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Math Help - closed and open points!!

  1. #1
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    closed and open points!!

    1) Let S and T be subsets of R. Find a counter example for each of the following.

    a) If P is the set of all isolated points of S, then P is a closed set
    b) If S is closed, then cl (int S) = S
    c) if S is open, then int (cl S) = S
    d) bd (cl S) = bd S
    e) bd (bd S) = bd S
    f) bd (S U T) = (bd S) U (bd T)
    g) bd ( S-intersect-T) = (bd S)-intersect-( bd T)


    2) Prove:

    a) S is closed iff S = cl S
    b) cl S = S U bd S
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  2. #2
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    Can you give the definition of:
    Closed Set.
    Closure of a set.
    Open set
    Interior of a set?
    Boundary of a set.
    Isolated Point.

    Can you tell us what tools you have:
    To prove a set is closed.
    To prove a set is open
    To prove a point is an interior point.
    To prove a point is a boundary point.
    To prove a point is an isolated point.

    Until you answer those questions, it is almost impossible to help you.
    Definitions vary so much from textbook to textbook.
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  3. #3
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    closed: Let S be a subset of R. If bd S subset of S, then S is said to be closed.

    Open: If bd S is a subset of R\S, then S is open

    Let S be a subset of R. A point x in R is an interior point of S if there exists a neighborhood N of x such that N is an element of S.

    If for every neighborhood N of x, N-intersects-S is not equal to the empty set and N-intersects-(R\S) is not equal to empty set, then x is called a boundary point of S

    Let S be a subet of R. Then the closure of S, denoted cl S, is defined by
    cl S = S U S'

    If x is and element of S and x is not an element of S', then x is called an isolated point of S
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  4. #4
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    A set S is open iff S = int S. Equivalently, S is open iff every point in S is an interior point
    of S.
    A set S is closed iff its complement R\S is open
    Problem one here is the hint I have:
    parts a-e, think about the setes {5}, (3,5) U (5,7), Q, {1/n: n is an
    element of N}
    To prove B, you have to show that cl S is a subset S U bd S and that S U bd S is a subset cl S.
    To show that cl S is a subset S U bd S, take an arbitrary point x is an element cl S
    and deduce that x is an element of S U bd S – one way to do this is the following:
    since by definition cl S = S U S',
    x is an element of cl S implies that either x is an element S
    or x is an element of S' \ S consider each of these two possibilities
    separately and show that in each case you can conclude that
    x is an element of S U bd S by using
    only the definitions of bd S and S'. Note that the two possibilities x is and element of S
    and x is an element of S'\S
    are mutually exclusive, i.e., cannot occur simultaneously.
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  5. #5
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    I have taught courses in topology for over thirty years. That is the strangest set of definitions I have ever seen. Of course they are, in fact, equivalent to other “standard” definitions.

    Let’s see how to use them to prove “a) If P is the set of all isolated points of S, then P is a closed set.”
    If x is an isolated point of S then because x is not in S’, there is a neighborhood N of x that contains no other point of S. That means that x is a boundary point of P. So bd(P)=P, which proves that P is closed because its boundary is a subset of P.
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  6. #6
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    Right, I am trying to figure out the rest of them, I am just drawing a blank, not sure how to do them...This is just for review as I have a test coming friday.
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