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Topology question; derived pts and closure

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data
    If A is a discrete subset of the reals,

    prove that

    [tex]A'=cl_x A \backslash A[/tex]

    is a closed set.

    2. Relevant equations
    A' = the derived set of A
    x is a derived pt of A if [tex]U \cap (A \backslash \{x\}) \neq \emptyset[/tex] for every open U such that x is in U.

    Thrm1. A is closed iff A=cl(A)
    Thrm2. cl(emptyset)=emptyset

    3. The attempt at a solution

    "Proof". Using Thrm1 with A' we can see that A' is closed iff A'=cl(A'). Since A is a discrete subset of the Reals we know that the set consists of isolated pts. Since there are no derived pts in A, then A'= emptyset. Using Thrm2 we know that cl(emptyset)=emptyset. Therefore A'=cl(A'). And thus A' is closed.

    Does this proof work?
  2. jcsd
  3. Oct 26, 2008 #2


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    I think so. Is there any difference between cl_x(A) and cl(A)? A' is closed because it's empty.
  4. Oct 27, 2008 #3
    Well, a classmate pointed out to me that the set B={1/n : n in Naturals}, has a derived point of {0}. I feel like this set is discrete, yet it's derived points are not the empty set.

    I thought I had this one. Any ideas of where I have gone wrong? My guess would be assuming that discrete subsets have no derived points but then what direction do I take?
  5. Oct 27, 2008 #4


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    Whoa. You're right. You caught me. I thought the proof sounded a little vacuous. So you want to prove that if x is not in A', then there is a neighborhood of x that does not intersect A'. If x is in A, use that A is discrete. If x is not in A, then use that A is closed. Can you take it from there?
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