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Open and closed sets in metric spaces

  1. Jul 21, 2011 #1
    From the definition of an open set as a set containing at least one neighborhood of each of its points, and a closed set being a set containing all its limit points, how can we show that the complement of an open set is a closed set (and vice versa)? Usually this is taken as a definition, but I'm sure it can be shown under these assumptions.

    An open set is usually defined as a set S ⊆ U for which, for all x ∈ S, there exists a real number ϵ > 0 such that {y ∈ U : d(x, y) < ϵ} ⊆ S, i.e. there is a neighborhood of x entirely contained in S. Of course d: U2 → ℝ is a metric on U. In some books, I've read the complement of an open set is called a closed set, simply enough. Elsewhere, a closed set is defined as a set containing all its limit points: if p ∈ U is such that the set {t ∈ U : d(p, t) < ϵ} contains infinitely many points of S, then p ∈ S. I want to use this second definition of a closed set and the stated definition of an open set to show that if S is open, (U / S) is closed.
     
    Last edited: Jul 21, 2011
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  3. Jul 21, 2011 #2

    HallsofIvy

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    The way I prefer to do this is to define "interior" point, p, of a set, A, in usual way for metric spaces- there exist a [itex]\delta[/itex] neighborhood of p which is a subset of A. Then define "exterior point" of A by "p is an exterior point of A if and only if x is an interior point of the complement of A. Finally, define "boundary point" of A to be any point that is neither an interior nor an exterior point of A.

    It is easy to see that the interior points of the complement of A are precisely the exterior points of A and vice versa, and that A and its complement have the same boundary points.

    Now, define a set, A, to be "closed" if it contains all of its boundary points and "open" if it contains none of them. It should be easy to see that a set is open in this sense if and only if it is open in your sense (obviously a set does not contain any of its exterior points so if it contains none of its boundary points then it consists only of interior point- your "usual" definition of "open"). Also it is easy to see that the "limit points" of a set are either interior points or boundary points of that set so if a set contains all of its boundary points it contains all of its limit points and is closed in your sense. Finally, it is easy to see that if a set is open, so it contains none of its boundary points, then those boundary points are in its complement. Since a set and its complement have the same boundary points, it follows that the complement of an open set is closed and vice-versa.

    This also makes it remarkably easy to deal with sets that are both open and closed- if a set is open it contains none of its boundary points and if it is closed it contains all of them. To be both open and closed, we must have "all= none"- that is, as set is both open and closed if and only if it has no boundary points.
     
  4. Jul 21, 2011 #3

    Fredrik

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    The set {y ∈ U : d(x, y) < ϵ} is called the open ball around x with radius ϵ. Most books have a special notation for it. I use B(x,ϵ). Your definition of "limit point" states that a point x is said to be a limit point of S if every open ball around S contains infinitely many members of S. Use this to find a sequence in S that converges to x.

    You want to prove the following theorem: Suppose that X is a metric space. The following two conditions on a set [itex]E\subset X[/itex] are equivalent.

    (a) E contains all its limit points.
    (b) Ec is open.

    To prove (a) [itex]\Rightarrow[/itex] (b), suppose that (a) holds and derive a contradiction from the assumption that (b) is false. Let x be an arbitrary member of [itex]E^c[/itex] and suppose that every open ball around x contains a member of E. Explain why this contradicts the assumption (a).

    To prove (b) [itex]\Rightarrow[/itex] (a), let x be an arbitrary member of [itex]E^c[/itex] and show that no sequence in E can converge to x. Explain why this means that (a) must be true.
     
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