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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 $$ \sup_{x \in A}f(x)\leq \sup_{n \in \mathbb{N}}\sup_{x \in A^n}f(x) ? $$

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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.

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  • $\begingroup$ Oh my, I had placed the inequality in the wrong direction. I edited to fix it, sorry about that you're correct, it wasn't the wright question. $\endgroup$
    – ABIM
    Apr 29, 2019 at 16:46
  • $\begingroup$ Perfect, thanks for the edit and very complete answer :) $\endgroup$
    – ABIM
    Apr 29, 2019 at 17:45
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    $\begingroup$ @NikWeaver, since the first sentence only applies to the old version of the problem, maybe it could be deleted and "No" replaced by "No to the original version of the problem (with the opposite inequality)" or similar? $\endgroup$
    – LSpice
    Apr 29, 2019 at 19:45
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    $\begingroup$ @LSpice: sure, good idea. $\endgroup$
    – Nik Weaver
    Apr 29, 2019 at 20:48
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    $\begingroup$ This is a good exercise for you. $\endgroup$
    – Nik Weaver
    May 2, 2019 at 11:30

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