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First countability

  1. Aug 16, 2007 #1
    I have come up with an example when I trying to learn what first countability means
    It says(from wikipedia)
    In a metric space, for any point x, the sequence of open balls around x with radius 1/n form a countable neighbourhood basis [tex]\mathcal{B}(x) = \{ B_{1/n}(x) ; n \in \mathbb N^* \}.[/tex] This means every metric space is first-countable.

    According to definition of nhd basis we can express every open set can be expressed as a union of elts of nhd basis But for example [tex] \{ B_{ sqrt{2} }(x) [/tex] can not be expressed in this way

    Here is the question
    which point did i miss?
     
  2. jcsd
  3. Aug 16, 2007 #2

    George Jones

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    The union can be of an infinite number of sets.
     
  4. Aug 16, 2007 #3
    But the ball is given as [tex] \{B_{1/n}(x) [/tex] i.e. center is x , meaning the biggest ball( i.e. ball with smallest n) will be the union)
    and there is always a real number between 1/n and 1/n+1

    definition says every open ball of x can be written as a union of elts of basis .If it has said every open ball is contained in the union it would be easier to understand .

    So maybe I should ask first whether the definition i learned is correct or not
     
  5. Aug 16, 2007 #4

    George Jones

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    Forget my last post. Without thinking, I thought the answer had something to do with writing [itex]\sqrt{2}[/itex] as the limit of a sequence of rational numbers.

    I don't know what "union of elts of nhd basis" means. In particular, elt is not in any of my topology references.

    The definitions I see say that for any x, there a countable neighbourhood basis of x such any neigbourhood of x contains at least one member of the neighbourhood basis. In your example, every member of the basis in contained in the neighbourhood that you chose.
     
  6. Aug 16, 2007 #5

    matt grime

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    elt means element, a common shorthand.

    You can express the ball of radius sqrt(2) about a point as the union of elements in the given basis: just take a union over all possible sets of the form B_{1/n}(y) that are contained in B{\sqrt(2)}(x).

    A clearer countable set would be the set of balls of rational radii.
     
  7. Aug 16, 2007 #6

    George Jones

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    Yikes! I feel a bit silly now. It has benn some years since I was a student, but I don't recall hearing or seeing this term in any of the lectures for the analysis and topoplogy courses that I took.

    I also had something like this in mind, but I thought matness wanted to stick to {B_{1/n}(x)}.
     
  8. Aug 16, 2007 #7

    matt grime

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    I didn't go beyond sets of the form B_{1/n}(x). matness's problem is, I think, that he assumes that when expressing a set as the union of open nbd basis elements that they all have to have the same centre. This is not the case.
     
  9. Aug 16, 2007 #8
    yes the problem is this
    and the source of problem is the example in wikpedia that is included in my first post
    [tex]\mathcal{B}(x) = \{ B_{1/n}(x) ; n \in \mathbb N^* \}.[/tex]

    On the other hand I mostly agree with you
    using basis like a cover (roughly) seems more reasonable
     
    Last edited: Aug 16, 2007
  10. Aug 16, 2007 #9
    and one more thing
    in my lecture notes the definition is given as:
    (X,T) a topological space and B is a subset of T
    if every open nhd of a point p is union of a collection of elements of B then B is called a nhd basis for p.

    Is there a problem?
    I just wanted to be sure whether I have taken the notes correctly or not
     
    Last edited: Aug 16, 2007
  11. Aug 16, 2007 #10

    CompuChip

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    Nope, that's the correct definition.
    But note that it doesn't say that the elements in the union should all contain p (just at least one of them, of course, otherwise p is not in the union).
    Also note, that usually there isn't much to construct. For example, in your original question, the best we can do is:
    [tex]B_{\sqrt{2}}(x) = \bigcup_{B \in \mathcal{B}} B[/tex],
    where [itex]\mathcal{B}[/itex] consists of all sets in the basis which are inside the left hand side (so all [itex]B_{1/n}(y) \subseteq B_{\sqrt{2}}(x)[/itex], which is of course true but not very practical if you'd want to draw them or write them down.
     
    Last edited: Aug 17, 2007
  12. Aug 17, 2007 #11
    [tex]B_{\sqrt{2}}(x) = \bigcap_{B \in \mathcal{B}} B[/tex]
    I assume you mean union( if there is anything I missed please warn me )
    On the other hand I started to think the example in wiki wasn't that correct after your posts
    thanks
     
  13. Aug 17, 2007 #12

    CompuChip

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    Indeed, I meant union, apparently I can't tell the difference between \cup and \cap :smile:. My apologies, it's fixed.
     
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