&\text{ if }V\subseteq U \\ \emptyset & \text{ otherwise}\end{cases}$ for some open $U\subseteq X$.  This correspondence $U\mapsto F_U$ gives an isomorphism between the poset of open subsets of $X$ and the poset of subobjects of $1$ in $Sh(X)$.
Now $\Gamma$ acts on these subobjects $F_U$ by $\Gamma(F_U)=\emptyset$ unless $U=X$, in which case $\Gamma(F_U)=\{*\}$.  So to say $\Gamma$ maps some cover of $X$ to a cover of the $1$-point set just means that $F_X$ is one of the subobjects in the cover.  So condition (2) just says that any directed cover of $1$ must contain $F_X$ as one of its elements.  Or, translating along the correspondence $U\leftrightarrow F_U$, any directed open cover of $X$ must contain $X$ as one of its elements.
Now, why is this equivalent to compactness?  If $X$ is compact, every open cover has a finite subcover, and if the open cover was directed, you can take an upper bound for the finitely many elements of the subcover to conclude that $X$ itself is in the cover.  Conversely, if $X$ is not compact, it has some open cover $\mathcal{U}$ with no finite subcover.  Let $\mathcal{V}$ be the set of unions of finitely many elements of $\mathcal{U}$.  Then $\mathcal{V}$ is a directed open cover of $X$, and by hypothesis it does not contain $X$.
