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Topological spaces and basis

  1. Sep 20, 2010 #1
    I am having trouble proving this statement. Please help as I am trying to study for my exam, which is tomorrow

    Prove that the standard topology on RxR is equivalent to the one generated by the basis consisting of open disks.

    Thanks :)
  2. jcsd
  3. Sep 20, 2010 #2
    What are you having trouble understanding? What is the natural metric topology of R? This is essentially the same that is put on RxR, pointwise. Can you see how this coincides with open disks? It is not a hard problem.
  4. Sep 20, 2010 #3
    What is the topology generated by the basis consisting of open disks?
  5. Sep 21, 2010 #4
    What is the definition of a basis? What is the definition of the topology generated by a basis? Read carefully the definition and think if you completely understand every word and every statement there. Then close your book, write the definitions yourself on a blank piece of paper, as precisely as you can. Then open the book and compare - did you skip or added something? Once this is done, it will be easier to discuss the details and the difficulties you still may have.
  6. Sep 25, 2010 #5
    Suppose a U is an open set in R2 (consider the usual euclidean topology - we are drawing a distinction between R2 and RxR).

    Then for each point x in U, we can find an open disc Dx about x small enough so that it fits in U. If we like, we can also find a "rectangle" Rx small enough to fit inside Dx (top figure):


    Do you understand how the product topology in RxR naturally makes "open rectangles" open sets? It should follow very quickly from the definitions. What happens, now, if we take the union over all x in U of the rectangles Rx?

    Similarly, for any rectangle, we can find an open disc inside it (bottom figure). Is this clear?

    Attached Files:

    Last edited: Sep 25, 2010
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