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Math Help - compact set and convergent sequence

  1. #1
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    compact set and convergent sequence

    Using the definition of a compact set: "A subset K of R is compact if every open cover of K has a finite subcover of K, that is if {Oj}j in A is an open cover of K, then there exists j1, j2, j3.... , jn in A s.t. K is a subset of the Union (from i=1 to n) of Oji."

    I need to show that having a convergent sequence {Pn} in R, with lim = p, the set B = {p} U {pn: n in N} is a compact subset of R.

    Proof:
    Since {Pn} is a convergent sequence, it is nonempty. So for every e>0, the collection of of Ne(x):x in Pn} is an open cover of B. For every open cover, there is a finite subcover of B. Also, B is a subset of Union of open covers of B.

    I don't think my proof is clear at all. Please provide some hints.
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  2. #2
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    Quote Originally Posted by inthequestofproofs View Post
    Using the definition of a compact set: "A subset K of R is compact if every open cover of K has a finite subcover of K, that is if {Oj}j in A is an open cover of K, then there exists j1, j2, j3.... , jn in A s.t. K is a subset of the Union (from i=1 to n) of Oji."

    I need to show that having a convergent sequence {Pn} in R, with lim = p, the set B = {p} U {pn: n in N} is a compact subset of R.

    Proof:
    Since {Pn} is a convergent sequence, it is nonempty. So for every e>0, the collection of of Ne(x):x in Pn} is an open cover of B. For every open cover, there is a finite subcover of B. Also, B is a subset of Union of open covers of B.

    I don't think my proof is clear at all. Please provide some hints.
    Try this:

    Let O be an open cover of B. O = {O_j, j=1,2,3...} And Suppose p is on O_k.

    then there exists an e>0 such that the open ball of radius e around p is in O_k (by defintion of an open set)

    but since (pn) --> p, there exists a number N such that for all n > N, we have l pn - p l < e

    that is, for all n>N, pn is in the ball of radius e around p, and thus O_k is a cover for all pn, n>N.

    Now we only have to worry about the points pn, n = 1,2,3...N. But, this is a finite number of points which is cleary covered by a finite number of open sets...and the rest is easy.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    This is of course true in any topological space.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    This is of course true in any topological space.
    Yes it is true in general.
    If (p_n)\to p then for any open set \mathcal{O} containing p also contains almost all (all but a finite collection) points of p_n.
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