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Thread: metric spaces and equivalence classes

  1. #1
    Apr 2008
    Seoul, South Korea

    metric spaces and equivalence classes

    i think im pretty stuck in this class...

    here's the problem:

    Let (X,d) be a metric space and let S be the set of cauchy sequences in S. define a relation "~" in X by saying "{s_k} ~ {t_k}" to mean d(s_k,t_k) goes to 0 as k goes to infinite.

    i did the first problem which is to show that "~" is an equivalence relation, and i can do thesecond problem says:

    let X_tilda denote the set of equivalence classes of S and s_tilda denote the equivalence classes of s={s_k}k=1 to infinite. show that the function p(s_tilda, t_tilda)= lim (k approaches infinite) d(s_k,t_k), s_tilda, t_tilda in X_tilda defines a metric.

    but i am stuck on the third/fourth/fifth part which is
    3. show that (X_tilda, p) is complete
    4. for x in X define x_tilda to be the equivalence class of constant sequence {x,x,....}. show that the function going from x to x_tilda is an isometry of X onto a dense subset of X_tilda (isometry means d(x,y)=p(x_tilda,y_tilda))
    5. show that when Y is a completion of X, then the inclusion map X going to Y extends to an isometry of X_tilda onto Y (Note: Y is a completion of X if a complete metric space Y contains X as a dense subspace) according to hint given, this problem is supposed to show that completion of X is unique, up to isometry.

    i think like many other topology questions, i'm having trouble visualizing the problem (especially #4 and #5), and for #3 i am aware of the definition of complete (which means that every cauchy sequence in the metric space converges), but i'm having hard time applying the definition to prove this.

    thanks in advance for any inputs!
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    The approach taken by Kelley in his classic book General Topology is to prove 4. first, because that is comparatively easy. If \tilde{x} denotes the equivalence class of the constant sequence (x,x,x,...), and similarly for \tilde{y}, then it's more or less obvious that p(\tilde{x},\tilde{y}) = d(x,y). So the embedding X\subseteq\tilde{X} is isometric. To see that its range is dense, suppose that z is an element of \tilde{X} and let (z_1,z_2,z_3,\ldots) be a sequence in the equivalence class z. Then the sequence (\tilde{z_1},\tilde{z_2},\tilde{z_3},\ldots) converges to z in the metric given by p. (That's not difficult, once you've got your head around the idea of that construction.)

    Let f:x\mapsto \tilde{x} be the isometric embedding X\subseteq\tilde{X}. To prove that \tilde{X} is complete (problem 3.), I'll quote Kelley's argument practically word for word (it's short but clever). Observe first that it is sufficient to show that each Cauchy sequence in f(X) converges to a point of \tilde{X} (because we have already proved that f(X) is dense in \tilde{X}). But each Cauchy sequence in f(X) is of the form (\tilde{x_1},\tilde{x_2},\tilde{x_3},\ldots), for some sequence (x_1,x_2,x_3,\ldots) in X. Let y\in\tilde{X} be the equivalence class of the sequence (x_1,x_2,x_3,\ldots). Then \tilde{x_n}\to y as n→∞. (Again, this is almost immediate, once you see what's going on here.)

    Finally, for 5., suppose that g:X→Y is another isometric map with dense range. Then the map f^{-1}g is an isometry between dense subspaces of \tilde{X} and Y, which extends by continuity to an isometry from the whole of \tilde{X} to the whole of Y (that's Kelley's argument again).
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