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Uniqueness of a completion space

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    This question arises from Chapter ONE, section 2, ex. 7 (d) and (e) of "Introduction to Topology" by Gamelin and Greene.

    Xbar may be regarded as the completion of a metric space X by identifying each X with the constant sequence {x,x,...}. Show that when Y is the completion of X, then the inclusion map X -> Y extends to an isometry of Xbar onto Y. In the present context, S is the set of Cauchy sequences in X. Xbar is the set of equivalence classes of S, classes which in the limit have a vanishing metric: lim d(s_k, t_k) = 0 <-> {s} eq {t}.

    It should be noted that the completion Y of X means that Y is a complete space means that Cauchy sequences in Y have a limit AND X is dense in Y.

    A hint is offered: Define a map of Xbar into Y by sending the equivalence class of {s_k} to the lim{s_K). The authors then state that this (the mapping) is an isometry of Xbar onto Y.




    2. Relevant equations

    Isometry of Xbar onto Y: rho(sbar,tbar) = d(s,t)

    3. The attempt at a solution

    1. My effort is first to understand what the authors write. For example, though the sequences in S have the Cauchy property, nowhere is it stated that the space X is complete. If X is the space of reals, then we know it is complete. All that is given about X is that it is a metric space. The metric has a range in R, but we don't know that X is in R.

    2. A subset of Xbar, the constant sequences, may be identified with the points of X. All of the sequence of Xbar may be identified with their limits, but not all the sequences of Xbar have limits (see 1.) Thus, I do not see that the map of X into Y is defined. If X were a complete space, then one could have a mapping into but not onto, because each point of Y could come from more than one sequence of Xbar. Every point of Y would come from every sequence contained in an equivalence class; they all would have the same limit, y.

    I believe if I understood the above two points clearly, which I apparently do not, then I would see the isometry. I gather that the uniqueness means that any other mapping would produce an isometry.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 2, 2012 #2

    micromass

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    The thing in bold is very confusing. See below.

    X is not complete. That is the entire point. X is not complete and we show that X has a completion. We claim that [itex]\overline{X}[/itex] is that (unique) completion.

    True.

    No, we don't do this. All we know is that a Cauchy sequence in [itex]X[/itex] determines (by definition) a point in [itex]\overline{X}[/itex]. So let [itex](x_n)_n[/itex] be a Cauchy sequence. Then the equivalence class of this sequence determines a point y in [itex]\overline{X}[/itex]. One can prove that [itex]x_n\rightarrow y[/itex] in [itex]\overline{X}[/itex] (where we have identified [itex]x_n[/itex] with its constant sequence).

    We have such a map by hypothesis. We ASSUME that Y is a completion of X. This means by definition there Y is complete and that there is an isometry map [itex]\varphi:X\rightarrow Y[/itex] such that [itex]\varphi(X)[/itex] is dense in Y.

    What we wish to do is to find an isometry [itex]\psi:\overline{X}\rightarrow Y[/itex] such that [itex]\varphi(x)=\psi(x)[/itex] (where we identify x with its constant sequence).
     
  4. Feb 2, 2012 #3
    To micromass: Thanks for your reply! I shall benefit more from it when I am less sleepy, tomorrow.

    I did garble, quoting myself: "It should be noted that the completion Y of X means that Y is a complete space means that Cauchy sequences in Y have a limit AND X is dense in Y."

    What I understand is that Y as the completion space of X means: 1. Y is a complete space, all Cauchy sequences in Y have limits; and 2. X is dense in Y. Thus Xbar is a completion space of X, and, as was shown in part (c) of the exercise, (Xbar, rho) is a complete metric space. (By X is dense in Xbar is meant that one may find a constant sequence arbitrarily close to any given point of Xbar.)

    Those, (1) and (2), are, to my understanding, separate properties of the completion space.

    I shall have a fresh look tomorrow. Thanks!
     
  5. Feb 5, 2012 #4
    To micromass: Sorry I have not replied sooner! There seems to be misunderstandings in our correspondence, probably my fault for not having learned Tex yet; also, perhaps due to the text I am reading; then there is me!

    In the hint on p. 197, the authors state: "Define a map of Xbar into Y by sending the equivalence class of {s_k} to the limit of {s_k}." I find this confusing. The limit of {s_k} is an element of Y because the mapping is "into Y". Here {s_k} cannot be a c.s. in S, because these sequences do not have limits, necessarily. Each of the constant sequences (for a particular k) does correspond to a point in Xbar. These points represent equivalence classes for the constant sequences. Because {s_k} is a c.s. and the metric of the points of Xbar is an isometry on the metric of X, the points of Xbar accumulate on a limit point in Xbar; or so it seems to me. This is all I can make of the statement I have quoted above: "Define a map ..."

