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Set Question

  1. Aug 2, 2008 #1
    Working through Spivak "Calculus on Manifolds."

    On p. 7, he explains that "the interior of any set A is open, and the same is true for the exterior of A, which is, in fact, the interior of R[tex]\overline{}n[/tex]-A."

    Later, he says "Clearly no finite number of the open sets in [tex]O[/tex] wil cover R or, for that matter, any unbounded subset of R"

    My confusion: given some interval say A = [a,b] [tex]\subset R[/tex], then R-A and (a-1,b+1) would seem to be two open sets covering R.

    I clearly have misunderstood a definition.

    Thanks, in advance.

    Ken Cohen
  2. jcsd
  3. Aug 2, 2008 #2


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    Your example seems to be valid. But maybe the problem is in the definition of O (like: O is the set of all finite, connected open intervals)?
  4. Aug 2, 2008 #3
    Thank you! It was a context confusion, as you suggested.

    Is R compact?
  5. Aug 3, 2008 #4


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    By definition, a topological space T is compact if every open cover has a finite subcover. So let me cover R by:
    [tex]\{ {]n, n+1[} \mid n \in \mathbb{Z} \} [/tex]
    It's easy to see (and probably to prove) that no finite subcover also cover R.

    Alternatively, you can first try to prove this theorem:
    Theorem: A metric space is compact if and only if it is complete and totally bounded.​
    For a subset of finite-dimensional Euclidean space, totally bounded is just equivalent to bounded (as in: there exist [itex]a < b \in \mathbb{R}[/itex] such that the set is contained in [itex][a, b]^n[/itex]).
  6. Aug 4, 2008 #5
    Thank you!

    so it is not acceptable to count a particular (unbounded) open set as one of the covering open sets.
  7. Aug 4, 2008 #6


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    It is. But the point of being compact is that any open covering has a finite subcover. In particular, also that any covering in finite sets, such as the one I presented, has a finite subcover. So if you can construct some covering, of which no finite number of members is a cover itself, then it is not compact. (Note, that this is easier than proving that it is compact, for which you would have to show that no matter what cover you take, there is a finite subcover).

    In my example, you cannot make an unbounded set unless you take an infinite union.
  8. Aug 9, 2008 #7
    Thank you.
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