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 $x$?
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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general-topology
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0Is $B_x$ only a subset of $X$ or an open set? – 2012-08-18
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0Seems 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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0@SeanEberhard, in this case, the set would be $(0,\infty)$. – 2012-08-18
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0@Sigur It would be $[0,\infty)$. – 2012-08-18
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0@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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0If 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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0Indeed, you need to make precise what it means for $B_x$ to "vary continuously" with $x$. – 2012-08-18
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0Ditto 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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0I'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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0@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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0You're right Scott. I'll try to look if there is some definition of what varies continuously means. – 2012-08-18