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Do open sets in R^2 always have continuous boundaries?

  1. Aug 8, 2013 #1
    My basic question is this: does an arbitrary open set in ℝ2 look like a bunch of regions bounded by continuous curves, or are there open sets with weirder boundaries than that? Let me state my question more formally.

    A Jordan curve is a continuous closed curve in ℝ2 without self-intersections. The Jordan curve theorem states that the complement of any Jordan curve has two connected components, an interior and an exterior.

    Now let's define a simple unbounded curve to be a continuous map f: (-∞,∞) → ℝ2 such that f((-∞,0)) and f((0,∞)) are both unbounded. Then does the complement of a simple unbounded curve always have two connected components? It seems intuitively true, since you'd expect curve to have two sides, but considering how long it took to prove the Jordan curve theorem, things may not be as straightforward as they appear.

    Assuming it is true, let us call a "side" any connected component of a complement of a continuous curve in ℝ2. (Note that some continuous curves in ℝ2 have a connected complement.) Then my question is, can any open set U in ℝ2 be written as a countable union of disjoint open sets U_1, U_2, ..., such that each U_i is either a side or an intersection of two sides? If not, what if we let each U_i be a finite intersection of sides, rather than just an intersection of two sides?

    Can this be generalized to higher dimensions?

    Any help would be greatly appreciated.

    Thank You in Advance.
     
  2. jcsd
  3. Aug 9, 2013 #2
    I think my statement, as posed, is trivially false. For instance, R^2 with one or more isolated points taken out. So let's restrict the question to *simply connected* open sets U.

    Also, it looks like my proposed generalization of the Jordan curve theorem for simple unbounded curves is wrong. So let me modify it: let's assume that the curve goes to infinity in both directions, i.e. the limit of |f(t)| as to goes to plus or minus infinity is infinity.

    Given these modifications, is it true that any simply connected open set U can be written as a countable union of disjoint open sets U_1, U_2, ..., such that each U_i is one of two things:
    1. A connected component of the complement of either a Jordan curve or a simple unbounded curve going to infinity in both directions
    2. An intersection of two sets of type 1.
     
  4. Aug 11, 2013 #3
    An open set can be written as the union of multiple disjoint non-empty open sets if and only if it is disconnected. Therefore another way of phrasing this statement is:

    Any simply-connected connected open set U can be written as an open set of type 1 or type 2.

    To see that there exists counterexamples to this just let U be the open set obtained as the set of all points that are <1 from one the coordinate axes.

    If we allow all finite intersection in 2, then that is still not enough as we could generalize the previous example by choosing countable many distinct lines going through 0 and defining U to be all points that are close to one of these lines (say of a distance of <1 from one of the lines).
     
  5. Sep 5, 2013 #4
    I like this example, but do you mean U is the set of all points not on any of the lines of distance < 1 from some line? I don't see how the set could be open if include pieces of the lines.

    And wouldn't the set of all points not on any of the lines be an open set?
     
  6. Sep 6, 2013 #5

    mathwonk

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    did you mean for f to be injective?
     
  7. Sep 6, 2013 #6
    Yes, I do want f to be injective. I want no self-intersections.
     
  8. Sep 12, 2013 #7
    Im not sure if I completely understand what you're asking. But consider the set S = [(-1,1)X(-1,-1/2)]UB[(-1,0),1]UB[(-1,0),1]:

    2.png

    Where B[(x,y),r] is the open ball at (x,y) with radius r. It's open, and it's simply connected since the origin is not in the set. But the origin is a self-intersection point of the boundary, and the compliment of the boundary is the union of three "sides".

    We could extend the boundary out to infinity by considering A = SU[(1,infty)X(-1,-1/2)]
     
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