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Homework Help: Exterior of a set

  1. Jun 30, 2010 #1
    1. The problem statement, all variables and given/known data

    (R,dPO) - i.e. with post-office metric

    A=(0,1)

    My question is how to find ext(A)

    2. Relevant equations

    ext(A)=int(Ac)

    dPO(x,y)= |x|+|y| for x[tex]\neq[/tex]y and 0 for x=y

    3. The attempt at a solution

    As from the definition of an interior point, st.

    For every a[tex]\in[/tex]A and [tex]\epsilon[/tex]>0, there exists an open set B st. a[tex]\in[/tex]B(a,[tex]\epsilon[/tex])[tex]\subset[/tex]A then a is called an interior point of A.

    Ac = (-[tex]\infty[/tex],0][tex]\cup[/tex][1,+[tex]\infty[/tex])

    So we can create an open set st.;

    Take [tex]\epsilon[/tex]=1/2 and a=0 and a=1, respectively;

    B(0,1/2)={x [tex]\in[/tex] Ac : |x|< 1/2 for x [tex]\neq[/tex] 0 and 0< 1/2 for x=0}=(-1/2,1/2)

    for this I am not sure wheter I should take x not equal to 0 or x=0. But for x not equal to 0, Ac obviously covers the open set, so I take x [tex]\neq[/tex] 0


    and

    B(1,1/2)={x in Ac : |x|+|1|< 1/2 for x not equal to 1 and 0< 1/2 for x=1} = {1} since |x|+|1|< 1/2 ---> |x|< -1/2 isn't correct. So 1 is included.

    So, Ac covers each ball. That is, both 1 and 0 are the interior points of Ac and as a result;

    My answer is;

    int(Ac)= (-infty,0] (union) [1,+infty )

    But the given answer is;

    int(Ac)= (- infty ,0) (union) [1,+ infty )

    I probably set some wrong logic here, could you help me why 0 is not included in int(Ac) ?

    And for the open set;

    B(0,1/2)={x [tex]\in[/tex] Ac : |x|< 1/2 for x [tex]\neq[/tex] 0 and 0< 1/2 for x=0}=(-1/2,1/2)

    What if i take {0} instead of (-1/2,1/2)
     
    Last edited: Jun 30, 2010
  2. jcsd
  3. Jun 30, 2010 #2

    Dick

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    Homework Helper

    You are right that B(0,1/2)=(-1/2,1/2). But A^C doesn't cover that ball. It also contains points from A. Can you show for any r>0 B(0,r) also contains points from A? That would say 0 is NOT in int(A^C) wouldn't it?
     
  4. Jun 30, 2010 #3
    So you mean, since we cannot approximate zero by only the elements of Ac

    for any r>0;

    B(0,r)={x [tex]\in[/tex] Ac : |x|< r for x [tex]\neq[/tex] 0 and 0< r for x=0}=(-r,r)

    Then for the interval (-r,r) around 0, there exist many elements from A.

    Even though we can take r as small as we want, there exists elements from A. Thats why zero is not included, right?
     
  5. Jun 30, 2010 #4

    Dick

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    Right. And since B(1,1/2)={1} and 1 is not in A, 1 is included. Can you show any other point x of A^C has a ball around itself consisting only of x, just like 1?
     
  6. Jun 30, 2010 #5
    Actually it should work for and point in Ac except 0, for instance, for simplicity, we can take r=|a|/2 and we can take |a| instead of 1, then,

    B(|a|,|a|/2)={x [tex]\in[/tex] Ac : |x|+|a|< |a|/2 for x [tex]\neq[/tex] |a| and 0< |a|/2 for x=|a|}={|a|} since again |x|+|a|< |a|/2 is not correct.
     
  7. Jun 30, 2010 #6

    Dick

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    You've got the right idea. But you want B(a,|a|/2). B(|a|,|a|/2) isn't a ball around a, it's a ball around |a|.
     
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