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Simple Topology question

  1. Oct 17, 2005 #1
    I am not well 'accustomed' to these kind of proofs so please bear with my stupidity
    Suppose U and V are both open subsets in Rn. Prove that U intersection V and U union V are open as well.

    Dfeinition of open is that you cna center a ball about a point a in a set such that that ball is completely contained in the set.

    So let there be a ball with center a radius delta in U such that
    [tex] B(a,\delta_{1}) = { x: ||x-a||< \delta_{1}} [/tex] for U and
    [tex] B(b,\delta_{2}) = { y: ||y-b||< \delta_{2}} [/tex] for V

    now the for the union
    [tex] B(c, \delta) = {z: ||z-c|| < \delta} [/tex]
    and c belongs to the union of U and V. and picking delta = min (delta 1, delta 2) we can say that the ball is completely contained in U union V and the union is open?

    If this correct we can move to the intersection...
    do i do a similar procedure? Please help?

    Thank you for you help and advice!
  2. jcsd
  3. Oct 17, 2005 #2


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    If [tex] B(a,\delta_{1}) = { x: ||x-a||< \delta_{1}} [/tex] is in U and
    [tex] B(b,\delta_{2}) = { y: ||y-b||< \delta_{2}} [/tex] is in V, then
    [tex] B(c, \delta) = {z: ||z-c|| < \delta} [/tex]
    with [tex]\delta= min(\delta_1, \delta_2)[/tex]
    is in both U and V and therefore is in both U union V and U intersect V- you don't have to do it two different ways!
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