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Closed and Open sets in R (or 'clopen')

  1. Sep 27, 2010 #1
    I'm sure this has been asked before, but the proofs I've seen use the fact R is connected or continuous functions is some way. I'm trying to prove it with the things that have only been presented in the book so far (Mathematical Analysis by Apostol).

    So, let A be a subset of R which is both open and closed. Assume A is non-empty. I want to end up showing A = R.
    Now A is open so by the Representation theorem for open sets in R, A is the union of a countable collection of disjoint component intervals of A. Denote these intervals by [itex] I_{k} = (a_{k},b_{k}) [/itex]
    But both [itex]a_{k}[/itex] and [itex]b_{k}[/itex] are accumulation points of A, and so must belong to A (for A is closed).
    I'm not really sure how to proceed, or if I'm even using the right path. Would I then say, since A is open there is an r>0 such that [itex] (b_{k}-r,b_{k}+r)[/itex] is in A. Then use the representation again on this, and reiterate to conclude the whole real line is in A?
     
  2. jcsd
  3. Sep 28, 2010 #2

    HallsofIvy

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    A point, p, is an "interior point" of set A, in a metric space. if and only if there exist [itex]\delta> 0[/itex] such that the [itex]\delta[/itex] neighborhood of p ([itex]\{y | d(p, y)< \delta[/itex]) is a subset of A.

    A point, p, is an "exterior point" of set A if and only if it is an interior point of the complement of A.

    A point, p, is a "boundary point" of set A if and only if it is neither an interior point nor an exterior point of A.

    Clearly a set contains all of its interior points and none of its exterior points. A boundary point of set A may or may not be in A. Also the interior points of A are the exterior points of complement of A and vice-versa. A set and its complement have the same boundary points.

    One can show that a set is open if and only if it contains none of its boundary points and closed if and only if it contains all of its boundary points. In order to be both open and closed, A set would have to contain all of its boundary points and none of its boundary points. That is possible if and only if the set has no boundary points. For the set of real numbers with the "usual" topology, on R and the empty set have no boundary points.
     
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