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Lebesgue topological dimension

  1. Sep 4, 2012 #1

    I was reading the definition of dimension from the book: "Topology", Munkres, 2nd ed.
    Surely I don't understand, but I wonder how ℝ2 can have dimension 2.

    Take the open sets [tex] U_n=\{(x,y)\mid -\infty < x <\infty, n-1<y<n+1\} [/tex] for every integer n. It covers the plane but its order is 2, so the dimension should be less than 2.

    Shouldn't be the difinition with balls or "squares"?

    Thank you.
  2. jcsd
  3. Sep 5, 2012 #2


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    What makes you think the dimension is not 2? What is your understanding of dimension?
  4. Sep 6, 2012 #3


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    I presume that Munkres defines the "order" of such a collection of sets. What is that definition and how does it follow that this particular collection has "order 2"?

    Is there a theorem in Munkres that the dimension of a set is strictly less than the order of a covering collection of such sets?

    Definitions like this can be done in many different ways.
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