Let $(X,d)$ be a metric space. Suppose that $\{A^n\}\_{n \in \mathbb{N}}$ is a sequence of closed, non-empty subsets of $X$. 


Is there a Hausdorff topology on the space of closed subsets of $X$, guaranteeing that if $A^n$ converges in this space to a $A\subseteq X$, then for any continuous function $f:X \rightarrow \mathbb{R}$, we have that
$$
\sup\_{x \in A}f(x)\leq \sup\_{n \in \mathbb{N}}\sup\_{x \in A^n}f(x) ?
$$



Topology on closed subsets characterized by sup on continuous functions?
The answer to the original version of the problem, with the opposite inequality, is clearly "no": if $X$ contains more than one point then there is no such topology. Let $A^1 = \{x\}$ and $A^n = \{y\}$ for $n \geq 2$, where $x,y \in X$ are distinct. Define $f(z) = d(y,z)$. Since the sequence $(A^n)$ is eventually constant, it must converge to $\{y\}$ for any topology on the space of closed subsets. But the supremum over $\{y\}$ is $0$ and the supremum over $\{x\}$ is $d(x,y) > 0$.


For the revised question, with reversed inequality, the obvious answer is Hausdorff distance. If $A^n \to A$ in Hausdorff distance then for every $x \in A$ and every $k > 0$ there exists $y\_k \in A^n$, for some $n$, satisfying $d(x,y\_k) \leq \frac{1}{k}$. Thus $y\_k \to x$ and so $f(x)$ is $\leq$ the sup of sups.


