2
$\begingroup$

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 $x$?

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

2 Answers 2