$\{{(x,K) : x \in K}\}$ is closed in $X \times \{\text{closed subsets of }X\}$ Let $X$ be a locally compact (and Hausdorff, if needed) space. Let's denote the closed subsets of $X$ by $C(X)$, equipped with the Hausdorff topology.
I want to prove that $F=\{{(x,K) : x \in K , K \in C(X)}\}$ is a closed subset of $X \times C(X)$.
Usually I attack these kind of problems by drawing but somehow this time $X$ and $C(X)$ are too related. It is quite difficult to see how F looks like.
I really appreciate your help, thanks a lot.
 A: Let's assume you mean the $H$-topology as defined by Fell here, usually now called the Fell-topology on $C(X)$. With this I can show $F$ to be closed quite easily:
Recall that a base for this topology on $C(X)$ is provided by all sets of the form $$\langle C; \{U_1,\ldots,U_n\}\rangle = \{A \in C(X): A \cap C = \emptyset, \forall i \in \{1,\ldots n\} U_i \cap A \neq \emptyset\}$$
where $C \subset X$ is a compact subset, and all $U_i$ are (non-empty) open subsets of $X$. The family of open sets is also allowed to be empty. So a basic open set consists of all closed sets that miss a given compact set and hit all open sets from a finite family.
Now let $(x,K) \in X \times C(X)$ such that $(x,K) \notin F$, as defined above. This means that $x \notin K$. By local compactness of $X$ and closedness of $K$, we find a compact $C$ such that $x \in \operatorname{Int}(C)$ and $C \cap K = \emptyset$. 
Then $O = \operatorname{Int}(C) \times \langle C; \emptyset \rangle$ is an open set of $X \times C(K)$ such that for all $(p,A)$ in it we have that $p \in \operatorname{Int}(C)$ and $A \cap C = \emptyset$, so certainly $p \notin A$, so $(x,K) \in O \subset (X \times C(X)) \setminus F$. This shows that $F$ is closed.  
