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Completion of a metric space

  1. Apr 5, 2006 #1
    We recently discussed completion in my analysis class and I have a brief question on the subject. The completion X* of the metric space X is defined to be the set of Cauchy sequences of X with a defined equivalence relation ({xn}~{yn} if lim d(xn,yn)=0) and metric (D([xn],[yn])=lim d(xn,yn)). I understand the proof of this being a well-defined metric space, but how is it known that X* itself is complete?

    Thanks in advance.
  2. jcsd
  3. Apr 6, 2006 #2


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    Take an arbitrary Cauchy sequence in X*, and see if it converges. I haven't actually done this, but maybe you can prove that if:

    [[x1,1, x1,2, x1,3...], [x2,1, x2,2, x2,3, ...], ...] is Cauchy, then it converges to [x1,1, x2,2, x3,3, ...]
  4. Apr 6, 2006 #3
    Yes, that does work. Thanks a bunch!
  5. Nov 2, 2007 #4
    I was working on that problem and browsed this page and noticed it was wrong.
    For example, take any nonzero p in Q, the rationals. Then let x1 = (p,p,p,...), x2 = (0,p,p,p,...), x3 = (0,0,p,p,p,...), ..., xn = (0,0,...,0,p,p,p,p,p,...) (p starts in the nth position.

    Then [x1]=[x2]=...., so clearly [x1],[x2],[x3], ... is Cauchy, but it converges to (p,p,p,..).

    The (x11,x22,x33,...) construction above would give you (0,0,0,..), but that is not the limit of this sequence.

    I just wanted to point that out, though I won't go into the proof that X* is complete. I believe it's on other web pages.
  6. Nov 2, 2007 #5
    In the above construction let x1 = (0,p,p,p,...), x2 = (0,0,p,p,p,..), etc, where the p starts in the n + 1 position. That would yield (x11,x22,x33,...) = (0,0,0,0,...).
  7. Nov 2, 2007 #6
    OK here's a proof. I translated the proof from http://www.mathreference.com/top-ms,rcomp.html which seemed a bit hand wavey but nonetheless correct.

    Since this is plain text, here's a notation reference:

    N = {1,2,...} is the set of natural numbers.
    N_k will represent a natural number, not to be confused with N, the natural numbers.
    s = (s[1],s[2],...) will be a sequence in X. p_1, p_2, ... will be a sequence in X*,
    so p_i = (p_i[1],p_i[2], ...).
    d(x,y) is the distance between points in X.
    D(p,q) is the distance between points in X* (Cauchy sequences of points in X),
    i.e. D(p,q) = lim{k->infinity}d(p[k],q[k])

    Let {p_n} be a Cauchy sequence in X*. We need to find c in X* such that p_n -> c.

    Construct a sequence c = (c[1], c[2], ...) of points in X as follows:

    For k in N, p_k is a Cauchy sequence of points in X,
    so there is a N_k in N such that m,n >= N_k implies d(p_k[n],p_k[m]) < 1/k.
    Define c[k] = p_k[N_k].
    We inductively get c = (p_1[N_1], p_2[N_2], ...).

    We have to show two things:
    1) c is a Cauchy sequence of points in X, and
    2) p_n -> c.

    Proof of the first part, c is a Cauchy sequence of points in X:

    Fix e > 0.
    Choose M in N such that
    1/M < e,
    i,j >= M implies D(p_i,p_j) < e.

    So fix i,j >= M. It will suffice to prove that d(c,c[j]) < 3e.

    Since lim{q}d(p_i[q],p_j[q]) = D(p_i,p_j) < e,
    there exists K >= N_i,N_j such that q >= K implies d(p_i[q],p_j[q]) < e.

    Then we have
    d(c,c[j]) = d(p_i[N_i],p_j[N_j])
    <= d(p_i[N_i],p_i[K]) + d(p_i[K],p_j[K]) + d(p_j[K],p_j[N_j])
    < 1/i + e + 1/j
    < 3e.

    Since i,j >= M were arbitrary, it follows that c = (c[1],c[2],...) is Cauchy,
    hence c is in X*.

    Proof of the second part, p_n -> c:

    Fix e > 0. Choose M in N such that
    1/M < e,
    i,j >= M implies d(c,c[j]) < e.

    Fix i >= M.

    Then for any n >= max{M,N_i}, we have
    d(p_i[n], c[n]) <= d(p_i[n], p_i[N_i]) + d(p_i[N_i], c) + d(c, c[n])
    < 1/i + 0 + e
    < 2e

    This shows D(p_i,c) = lim d(p_i[n],c[n]) < 2e.

    But since i >= M was arbitrary, it follows that for all e > 0,
    there exists M such that i >= M implies D(p_i,c) <= 2e.

    Hence D(p_i,c) -> 0 as n -> infinity, i.e. p_n -> c.

  8. Nov 3, 2007 #7
    The main idea is that you can make your nth sequence very 'scrunched up' so that your 'diagonal' sequence is indeed cauchy. To prove this, as you can see above, is triangle inequality overkill.
  9. Nov 3, 2007 #8
    This theorem is useless if it doesn't go with the uniqueness up to isometric isomorphism!
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