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Equivalent conditions on a metric space

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Let X be a metric space and A a subset of X. Prove that the following are equivalent:
    i. A is dense in X
    ii. The only closed set containing A is X
    iii. The only open set disjoint from A is the empty set


    2. Relevant equations
    N/A


    3. The attempt at a solution
    I can prove that i implies ii: assume that there is a closed set B containing A other than X, and show that B must equal X since the closure of A is X.

    Presumably I can prove the other implications (ii -> iii and iii -> i) in a similar way but I'm not sure how to get started. Is there a certain property or fact I should be using?
     
  2. jcsd
  3. Mar 20, 2012 #2

    micromass

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    So I take it that your definition of dense is "the closure is the entire space"?

    OK, but what is the definition of closure?
     
  4. Mar 20, 2012 #3
    Yes, that's the definition of dense I'm using.

    The definition of closure of A, as far as I know, is the union of A with its limit points.
     
  5. Mar 20, 2012 #4

    micromass

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    Good, now what is a limit point??

    I want you to unwind each definition until we see something we can use. Something with open sets and with disjointness...
     
  6. Mar 20, 2012 #5
    Ok, so a limit point of A is a point x in X such that every open ball centred at x contains another point of A, right?

    I'm not sure where this is going. :tongue:
     
  7. Mar 20, 2012 #6

    micromass

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    OK, so a set A is dense iff all points not in A are limit points, right??

    Now, take a ball disjoint from A. Are the points in this ball limit points??
     
  8. Mar 20, 2012 #7
    Right.

    So if you take an open ball disjoint from A, all the points in the ball are limit points... right?
     
  9. Mar 20, 2012 #8

    micromass

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    Yes. So can the ball be disjoint from A?? The center of the ball is a limit point, so by definition...
     
  10. Mar 20, 2012 #9
    So it contains points of A, and therefore can't be disjoint from A. I see.

    But shouldn't I be proving that ii -> iii and iii -> i?
     
  11. Mar 20, 2012 #10

    micromass

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    Right. But it's maybe easier to prove ii -> i first. This is extremely easy.

    Now, iii -> i shouldn't be too hard if you unwind all the definitions.
     
  12. Mar 20, 2012 #11
    Ok, I'll see how it goes.

    My/your/our reasoning proves i -> iii doesn't it? The only problem is the question talks about "open sets", but we were talking about (open) balls. Or does the fact that there are no balls mean that there are no open sets? I'm a bit confused about that
     
  13. Mar 20, 2012 #12

    micromass

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    Aha, yes. I forgot to mention that. You're paying attention! :tongue2:

    Indeed, we only proved now that there are no open balls disjoint from A. But we wish to prove it for open sets.

    Therefore, we will need to unwind some definitions again. What is an open set?

    Eventually, I want to show that each nonempty open set contains an open ball.
     
  14. Mar 20, 2012 #13
    Ok, well my definition of an open set is a set in which every point is an interior point.

    I think I see now. If there is a nonempty open set, it contains an interior point, and thus there is an open ball within the set, by definition of interior point. So if there are no open balls there can be no open sets. Right?
     
  15. Mar 20, 2012 #14

    micromass

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    Right!! That is perfect!
     
  16. Mar 20, 2012 #15
    Cool, thanks for your help! I think I've worked out the other parts now.
     
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