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A metric space of equivalent sequence classes

  1. Jan 22, 2012 #1
    A metric space of equivalent Cauchy sequence classes (Z, rho) is defined using a metric of the sequence elements in the space (X,d), where d is from XX to R (real numbers). The metric of the sequence classes is rho = lim d(S, T), where S and T are the elements of the respective sequences. To show that the limit exists, the authors* apply the triangle inequality in the metric space of the sequence elements and state:

    (1) /d(S, T) - d(S, T)/ < d(S, S) + d(T, T)

    The argument continues from (1) that if the sequences are Cauchy sequences (rt side -> 0), then the left side goes to zero too and lim d(S, T) exists in the complete space which is the range of d.

    I am having difficulty obtaining (1) from the triangle inequality, which I take to be:

    (2) d(S, T) < d(S, S) + d(T, T)

    In (1) and (2), I have omitted the subscripts, but you may surmise that on the left sides of the two inequalities they are mixed; whereas on the right sides they are not.

    In my naive effort to see through this derivation, I considered the space RxR and a quadrilateral with corners, S, S, T, T and subscripts m, n. In this simple representation, one can see that (1) is says that the sum of two opposite sides of the quad is greater than the difference of the other two sides; or that the sum of the diagonals is greater than the difference of two opposite sides (depending on how the vertices are ordered.

    But this is not a general proof. That is what I should like some help to understand.

    Not an exciting problem, I know, but I would appreciate the help greatly.

    -------
    * "Into. to Topology", by Gamelin and Greene (Dover, 1999), p 13, ex 7 (b) with hints on p197.
     
    Last edited: Jan 22, 2012
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  3. Jan 23, 2012 #2

    micromass

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    First of all, (2) is not a valid form of the triangle inequality.

    The triangle inequality is [itex]d(x,y)\leq d(x,z)+d(z,w)[/itex]. So in the left hand side, z must appear twice.

    Now, what do you really mean for (1). You wrote

    [tex]d(S,T)-d(S,T)[/tex]

    but this is just 0, right?
     
  4. Jan 23, 2012 #3
    Yes, you are correct; I made a typo in stating the triangle inequality. It should be, as you point out, of the form:
    d(S, T) < d(S, S) + d(S, T).

    The subscripts are not shown, as I explained in my first message. I do state what the subscripts should be. They are in ineq (1). left side, k,k in the first metric and l,l in the second. This is of the form of an estimate on the Cauchy difference of d(S, T), which the author demonstates to be zero by means of ineq (1). My problem is in justifying ineq (1) by means of the triangle inequality (2). Can you assist with that?
     
  5. Jan 23, 2012 #4

    micromass

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    OK, so to be clear, you need to show:

    [tex]|d(S_n,T_n)-d(S_m,T_m)|\leq d(S_n,S_m)+d(T_n,T_m)[/tex]

    This is equivalent with

    [tex]d(S_n,T_n)-d(S_m,T_m)\leq d(S_n,S_m)+d(T_n,T_m) ~\text{and}~-d(S_n,S_m)-d(T_n,T_m)\leq d(S_n,T_n)-d(S_m,T_m)[/tex]

    So you must show two inequalities now. Rearrange those to get something you can easily prove from the triangle inequality.
     
  6. Jan 23, 2012 #5
    Yes, I reached this point, the point of the two inequalities. I tried transposing a term from the central difference to either side. I found pairs of terms, after transposition, that look like one side of the t.i., but I could not use the t.i. to combine them because they seem to be on the wrong side of the inequality to permit so doing.

    I should mention that in the RxR metric space, I drew a quadrilateral to represent four sequence elements as vertices S, S, T, T. They can be arranged either S, S, T, T or S, T, S, T. In both arrangements, the inequality in question is a clear consequence of properties of triangles. But this representation does not help me in the general case.
     
  7. Jan 23, 2012 #6

    micromass

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    Certainly you do know that

    [tex]d(S_n,T_n)\leq d(S_n,S_m)+d(S_m,T_m)+d(T_m,T_n)[/tex]

    ??
     
  8. Jan 23, 2012 #7
    Yes, certainly! I was just starting the "triangulation" from the wrong end! Thanks!

    Thanks, too, for toleration my notational difficulties.
     
  9. Jan 23, 2012 #8
  10. Jan 23, 2012 #9
    To Tarantinism: Thank you for the referral. "Foundations..." does seem a book I could profit from reading. I have bookmarked the link. When I have completed Gamelin and Greene's book on topology, I shall have another look. Regards, goedelite.
     
  11. Jan 24, 2012 #10

    Fredrik

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    I also like Friedman's book. It was the first thing I thought of when I saw the thread title. I'm currently using it to get better at integration theory. It has several sections that I really like, e.g. the short section on projection operators.
     
  12. Jan 24, 2012 #11
    Yes, and it's only 10 $, the same as Gamelin-Greene's good book :smile:
     
  13. Jan 24, 2012 #12
    OK, thanks! I'd better buy it before the price rises.
     
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