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Isolated points must be boundary points?

  1. Nov 10, 2013 #1
    In the textbook I am working with, an isolated point of A is defined to be a point X in A such that there exists a neighborhood (open ε-ball) centered on X containing no point in A other than X itself.

    A boundary point of A (which need not be in A) is defined as a point X in A such that every open ε-ball centered on X contains at least one point in A and at least one point not in A.

    Is it the case that isolated points must be boundary points? Most textbooks use definitions other than the one I'm using, hence I am very confused.

    Also, maybe this is true for metric spaces but not for topological spaces?

  2. jcsd
  3. Nov 11, 2013 #2
    That is how I first saw these concepts defined, and are probably the most practical definitions when dealing with R^n. Isolated points are indeed boundary points when working with R^n.

    A quick way to see that isolated points need not be boundary points is to take any non-empty set, X, and equip it with the discrete topology (all subsets are open). Then for any subset A of X, all points of A are isolated but A has no boundary points (since singletons are open in the discrete topology).

    This can also fail for metric spaces, which can be seen by considering the integers with the metric topology. The metric topology is the discrete topology in this case, hence the above argument holds.
  4. Nov 11, 2013 #3
    I see. But on the standard Euclidean space with Euclidean metric, it is true? What is the most general metric space for which this is true?

  5. Nov 11, 2013 #4
    Yes. Let ##A\subset ℝ^n##, and let ##x\in A## be an isolated point of A. By definition of isolated point, there is some ##r>0## s.t. ##B(x,r)\cap A = \{x\}##. Given any ##ε>0##, ##x\in B(x,ε)\Rightarrow B(x,ε)\cap A≠∅##. Moreover, ##B(x,r)\cap A = \{x\}\Rightarrow x+(\frac{1}{2}min\{ε,r\}, 0,..., 0) \in A^c\Rightarrow B(x,ε)\cap A^c≠∅##. We've shown that ##B(x,ε)\cap A≠∅## and ##B(x,ε)\cap A^c≠∅##, by definition this shows that ##x## is a boundary point of A. Thus any isolated point of A is a boundary point of A.

    This property holds in a metric space if and only if the metric space does not have any isolated points to begin with. A topological space as a whole does not have any boundary points, so the condition clearly fails if we have an isolated point. Conversely, if we have a metric space ##X## s.t. there is a subset ##A\subset X## and a point ##x\in A## which is an isolated point but not a boundary point, then we can show that x must be an isolated point of ##X## (sketch: x isolated in A => no points of A are nearby, x not a boundary pt. of A=> no points of A^c are nearby, hence no points of X are nearby).

    The other common way to define the boundary of A is to let it be the points in the closure of A that are not in the interior of A, where the closure is defined as the intersection of all closed sets containing A, and the interior is defined as the union of all open sets contained in A. Starting with the definition of boundary that you have been using, you can alternatively define the interior of A to be the set A with boundary points removed. There are other ways to define these concepts, but they all end up being equivalent in the end.
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