    If my understanding is correct, then Y is the space of all the limit points of constant sequences of Xbar The mapping is into, because not all of those limit points come from the constant sequences derived from X.

    To show that the above mapping is an isometry of Xbar onto Y. I must show that rho({s_k},{t_k}) = lim d(s_k,t_k), where the curly brackets represent limits of the constant sequences' equivalence classes.

    This is what I can understand of the statements in the text. My use of the word understand is very free, because I don't feel that I understand very much. The authors may be too terse for me. Perhaps, I should go on to another text. If you know of another text that might be of help, please let me know.

    Thanks for you assistance to date.
     
  6. Feb 5, 2012 #5

    Deveno

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    it might help to think of X as being "like the rationals" and Y as being "like the reals".

    √2 is not a rational number. how do we define √2, as a real number (using Cauchy sequences of rational numbers)?

    we define √2 as the limit of every rational sequence that converges to √2. this sounds a bit circular. so let's look a bit closer:

    consider the following two Cauchy sequences of rationals:

    (1,1.4,1.41,1.414,1.4142,.....) and

    (0,1.3,1.40,1.413,1.4141,.....)

    these are two different Cauchy sequences, but it appears they would have the same limit, if that limit existed. which sequence should we take as being the one that defines √2? either one would work, right?

    so instead of just having ONE sequence that has the limit √2, you can see we have LOTS. so we need to define an "equivalence class" of sequences to get just a single real number.

    now in the rationals (the X), neither of the sequences above HAS a limit. but in [itex]\overline{Q}[/itex], (the set of equivalence classes of Cauchy sequences in Q), we by definition automatically get a limit for every Cauchy sequence, and equivalent Cauchy sequences get sent to the same element of [itex]\overline{Q}[/itex].

    now if we already have the real numbers R (our Y) sitting "over there", we have the inclusion of Q→R, which is an isometry (the distance between a and b as real numbers is the same real number as: the image of the rational number which is the distance between a and b as rational numbers (boy, THAT's a mouthful)).

    when we extend this inclusion to send {sn}→lim{sn} (note that the equivalence classes of constant sequences correspond to rational numbers), we preserve distances. so √2 goes in-between 1 and 2, and in-between 1.4 and 1.5, and in-between 1.41 and 1.42, and so forth, right where it belongs. the extension is not only order-preserving, it's scale-preserving (that's where our metric comes in...we use the metric on X, to get a metric on Xbar that IS the metric on X when restricted to X).

    and we know this, because since we are dealing with equivalence classes of sequences, for rational numbers, we can just use "the constant sequence representative" (Cauchy because the difference between ANY 2 terms is 0 < ε, for any ε > 0), and for non-rationals (which don't have any constant sequences in their equivalence classes) we use the limit of the value of the (rational) metric on the rational terms of a representative sequence.

    it's kind of a delicate balancing act: and it only works because the terms of a Cauchy sequence get "arbitrarily close together", so we're justified in thinking they are "getting close to some value". "close" in this case, means: a small value for the metric.

    now, for a real mind-blower, the definition of a metric usually depends on the real numbers, so imagine the difficulty involved in making the completion of the rationals for the first time, where you have to use "something else besides d(x,y)" (and the winner of that race turned out the be the absolute value of the difference, which could be defined solely in terms of the rational numbers).
     
  7. Feb 6, 2012 #6
    Micromass: I am much in your debt for the time you took to explain the completion of the space of C.S. by exemplifying with the space of reals as the completion space. In my own efforts, I thought of this in terms of Dedekind cuts (the way I learned the extension from Q to R). Your example is better for my understanding because it uses C.S. of rationals and C.S. are what the exercise is about. I see the limit process in Qbar as comparable to the limit process in the real-rationals towards the real which is not a rational. The real-rationals are comparable to the constant sequences' equivalence classes, each of which contains only one sequence. What I had difficulty with was recognizing the place of Y in this comparison. Your reply has satisfied me on this. Thanks very much!
     
    Last edited: Feb 6, 2012
  8. Feb 6, 2012 #7

    micromass

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    You may want to thank Deveno instead! :smile:
     
  9. Feb 6, 2012 #8
    I don't know why I ought thank Deveno "instead" of you. Thinking that you are distinct people, I thank you both on your recommendation. If you are not distinct, then no harm is done in thanking you twice.
     
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