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Real Analysis -Open/Closed sets of Metric Spaces

  1. Dec 10, 2006 #1
    So I have an exam in Real Analysis I coming up next week and I was hoping if someone can help me out.

    I hope my question makes sense because I think I might be confused with defining the metric space or so...

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



    a)Suppose that we have a metric space M with the discrete metric

    d(x,y) = 1 if x = y
    d(x,y) = 0 if x =/= y

    Is this open or closed?



    b)Suppose that we are in R (the real line) and the metric is define as

    d(x,y) = 1 if x = y
    d(x,y) = 0 if x =/= y

    Is this open or closed?



    2. Relevant equations



    Definition:

    A set is Y open if every point in Y is an interior point
    A set is Y closed if every point in Y is an limit point



    3. The attempt at a solution



    a)Im not even sure if question a makes sense because I didn't define the metric.

    b) I'm pretty sure it is open and closed because both the definitions work.
     
  2. jcsd
  3. Dec 10, 2006 #2

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    Is what open/closed? You haven't identified a set, only the metric. Also I think you have the definiton of the discrete metric backwards.
     
  4. Dec 10, 2006 #3
    oh darn...
     
  5. Dec 10, 2006 #4
    Can we focus on part b) only.

    What if the set was just R (the entire real line)

    Then it is open and closed?
     
  6. Dec 10, 2006 #5

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    Remember, a set is only open or closed relative to a given topology. For a metric space, there is a natural induced topology from the metric. But for your last question the topology doesn't matter: for any topology on a space X, X is, by definition, open (and closed) in the topology.

    In fact, the discrete metric induces the discrete topology, in which every subset is open (and closed).
     
  7. Dec 10, 2006 #6
    Oh okay...

    I find the discrete metric very unusual
     
  8. Dec 10, 2006 #7

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    It's not that complicated. Every point has a ball containing it and no other points (eg, of radius 1/2), which just means points are open sets. Since unions of open sets are open, this means all sets are open.

    The picture I get is sort of a lattice of isolated points. Also, don't get too hung up on the distances actually all being 1. This may be hard to visualize (if there are more than 4 points it's impossible to embed them in 3D space so they're all a distance 1 from every other point). But all that matters for most purposes is that it induces the discrete topology.
     
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