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Homework Help: Question concerning a topology induced by a particular metric.

  1. Dec 24, 2012 #1
    The question comes from the Munkres text, p. 133 #3.

    Let Xn be a metric space with metric dn, for n ε Z+.

    Part (a) defines a metric by the equation

    Then, the problem askes to show that ρ is a metric for the product space X1 x ... x Xn.

    When I originally started the problem, after showing that ρ is indeed a metric, I went on to try to show that the topology on X1 x ... x Xn induced by ρ is the same as the product topology (or we can say the box topology since our product is finite and therefore the two topologies coincide) on X1 x ... x Xn. I am fairly confident that I showed the ρ-topology is finer than the product toplogy, but I am having issues proving the converse. It boils down the following inequality;

    If I can fill in the blank with some upperbound, then I believe that for each i,
    Bdi(xi; ε/something) is a subset of Bρ(x;ε)​

    After re-reading the question, I believe all that Munkres wants me to do is show that ρ indeed defines a metric. But, I am curious about the topology ρ generates (Perhaps, it is finer and thats it). Thank you for the help!!! :)
  2. jcsd
  3. Dec 24, 2012 #2
    There are some weird things about your post:

    But [itex]x,y\in X_1\times ...\times X_n[/itex], right? So how does [itex]d_1(x,y)[/itex] even make sense?

    Why would this inequality help (even it would even make sense)? The product topology is defined by open sets, not by a specific metric. It is true that two metric topologies are equal if the metrics satisfy some inequality. But here you don't know that the product topology comes from a metric, right??
  4. Dec 24, 2012 #3
    Micro, that is a HUGE typo on my part.

    ρ(x,y)=max{d1(x1,y1), . . . , dn(xn,yn)}.​

    In fact, the book does not have x and y in boldface, which it should.

    I had no trouble proving ρ is metric (which I now know is all the exercise wanted). Unknowning, I continued. I let ∏Ui be a basis for the product topology on ∏Xi, where Ui is open in Xi for each i. Then, I let x be a point in ∏Ui. Since Ui is open in Xi and contains xi, we can find an εi-ball Bdi(xii) centered at the ith coordinate of x and contained in Ui. Let ε = min {ε1,...,εn}. Then, I showed for each i, Bρ(xi; ε) is a subset of Bdi(xii); which implies x is a point in Bρ(x;ε) which is a subset of ∏Ui. Hence, the ρ-topology is finer.

    Edit: I now realize this is nonsense because ρ is defined for the product space. Major brain meltdown, sorry! :)
    Last edited: Dec 24, 2012
  5. Dec 24, 2012 #4
    It really is true that the topology of [itex]\rho[/itex] is the product topology, but you seem to be confused by the notations.

    Try to prove this:

    [tex]B_\rho((x_1,...,x_n), \varepsilon) = B_{d_1}(x_1,\varepsilon) \times ...\times B_{d_n}(x_n,\varepsilon)[/tex]
  6. Dec 24, 2012 #5
    Wow, that is tremendously easy and illustrating! :)

    Thank you so much!

    To show two topologies are the same, I keep trying to use the method of showing one is finer than the other and vice versa. But, I now see how easy it can be to simply show that their bases are the same. I think the text (Munkres) did that once, yet for some reason I have abandoned that approach. Thanks again Micro; you're are always extremely helpful!
  7. Dec 24, 2012 #6
    This is a special case of course. You won't always be in the situation that the bases are the same. The usual technique of showing that one is finer that the other (and vice versa) works more generally.
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