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Question on Basic Topology, open sets

  1. Oct 5, 2006 #1
    In a euclidean space X with two subsets E and F, the subset E+F is defined as the collection of all x+y, where x E and y F. “+” denotes the addition in the euclidean space. Prove that if E and F are open, then E + F is open.

    I'd really appreciate your help. Thanks so much!
    Last edited: Oct 5, 2006
  2. jcsd
  3. Oct 5, 2006 #2
    Given that E and F are open for the remainder of this.

    E + F = {x | x is an element of E or x is an element of F}

    The definition the E + F given is also the definition of E union F.

    Now it needs to be proved that the union of open sets is open. Your book should have a theorem and a proof of this so I will not prove it here due to lack of ability to write in proper notation.
  4. Oct 5, 2006 #3
    That's what I thought at first, but E+F isn't quite the union of E and F. In other words, if E is something like {1,3,5} and F is {20,40,60}, E+F does not contain any point in E or F. So I'm not sure that's the right proof.
  5. Oct 5, 2006 #4
    So, are you saying given those sets you defined E + F = {21, 43, 65} ?
  6. Oct 5, 2006 #5
    well yes, but also 23, 25, 41, etc
  7. Oct 6, 2006 #6


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    Let p be in E+ F. Then p= x+ y for some x in E, y in F. Since E is open, there exist [itex]\delta_1[/itex] such that the [itex]\delta_1[/itex] neighborhood of x is as subset of E. Since F is open, there exist [itex]\delta_2[/itex] such that the [itex]\delta_2[/itex] neighborhood of y is a subset of F. Now let [itex]\delta= \delta_1+ \delta_2[/itex] and show that the [itex]\delta[/itex] neighborhood of p= x+y is in E+ F.
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