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How to prove that $\{x \in X: A \cap B_x = \emptyset \}$ is open, where $A$ is a closed set and $B_x$ varies continuously with $x$?

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    You're right Scott. I'll try to look if there is some definition of what varies continuously means.2012-08-18

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Note: I modified the answer to make $X$ compact.

Let $X = [0,2]$, $B_x = (0,x)$ and $A = \{1\}$. Then $\{x \in X: A \cap B_x = \emptyset \} = [0, 1]$.

$A$ is closed, and $B_x$ varies continuously with $x$ by many definitions, but $[0, 1]$ is not open in the usual topology.

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    Michael Shub, Global Stability of Dynamical Systems. My problem is in the proof of proposition 10.14.2012-08-20
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With some additional structure, in a suitable sense, for example in topological groups or topological vector spaces (maybe uniform spaces more generally) compact families of opens have open intersection, and compact families of closeds have closed union. I had an old essay about this, now at http://www.math.umn.edu/~garrett/m/fun/cpt_families.pdf