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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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Is $B_x$ only a subset of $X$ or an open set? – Sigur Aug 18 '12 at 17:59
Seems dubious, depending on what "$B_x$ varies continuously with $x$" means. Consider for instance $X=\mathbf{R}$, $A=(-\infty,0]$, $B=(x,\infty)$. – Sean Eberhard Aug 18 '12 at 18:03
@SeanEberhard, in this case, the set would be $(0,\infty)$. – Sigur Aug 18 '12 at 18:05
@Sigur It would be $[0,\infty)$. – Sean Eberhard Aug 18 '12 at 18:07
@SeanEberhard, sorry, you are right. $x$ does not need to be an element of $B_x$. So, my first question remains. – Sigur Aug 18 '12 at 18:09

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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I can't precise what "vary continuously" means.I'll try to look if there is some definition of this. thanks – cadreno Aug 18 '12 at 19:05
Here is one What book is your defintion in? – copper.hat Aug 18 '12 at 19:07
Michael Shub, Global Stability of Dynamical Systems. My problem is in the proof of proposition 10.14. – cadreno Aug 20 '12 at 12:11

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

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