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Can open sets be written as unions of intervals?

  1. Sep 12, 2011 #1
    A theorem of real analysis states that any open set in [itex]\Re^{n}[/itex] can be written as the countable union of nonoverlapping intervals, where "interval" means a parallelopiped in [itex]\Re^{n}[/itex], and nonoverlapping means the interiors of the intervals are disjoint. Well, what about something as simple as an open ball in [itex]\Re^{2}[/itex] or [itex]\Re^{3}[/itex]? Intuitively, I can't visualize how non-overlapping rectangles, even a countably infinite number of them, could ever make a circle. If you could write it as such a union, then just pick a point on the circumference of the circle: it is either a corner of a rectangle or on the side of a rectangle. Either way, it would not look like a circle near that point.

    Any help would be greatly appreciated.

    Thank You in Advance.
     
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  3. Sep 12, 2011 #2

    micromass

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    The point on the circumference of the circle does not belong to the open set. This is crucial. An open set does not contain its boundary points. So points on the circumference is something we do not need to worry about.

    I agree that the theorem is counterintuitive, and I find it difficult to visualize it myself. But the theorem is true.
     
  4. Sep 12, 2011 #3
    First of all, are all the intervals in the theorem closed intervals? If they are, then I find it bizarre that the points on the circumference do not belong to the union of intervals.
     
  5. Sep 12, 2011 #4

    micromass

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    Open intervals can be written as the union of open intervals, that is what the theorem says. So we're dealing with open sets here...
     
  6. Sep 12, 2011 #5
    OK, Wheeden and Zymund say every open set can be written as the union of nonoverlapping closed cubes, and they might have made a mistake, but in the interior of the circle, if you exclude the edges of rectangles, then aren't there points missing when you look at the boundary between two rectangles?
     
  7. Sep 12, 2011 #6

    micromass

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    OK, closed intervals work as well actually. And no, there aren't any points missing. It's really hard to visualize, though. The point is that the intervals can be made arbitrary small and close to each other.
     
  8. Sep 12, 2011 #7
    OK, I can actually kind of visualize the open-interval version: within every gap between open intervals you can put an open interval, and within the new gaps you can put even smaller open intervals.

    But I'm trying to understand the closed interval version. In the proof of the theorem given by Wheeden and Zymund, the closed intervals are adjacent to one another. So my first question is, in this version are ALL the intervals closed? Or is it only the intervals that are entirely in the interior of the circle that are closed, but the intervals that touch the "open air" have (parts of) their boundaries excluded? If all the intervals are closed, that would be really counterintuitive.
     
  9. Sep 16, 2011 #8
    Just to clarify what theorem I'm talking about, attached is the statement and proof of the result from Wheeden and Zygmund.
     

    Attached Files:

  10. Sep 17, 2011 #9

    Stephen Tashi

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    What does "non overlapping" mean in that book? - no common points? - or no common volume?
     
  11. Sep 17, 2011 #10
    It means the interiors are disjoint. Two intervals are allowed to share a boundary.
     
  12. Sep 17, 2011 #11

    Hurkyl

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    Ah, this makes a big difference! I don't believe an arbitrary open set in Rn (for n > 1) can be written as a disjoint union of open n-cubes.

    In fact, I don't even believe an open interval in R1 can be written as the disjoint union of 2 or more open intervals.


    (aside: my gut thinks you might be able to do better than "non-overlapping" closed cubes -- that you can actually have "disjoint")

    You missed a third and very important case: it is not on any rectangle at all. (but there is an infinite sequence of rectangles approaching it)
     
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