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Homework Help: Proving R^2 is metrizable in the dictionary order topology

  1. Sep 25, 2010 #1


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    1. The problem statement, all variables and given/known data

    As the title suggests, I need to show that RxR is metrizable in the dictionary order topology.

    As a reminder, for two elements (a, b) and (c, d) of R^2, the dictionary order is defined as (a, b) < (c, d) if a < c, or if a = c and b < d.

    3. The attempt at a solution

    The typical basis elements for the dictionary order topology (dot from now on) are represented by either "infinite vertical regions" or "vertical intervals".

    If we define a metric on R^2 with:

    d(x, y) = |x2 - y2| , for x1 = y1 ; max{|xi - yi|, for x1 =/= y1}

    where x = (xi) and y = (yi) are points in R^2,

    we can prove that the dot equals the topology induced by this metric, for, if B is some basis element in the dot, we can find open balls which are either regions or open intervals contained in this basis element, whatever it looks like. The same vice versa.

    I hope this works.
  2. jcsd
  3. Sep 25, 2010 #2


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    This is the same as simply
    [tex] d(x,y)=max\{ |x_1-y_1|,|x_2-y_2| \}[/tex]

    since if [tex] x_1=y_1[/tex] then in the max you'll definitely be taking the difference between [tex] x_2[/tex] and [tex] y_2[/tex].

    What do you mean by "the dot"?
  4. Sep 25, 2010 #3


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    The dictionary order topology, as I stated in the first post.

    Yes, I'm aware of that. But in that case, if given a basis element of the dot of type "vertical intercal", for example, the interval <(a, c), (a, d)>, for any x in this interval, how can I find a basis element of the metric topology defined as you proposed above, if the basis elements in this case (open balls) are square regions?

    Btw, the concept of proving two topologies are equal is to show each one is finer than the other.
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