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Open/closed subsets of metric space

  1. Nov 25, 2011 #1
    1. The problem statement, all variables and given/known data

    2mpx05s.jpg

    3. The attempt at a solution

    I've got through this question up to the last bit.

    I've got [itex]B(0,1) = \{0\}[/itex] and [itex]B(0,2) = \{y\in\mathbb{R} : -1<y<1 \}[/itex] (i.e. the open interval (-1,1).)

    How do I show that every subset of [itex]\mathbb{R}[/itex] is open ([itex]A \subseteq X[/itex] is open if it contains none if its boundary)

    and then find which subsets are closed?
     
  2. jcsd
  3. Nov 25, 2011 #2

    vela

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    What definition of the boundary are you using?
     
  4. Nov 25, 2011 #3
    [itex]x\in\partial A[/itex] if for all [itex]r>0[/itex], [itex]B(x,r)[/itex] intersects both [itex]A[/itex] and [itex]A^c[/itex].
     
    Last edited: Nov 25, 2011
  5. Nov 25, 2011 #4

    vela

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    OK, so that means if A is open, for every x in A, you can find some r>0 such that B(x,r) is contained in A.
     
  6. Nov 25, 2011 #5
    OK, so if [itex]A \subseteq \mathbb{R}[/itex] and [itex]x\in A[/itex] then [itex]B(x,1)=\{x\} \subseteq A[/itex] so every subset [itex]A[/itex] of [itex]\mathbb{R}[/itex] is open.

    How do I find which subsets are closed?
     
  7. Nov 25, 2011 #6

    vela

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    The answer should be pretty obvious if you consider the definition of a closed set.
     
  8. Nov 25, 2011 #7
    Is every subset also closed?

    If [itex]x\in A^c[/itex] then [itex]B(x,1) = \{x\} \subseteq A^c[/itex]
     
  9. Nov 25, 2011 #8

    vela

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    Yup. If every subset is open, Ac is open; therefore, A is closed.
     
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