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Boundary union and intersection problems

  1. Sep 18, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Let A, C [itex]\subseteq[/itex] ℝn with boundaries B(A) and B(C) respectively. Prove or disprove :

    B(AUC) O B(A)UB(C)

    and

    B(A[itex]\cap[/itex]C) O B(A)[itex]\cap[/itex]B(C)

    Where O represents each of these symbols : [itex]\subseteq, \supseteq, =[/itex]

    2. Relevant equations
    I know that double inclusion is going to cut the work required by 33% :)?


    3. The attempt at a solution

    I guess I'll try to start with the case B(AUC) [itex]\subseteq[/itex] B(A)UB(C) ( Since intuitively I know the boundary simply can't get bigger when I union two sets, so I have a feeling that testing for [itex]\supseteq[/itex] is going to flop ).

    So suppose [itex]x \in B(A \cup C)[/itex] then we know x is a boundary point of AUC, that is : [itex]\forall δ>0, \exists P \in (A \cup C) \wedge Q \in (ℝ^n - A \cup C) | P, Q \in N_δ(x)[/itex]

    Now how to proceed from here I'm not sure, any pointers would be great!
     
  2. jcsd
  3. Sep 18, 2012 #2

    jbunniii

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    OK, you've established that every neighborhood [itex]N_\delta(x)[/itex] contains points of [itex]A \cup C[/itex].

    Is it possible for both of the following statements to be false?

    * Every neighborhood [itex]N_\delta(x)[/itex] contains points of [itex]A[/itex]
    * Every neighborhood [itex]N_\delta(x)[/itex] contains points of [itex]C[/itex]
     
  4. Sep 18, 2012 #3

    Zondrina

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    Case #1 : B(AUC) [itex]\subseteq[/itex] B(A)UB(C). So we know that x is a boundary point of AUC, that is [itex]\forall δ>0, \exists P_1 \in (A \cup C) \wedge P_2 \in (ℝ^n - A \cup C) | P_1, P_2 \in N_δ(x).[/itex]

    So we want to show either x is in B(A) or x is in B(C). To show this, consider [itex]\forall δ'>0, \exists Q_1 \in A \wedge Q_2 \in (ℝ^n - A) | Q_1, Q_2 \in N_δ(x).[/itex].

    Let Q1 = P1, if Q1 [itex]\in[/itex] AUC then Q1 [itex]\in[/itex]A.

    Let Q2 = P2, if Q2[itex]\in[/itex] (ℝn-A) ( By De Morgans laws of course ) then Q2[itex]\in[/itex] (ℝn-A)

    This is finally sufficient to show that x[itex]\in[/itex]B(A) or x[itex]\in[/itex]B(C) through similar argument.
     
    Last edited: Sep 18, 2012
  5. Sep 18, 2012 #4

    Zondrina

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    Is this good so far or?
     
  6. Sep 18, 2012 #5

    jbunniii

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    It's not necessarily true that every neighborhood of [itex]x[/itex] contains a point of A.

    Here is an example:

    [itex]A = [0,1][/itex]
    [itex]C = [2,3][/itex]
    [itex]A\cup C = [0,1]\cup [2,3][/itex]
    Let [itex]x = 3[/itex]. Then [itex]x \in B(A \cup C)[/itex], but [itex]N_1(x)[/itex] contains no point of A.

    However, consider my earlier hint. I claim that at least one of these must be true:

    * Every neighborhood [itex]N_\delta(x)[/itex] contains points of [itex]A[/itex]
    or
    * Every neighborhood [itex]N_\delta(x)[/itex] contains points of [itex]B[/itex]

    I recommend that you start by proving this claim. Then you can proceed knowing that one of the sets, A or B, has the desired property.
     
  7. Sep 18, 2012 #6

    Zondrina

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    Sorry about that, had to step out for a bit. So I want to prove no matter what point x is in the boundary, the neighborhood of that x will contain points from A and points from C?
     
  8. Sep 18, 2012 #7

    jbunniii

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    No, it won't necessarily contain points from both.

    What you need to be able to say is this:

    "Choose a point x in the boundary of the union. Every neighborhood of x contains points from A."

    or

    "Choose a point x in the boundary of the union. Every neighborhood of x contains points from B."

    Which one of these statements is true will depend on x and on the sets. And it doesn't matter which one is true. What's important is that at least one of them has to be true for any given x. See if you can prove that. I also have to take off for a while, will check in again later this evening. Good luck in the meantime!
     
  9. Sep 19, 2012 #8

    Zondrina

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    I managed to figure these out. Turns out the subset condition held, but the superset and therefore equivalence both failed.

    I actually have a slightly more interesting exercise the prof gave me and I have some of it done already, but it's the one I'm interested in and I'm stuck again :P.

    Would you care to have a look at it so I don't have to make ANOTHER thread for no reason.
     
  10. Sep 19, 2012 #9

    jbunniii

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    Sure, I'll take a look. But please create a new thread for it, if it's a new problem.
     
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