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Math Help - Connectivity

  1. #1
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    Connectivity

    Let A be a union of a countable number of "lines" in \mathbb{R}^3.
    Prove that B := \mathbb{R}^3 - A is connected.

    This seems intuitively true - take two points a, b \in B. Look at the line  [a, b] = \{ ta + (1-t)b \ : \ t \in [0,1] \}. If a point on the line intersects one of the lines in A, then simply deviate the connecting path a little bit, such that you don't intersect with the next closest line.

    Making this rigorous, however, seems a bit painful, moreso considering that it's a question from a previous exam. Any ideas/hints are welcome.

    Thanks, Defunkt.
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  2. #2
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    Quote Originally Posted by Defunkt View Post
    Let A be a union of a countable number of "lines" in \mathbb{R}^3.
    Prove that B := \mathbb{R}^3 - A is connected.

    This seems intuitively true - take two points a, b \in B. Look at the line  [a, b] = \{ ta + (1-t)b \ : \ t \in [0,1] \}. If a point on the line intersects one of the lines in A, then simply deviate the connecting path a little bit, such that you don't intersect with the next closest line.

    Making this rigorous, however, seems a bit painful, more so considering that it's a question from a previous exam. Any ideas/hints are welcome.
    There are uncountably many planes containing the two points a and b. They are "almost disjoint", in the sense that the intersection of any two of them is just the line joining a to b. Since there are only countably many lines in A, at least one of these planes does not contain any of the lines.

    Choose such a plane. It contains at most one point from each line in A. So that reduces the problem from three dimensions to two. The two-dimensional problem is this: Given a countable set of points B in a plane, show that the complement of B is connected.

    The proof of that is a similar counting argument. Given two points a and b in the plane, there are uncountably many disjoint circular arcs connecting a and b, with the property that any two of them are almost disjoint (in the sense that their intersection consists only of the points a and b). Only countably many of these can contain a point in B. Choose an arc containing no such points. That demonstrates that the complement of B is connected (in fact, path-connected).
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  3. #3
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    Thanks for the quick answer. I asked the prof. how to do it as well, at first he didn't come up with an answer but he proposed the same solution the next time I saw him.
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