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Open sets in metric spaces

  1. Mar 14, 2009 #1
    I have some topology notes here that claim that on any metric space (A,d), A is an open set

    But surely we can just take a closed set and define a metric on it, like [0,1] in R with normal metric?
     
  2. jcsd
  3. Mar 14, 2009 #2
    A set can be both open and closed. The entire space and the empty set are both open and closed in any topological (or metric) space. If these are the only open and closed subsets, then the space is said to be http://en.wikipedia.org/wiki/Connected_space" [Broken].
     
    Last edited by a moderator: May 4, 2017
  4. Mar 14, 2009 #3
    I just read that in a metric space (A,d) the set A is both open and closed but I dont understand why

    Further, in my notes it says that we want to define a topology on a set to be the set of all open subsets. It then formally defines a topology to be the set of subsets of A that satisfies:
    1. A and the empty set are in T (where T is the topology)
    2. Any union of elements of T are in T
    3. Any intersection of elements of T are in T

    But it seems to be the set of all closed subsets would equally well satisfy all those axioms, since A and the empty set are also closed, any union of closed sets is closed and any intersection of closed sets is closed.
     
  5. Mar 14, 2009 #4

    HallsofIvy

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    In a metric space, a set is closed if it contains all of its boundary points. A set in open if it contains none of its boundary points. Since the entire set itself has no boundary points, both of those are true: "all"= "none" so it both open and closed.
     
  6. Mar 14, 2009 #5
    It should be
    3. Any finite intersection of elements of T are in T.

    The closed sets satisfy similar axioms but the role of intersection and union is reversed: finite unions of closed sets are closed and any intersections of closed sets are closed.
     
  7. Mar 14, 2009 #6

    matt grime

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    Here's an open set in your space:

    { x : d(x,1/2) < 1 }

    The set of points strictly less than 1 unit distance from 0.5. It's open, surely you agree? It's also all of [0,1], so that set is an open subset in that metric space you just defined! Note that something like 1.1 is not in the interval [0,1] so isn't in the set I just defined.
     
  8. Mar 14, 2009 #7

    HallsofIvy

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    In particular [0, .1) and (.9, 1] are both open sets in [0,1]. they are open neighborhoods in [0,1].
     
  9. Mar 14, 2009 #8
    A is open because any open set in the metric space (A,d) is contained in A and this will imply that any open sphere centred on any point of the space (A, d) will surely be in A. On the other hand, A is closed because (A, d) is regarded as the full space w.r.t the metric d, thus all the necessary points are contained in A. for instance, If x is an arbtrary point of A where A is said to contain all the points needed, then x is a limit point of A or an isolated point of A and it is contained in A. By definition of a closed set, it must contain its limit point. (NB. If a set is considered as a metric space then it can be regarded as the fullspace containing all the points needed). But, If (A,d) is regarded as a subspace of a metric space say (X, d') where d has been restricted by d', then A may niether be closed nor open subset of X with respect to the metric d' restricting d, where (A,d)=(A,d') and (X,d) is not in general equal to (X,d'). Thus as a metric Space in its own right, A is both open and closed.
     
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