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Show set is open - technical difficulties

  1. Feb 15, 2008 #1

    quasar987

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    [SOLVED] Show set is open - technical difficulties

    1. The problem statement, all variables and given/known data
    Let C be a closed convex set in a Banach space with non empty interior. Show that cl(int(C))=C (show that the closure of the interior of C is S itself).

    3. The attempt at a solution

    Since int(C) is included in C, we have cl(int(C)) [itex]\subset[/itex] cl(C)=C.


    For the other inclusion, let x be in C. I'm going to try to find a sequence in int(C) that converges to x. This will prove that C [itex]\subset[/itex] cl(int(C)).

    Let y be in int(C) and B(y;r) be a ball centered on y and entirely contained in int(C). Since C is convex, the line d joining y and x is entirely contained in C. If I can show that d\{x} is entirely contained in int(C), then I will have won.

    My idea is that not only is d contained in C, but so is the line from any point in B(y;r) to x! The set of all these lines make up a set the form of an ice cream cone where x is the tip and the ball B(y;r) is the ice cream ball.

    So, I can either show that this ice cream cone minus {x} is open, or find explicitly for each point of d\{x} an open ball entirely contained in the ice cream cone.

    In either case, how do we do that?!?
     
  2. jcsd
  3. Feb 15, 2008 #2

    StatusX

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    You can use a little of both. For each point in d/{x}, use the ice cream cone to show it has a ball around it in C, so is in int(C).
     
  4. Feb 16, 2008 #3

    quasar987

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    I really don't see how! What would be the first step?
     
  5. Feb 17, 2008 #4

    Dick

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    How about considering s sequence of points along the axis of the cone, the line connecting y and x?
     
  6. Feb 17, 2008 #5

    StatusX

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    d/{x} is the set of points tx+(1-t)y with 0<=t<1. Label these points p(t). Let r be the radius of a ball B centered at y and contained in C. Then there is a ball B(t) of radius r(t) around p(t) that's contained in C. Because for any point z in B(t), consider the line passing through x and z. One convenient way to parameterize this is as:

    [tex] q(s) = p(s) + \frac{1-s}{1-t} (z-p(t)) [/tex]

    Then:

    [tex] ||y-q(0)|| = ||p(0)-q(0)|| = ||\frac{1}{1-t} (z-p(t))|| < \frac{r(t)}{1-t} [/tex]

    So we can pick r(t) so that q(0) lies in B, and so z must be in C.
     
  7. Feb 17, 2008 #6

    quasar987

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    This is the idea, but we need to show first that the axis of the cone d\{x} is indeed in the interior of C.


    I understand what you did but not quite how you did it. Specifically, how did you create that line q(s)? What were your guiding principles for coming up with this formula?
     
  8. Feb 17, 2008 #7

    StatusX

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    q(s) is just the line connecting x and z. Since z is close to p(t), this line should be "close" to p(s) (the line connecting x and p(t)), with the distance scaling with 1-s as you move farther away from x. The formula was written that way with this in mind so it'd be easy to find the actual point on q(s) that's close to p(0)=y.
     
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