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Closure of an open ball and separable space?

  1. Nov 25, 2004 #1
    1) In R3 closure of an open ball is a closed ball, but this may not hold in general metric spaces.
    Can somebody give an example explaining the above statement?

    2) what is a separable space?
    some good references are welcome.

    thanks for your help in advance.
    aditya.
    www.geocities.com/aditya_tatu
     
  2. jcsd
  3. Nov 27, 2004 #2
    a discrete metric space is just the example you're in need of.
    to be more precise the metric is defined as d(x,y) = 0 if x=y and
    1 if x is not equal to y.
    after that you just need to see any open ball in the space.
     
  4. Dec 28, 2004 #3
    2)A separable space is a metric space with a
    countable dense subset.
    Reference: "Dictionary of analysis, calculus and differential equations"
     
  5. Dec 28, 2004 #4

    mathwonk

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    the answer above is not quite precise enough to be correct, nor is the question. i.e. it is true that in a discrete metric space one can find open blass of form: {all points closer than r to p} whose closure is not the closed ball {all points at distance at most r from p}. nonetheless the closure may equal a closed ball defined by a different radius. for example although the closure of the open ball of radius one about p is not the closed ball of radius one, it is the closed ball of radius 1/2.

    one does not need any fancy space to get examples. just take the real line and omit say the interval (1,2). then the closure of the open ball of radius 2 about 0 is no longer the closed ball of radius 2 about 0.
     
  6. Dec 28, 2004 #5

    NateTG

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    Because the closed ball of radius 2 includes 2, and the closure of this open ball does not, this closure is not the closed ball of radius 2. However,the closure of the set is,for example, the closed ball of radius 1.5 about -0.5 -- an example of the problem that you described with discrete topologies.

    This is essentially the same solution but it resolves that issue.
    Consider the a plane, but remove the subset where [itex]x\in(1,2)[/itex]. Now, under the usual topology, the open ball of radius 2 about 0 has a closure that is not a closed ball.
     
  7. Dec 28, 2004 #6

    mathwonk

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    i never claimed i was giving an example of an open ball whose closure was not any closed ball. i was just giving an easier version of mansi's example of one whose closure was not the closed ball of the same radius. i was not interested enough in the problem to try to think of an open ball whose closure was not any closed ball.

    in any event i did not think of your nice example.
     
  8. Dec 29, 2004 #7

    NateTG

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    Ah, I see. The original post had 'not a closed ball' rather than 'not a closed ball of the same radius', and I thought you were referring to that with the comment about closed balls of a different radius.

    Thank you.
     
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