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Topology boundary problem

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Let X be a space. A[tex]\subseteq[/tex]X and U, V, W [tex]\in[/tex] topolgy(X). If W[tex]\subseteq[/tex] U[tex]\cup[/tex] V and U[tex]\cap[/tex] V[tex]\neq[/tex] emptyset,

    Prove bd(W) = bd(W[tex]\cap[/tex]U) [tex]\cup[/tex] bd (W[tex]\cap[/tex] V)

    2. Relevant equations
    bd(W) is the boundary of W...
    I think I have the "[tex]\supseteq[/tex]" part, but I am having trouble with the "[tex]\subseteq[/tex]" part.


    3. The attempt at a solution
    [tex]\supseteq[/tex]: Assume x [tex]\in[/tex] bd(W[tex]\cap[/tex]U) [tex]\cup[/tex] bd(W [tex]\cap[/tex]V). Show x[tex]\in[/tex] bd(W). Then x [tex]\in[/tex] bd(W[tex]\cap[/tex] U) or x[tex]\in[/tex] bd(W[tex]\cap[/tex] V). If x[tex]\in[/tex] bd (W[tex]\cap[/tex]U) means x is in bd(W) since W[tex]\cap[/tex]U[tex]\neq[/tex] emptyset and since W[tex]\subseteq[/tex] U[tex]\cup[/tex] V, some part of W[tex]\subseteq[/tex]U.
    If x[tex]\in[/tex] bd(W[tex]\cap[/tex] V) then x[tex]\in[/tex] bd(W) since W[tex]\cap[/tex] V[tex]\neq[/tex] emptyset and since W[tex]\subseteq[/tex]U[tex]\cap[/tex] V, some part of W[tex]\subseteq[/tex] V.

    Does this look ok for this part of the proof?

    [tex]\subseteq[/tex]: Assume x[tex]\in[/tex]bd(W). Show x[tex]\in[/tex] bd(W[tex]\cap[/tex] U)[tex]\cup[/tex] (bd(W[tex]\cap[/tex] V). Since U and V are disjoint and W[tex]\subseteq[/tex]U[tex]\cup[/tex], W_U[tex]\subseteq[/tex]U or W_V[tex]\subseteq[/tex]V. Suppose W_U[tex]\subseteq[/tex]U then x[tex]\in[/tex]bd(W_U)[tex]\subseteq[/tex]bd(W_U[tex]\cap[/tex]U).Suppose W_V[tex]\subseteq[/tex] V then x[tex]\in[/tex] bd(W_V)[tex]\subseteq[/tex]bd(W_V[tex]\cap[/tex] V). Since x[tex]\in[/tex]bd(W)[tex]\subseteq[/tex]bd(W_U[tex]\cap[/tex]U) or bd(W_V[tex]\cap[/tex] V) then x[tex]\in[/tex] bd(W[tex]\cap[/tex] U)[tex]\cup[/tex] (bd(W[tex]\cap[/tex] V).

    I am not sure if I need to specify what W_U and W_V are? or if this even works for this second part of the proof?
     
    Last edited: Oct 26, 2008
  2. jcsd
  3. Oct 27, 2008 #2

    morphism

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    Why is this the case?

    OK, I think I see why you're having a problem:
    [itex]U \cap V \neq \emptyset[/itex] says that U and V are not disjoint.
     
  4. Oct 27, 2008 #3
    Thanks morphism. I tried to fix it. Am I on the right track?


    [tex]\subseteq[/tex]: Assume x[tex]\in[/tex] bd(W). Show x[tex]\in[/tex] bd(W[tex]\cap[/tex] U)[tex]\cup[/tex] (bd(W[tex]\cap[/tex] V). Since U and V are not disjoint and W[tex]\subseteq[/tex]U[tex]\cup[/tex], W_U[tex]\subseteq[/tex]U or W_V[tex]\subseteq[/tex]V where W_U is the set of W in U and W_V is the set of W in V. Suppose W_U[tex]\subseteq[/tex]U then x[tex]\in[/tex]bd(W_U)[tex]\subseteq[/tex]bd(W_U[tex]\cap[/tex]U).Suppose W_V[tex]\subseteq[/tex] V then x[tex]\in[/tex] bd(W_V)[tex]\subseteq[/tex]bd(W_V[tex]\cap[/tex] V). Since x[tex]\in[/tex]bd(W)[tex]\subseteq[/tex]bd(W_U[tex]\cap[/tex]U) or bd(W_V[tex]\cap[/tex] V) then x[tex]\in[/tex] bd(W[tex]\cap[/tex] U)[tex]\cup[/tex] (bd(W[tex]\cap[/tex] V).
     
    Last edited: Oct 27, 2008
  5. Oct 27, 2008 #4

    morphism

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    I'm not sure I quite understand what W_U and W_V are supposed to denote.
     
  6. Oct 27, 2008 #5
    Ignore my last proof...I have been working on another approach to proving
    [tex]\subseteq[/tex].

    I am struggling but I think I am beginning to understand. Will you take a look?

    Assume x[text]\in[/tex] bd(W). Show
    x [tex]\in[/tex] bd(W[tex]\cap[/tex]U) [tex]\cup[/tex]
    bd(W[tex]\cap[/tex]V). So x[tex]\in[/tex]cl(W)-int(W) and we know that W[tex]\subseteq[/tex]U[tex]\cap[/tex]V[tex]\neq[/tex] emptyset. Then x[tex]\in[/tex](bd(W)[tex]\cap[/tex]U) [tex]\cup[/tex](bd(W)[tex]\cap[/tex]V). Since bd(W)[tex]\subseteq[/tex] cl(W), x[tex]\in[/tex](cl(W)[tex]\cap[/tex]U) [tex]\cup[/tex](cl(W)[tex]\cap[/tex]V).

    So x[tex]\in[/tex]cl(W[tex]\cap[/tex]U) [tex]\cup[/tex]cl(W[tex]\cap[/tex]V).
    Since x[tex]\in[/tex]bd(W), x[tex]\in[/tex]bd(W[tex]\cap[/tex]U) [tex]\cup[/tex]bd(W[tex]\cap[/tex]V).
     
    Last edited: Oct 27, 2008
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