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Multivariable proofs

  1. Sep 2, 2007 #1
    I have no luck with proofs...

    Prove that B[tex]_{r}[/tex] ((x[tex]_{0}[/tex], y[tex]_{0}[/tex])) = {(x,y) : || (x,y) - (x[tex]_{0}[/tex], y[tex]_{0}[/tex])|| < r} is an open set in R.

    Now I know that to be an open set if and only if each of its points is an interior point and if it contains no boundary points. I would consider trying to prove it for any (a,b) [tex]\in[/tex] B[tex]_{r}[/tex]

    Any ideas?
     
  2. jcsd
  3. Sep 3, 2007 #2

    HallsofIvy

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    You don't need to prove both of those. If all points are interior points, then none of them can be boundary points. Standard proofs that a given set are open show that every point is an interior point.

    You are starting correctly. let (a,b) be some point in Br. Then its distance from (x0,y0) is strictly less than r. You need to show that there exist some neighborhood of (a,b) consisting entirely of points that are in Br: that is, that their distance from (x0,y0) is also strictly less than r.

    It might be a good idea to draw a picture: Br is, of course, the disk inside the circle around (x0,y0) of radius r. Hint: the triangle inequality is very helpful here!

    By the way, I know a professor who says that it was being able to do precisely this proof as an undergraduate that convinced him he could be a mathematician!
     
    Last edited: Sep 3, 2007
  4. Sep 3, 2007 #3
    Let [tex]x\in B(a,r)[/tex]. Then let [tex]\epsilon = r - |x-a|[/tex]. Show that [tex]B(x,\epsilon) \subset B(a,r)[/tex]. To show this you need to show given any [tex]y\in B(x,\epsilon)\implies y\in B(a,r)[/tex].
     
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