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[topology] The metric topology is the coarsest that makes the metric continuous

  1. Feb 2, 2012 #1
    [topology] "The metric topology is the coarsest that makes the metric continuous"

    1. The problem statement, all variables and given/known data
    Let (X,d) be a metric space. Show that the topology on X induced by the metric d is the coarsest topology on X such that [itex]d: X \times X \to \mathbb R[/itex] is continuous (for the product topology on [itex]X \times X[/itex]).

    2. Relevant equations
    N.A.

    3. The attempt at a solution
    I can prove that d is continuous, but I'm having trouble proving that the topology is the coarsest. Let V be a subset of R, then denote [itex]U := d^{-1}(V)[/itex] and suppose U is open. I want to prove that V is open, but I'm not sure how.
     
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  3. Feb 2, 2012 #2

    micromass

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    Re: [topology] "The metric topology is the coarsest that makes the metric continuous"

    Try to prove that if d is continuous, then the open balls are open sets in that topology.
     
  4. Feb 2, 2012 #3
    Re: [topology] "The metric topology is the coarsest that makes the metric continuous"

    Thanks.

    Hm, I seem to be confused.

    The best thing I can do, is to say: take an open ball [itex]B(x, \varepsilon)[/itex], then we know that [itex]d^{-1}(]-\epsilon,\epsilon[)[/itex] is open, and hence [itex]d^{-1}(]-\epsilon,\epsilon[) \cap \left( \{x\} \times X \right)[/itex] is open in [itex]\{x\} \times X[/itex], which proves your statement if we identify
    [itex]B(x, \varepsilon) \leftrightarrow d^{-1}(]-\epsilon,\epsilon[) \cap \left( \{x\} \times X \right)[/itex]
    [itex]X \leftrightarrow \{x\} \times X [/itex]
    but I would doubt this round-about method is necessary...

    Part of my confusion seems to be about the following: I think I'm wrong about my initial assumption that the statement is equivalent to saying it's (also) the coarsest on [itex]X \times X[/itex] such that d is continuous? Because I'm confused by the fact that (take [itex]X = \mathbb R[/itex] for the moment) [itex]d^{-1}(]a,b[)[/itex] is always an unbounded region, consequently the initial topology on [itex]\mathbb R \times \mathbb R[/itex] induced by d cannot coincide with the product topology on [itex]X \times X[/itex] inherited from the metric topology on X (which would have to be the case if the above statement were true).

    I hope my confusion makes sense...
     
  5. Feb 2, 2012 #4

    micromass

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    Re: [topology] "The metric topology is the coarsest that makes the metric continuous"

    That is correct.

    Yes, I indeed do not think that is true. Note that the actual initial topology of d (on the set XxX) is not even Hausdorff. Indeed, the point (x,x) and (y,y) can never be separated by open sets. If (x,x) is in [itex]d^{-1}(]a,b[)[/itex] then [itex]0\in ]a,b[[/itex]. But then we also have [itex](y,y)\in ]a,b[[/itex].

    Some extra information:
    As you can see, the set [itex]\{d^{-1}(]a,b[~\vert~a,b\in \mathbb{R}\}[/itex] does not induce a nice topological space. But it does introduce a nice structure called a uniformity. Uniform spaces (spaces equipped with a uniformity) are a generalization of metric spaces which can be used to define notions such as uniform continuity, uniform convergence, totally boundedness and Cauchy sequences. Every uniform space induces a topological space. So we actually have "metric space --> uniform space --> topological space".
     
  6. Feb 2, 2012 #5
    Re: [topology] "The metric topology is the coarsest that makes the metric continuous"

    Ah, thank you :)

    And is the way I proved it the most efficient way/the way you had in mind? With the identifications? Or is it a round-about way for something shorter that I'm overlooking?
     
  7. Feb 2, 2012 #6

    micromass

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    Re: [topology] "The metric topology is the coarsest that makes the metric continuous"

    It was the way I had in mind. Perhaps there is a shorter way, but I doubt it.
     
  8. Feb 2, 2012 #7
    Re: [topology] "The metric topology is the coarsest that makes the metric continuous"

    Ah perfect :) You have settled my qualms.
     
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