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Metric spaces - Clopen property

  1. Oct 19, 2011 #1
    Why is it that a metric space (X,d) always has two clopen subsets; namely {0}, and X itself?

    Rudin calls it trivial, and so do about 15 other resources I've perused.

    What confuses me is that if we define some metric space to be the circle in ℝ2: x2+y2 ≤ r2, then points on the boundary of the circle don't have neighborhoods contained entirely in X, since for any radius > 0, the neighborhood will extend out of X.

    Thanks!
     
  2. jcsd
  3. Oct 19, 2011 #2

    Fredrik

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    {0} is the set that contains 0. The empty set can be written as {}, but is usually written as ∅.

    It's not hard to prove that ∅ and X are both open and closed. If that's what you need help with, then please post the definitions of "open" and "closed" that you would like to use.

    You may have misunderstood something about neighborhoods or open balls. They never extend out of the metric space. Consider e.g. the metric space [0,1] with d defined by d(x,y)=|x-y|. Denote the open ball around x with radius r by B(x,r). Then, for example,
    [tex]
    \begin{align}
    B(1/2,1) &= \{x\in[0,1]:|x-1/2|<1\}=[0,1]\\
    B(1/4,1/2) &= \{x\in[0,1]:|x-1/4|<1/2\}=[0,3/4)
    \end{align}
    [/tex]
     
    Last edited: Oct 19, 2011
  4. Oct 19, 2011 #3

    Bacle2

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    Actually if you meant exactly two, I think the OP's statement is true of _connected_ metric (or otherwise) spaces. A discrete metric space has every single subset as a clopen set. If you meant _at least_ two, then I think the total space is clearly both open --every point in the set has a basis element containing it, and closed, since every convergent sequence in the space converges to a point in the space. The empty set is also both closed and open by default, i.e., if you start with a false premise, anything else that follows is true: every point in the empty set has a neighborhood...., and the complement of the empty set is the whole space, which is open.
     
  5. Oct 19, 2011 #4

    Deveno

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    the reason why it is trivial is this:

    every point in a metric space is always contained in some ε-ball (primitive open set, which depends on the metric, d). the union of all these open sets is the entire space, so the entire space is open.

    since the complement of the entire space is X-X = Ø, and the entire space is open, Ø is closed.

    but Ø is by definition open, since it has no members, so any condition which starts "for every member x of U,...." is automatically true of Ø (including the statement, every element of U is contained in some ε-ball). since Ø is open, it's complement, X, must be closed.

    one can also show that no points in X are near the empty set (or equivalently, Ø has no limit points), so the closure of Ø, cl(Ø) = Ø.

    but every point in X is IN X, thus near X, so cl(X) = X (equivalently, the closure of X in X contains X, and is a subset of X, so IS X...an extreme case would be where X was all "isolated points" so has NO limit points). hence Ø and X are both closed, and being complements of each other, both open as well.

    (an interesting "bizzare" example in the real plane is the open disk....this has a completely different set of limit points as its own space than it does as a subset of R2, because the "boundary points" no longer have any neighborhoods in the open disk, so as a metric space in its own right, has only interior limit points, which are already in the open disk)

    if one uses an axiomatic definition of a topology, defined on 2X (the power set of X), then Ø and X are both automatically open, by definition (and thus closed, as well). so this statement isn't just true for metric spaces, but for arbitrary topological ones, as well.

    for a metric space which uses an induced metric from some larger space, such as the disk D2, the ε-balls are defined to be Nε(x) ∩ D2, that is the relative metric topology of R2, restricted to D2.

    for example, the set of all points in the unit interval [0,1] "less than 1/2 away from 3/4" does not include the real number 9/8, even though it is less than the specified distance away in the metric d(x,y) = |y-x| for R (because...it's not in [0,1]).
     
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