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Homework Help: Seemingly simple proof that I cannot conclude

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Let (X, d) be a metric space, and we have a sequence p (sub n) is a subset of X and p is an element in X. Prove that lim p (sub n) = p if and only if the sequence of real numbers satisfies lim d(p, pn) = 0.


    2. Relevant equations
    Theorems from my class.
    1. Let (X, d) be a metric space, {pn}, a subset of X,is a sequence in X and p and element of X. We say that the sequence {pn} converges to p and write lim pn=p, provided that for every e > 0, there is a real number N so that when n > N, then d(p, pn) < e.
    2. Let (X,d) be a metric space and let S be a subset of X. Then S is a closed subset if and only if whenever {pn} is a subset of S and a convergent sequence, we have lim {pn} is an element of S.
    3. Let (X, d) be a metric space and let E be a subset of X. Then the following are equivalent:
    1. there exists a point p which is an element of X and r1 >0 such that E is a subset of B(p;r1), 2. there exists a point q which is an element of X and r2 >0 such that E is a subset of B-(q;r2), 3. there exists M > 0 so that every x,y in E satisfies d(x,y) <= M.
    4. Let (X, d) be a metric space and let E is a subset of X. We say that E is a bounded set provided that it satisfies any of the three equivalent conditions of the above proposition.
    5. Let (X,d) be a metric space. If {pn} is a convergent sequence, then it is bounded.

    3. The attempt at a solution

    The forward implication: lim {pn} = p => convergent => bounded => for p in X and r > 0 then {pn} is a subset of B (p; r) ....I am not really sure where to go from here to get lim d(p, pn) = 0. I tried using a bunch of inequalities but thats just it. The are not EQUALities.

    The reverse implication: not sure if this is correct (actually I am pretty sure its wrong) lim (d(p, pn)) = 0 => d (p, pn) = 0. This is a metric space we know that d (x, y) = 0 if and only if x = y. So this implies p = pn. (From here on, i am sure I am incorrect). => lim pn = lim p = p. Hence lim pn = p. I think this is wrong because I didn't use an e, n or N anywhere.

    Any hints or help would be appreciated.
    Thanks
     
  2. jcsd
  3. Sep 14, 2011 #2
    OK, write down the following definitions:

    [tex]p_n\rightarrow p~\text{in}~X[/tex]

    and

    [tex]d(p_n,p)\rightarrow 0~\text{in}~\mathbb{R}[/tex]

    Don't you get exactly the same thing (or close)?
     
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