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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?2012-08-18
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    Seems dubious, depending on what "$B_x$ varies continuously with $x$" means. Consider for instance $X=\mathbf{R}$, $A=(-\infty,0]$, $B=(x,\infty)$.2012-08-18
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    @SeanEberhard, in this case, the set would be $(0,\infty)$.2012-08-18
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    @Sigur It would be $[0,\infty)$.2012-08-18
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    @SeanEberhard, sorry, you are right. $x$ does not need to be an element of $B_x$. So, my first question remains.2012-08-18
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    If for some reason you demand that $x$ be an element of $B_x$ you could instead take $B = (x-1,\infty)$ in my example above.2012-08-18
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    Indeed, you need to make precise what it means for $B_x$ to "vary continuously" with $x$.2012-08-18
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    Ditto from me on defining continuity. There are many notions of continuity for set values functions. The Hausdorff distance is one possibility.2012-08-18
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    I'm sorry. I'm reading a proof and I put here the information I thought relevant. In the book, X is a compact set (which makes your examples doesn't work) and I can't precise what "vary continuously" means.2012-08-18
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    @copper.hat’s example still works if you let $X=[0,1]$, say: $\{x\in[0,2]:\{1\}\cap(0,x)=\varnothing\}=[0,1]$. The result really will depend on exactly what’s meant by *varies continuously*.2012-08-18
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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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    I can't precise what "vary continuously" means.I'll try to look if there is some definition of this. thanks2012-08-18
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    Here is one http://en.wikipedia.org/wiki/Hausdorff_distance. What book is your defintion in?2012-08-18
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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