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Lines, points and contradiction

  1. Aug 9, 2005 #1
    this question may sound kind of dumb, so if it is, i apologize, but let say you have a line with a point taken out of it. (<------- -------->) since a point does not have a defined length, how could you prove that the line is now two rays, and not still a line, because it seems a bit contradictory to say that something isn't connected but yet has no distance between it
  2. jcsd
  3. Aug 9, 2005 #2
    It is very simple, the parametrization (Is this word English?) cannot be a continuous (Is this word Maths?) function, then by definition, it is not a curve.
  4. Aug 9, 2005 #3
    Topologically a set is called connected if it can not be devided in two disjoint open subsets. Now consider the set A of a line with one point removed from it (for instance [tex]A = \mathbf{R}- \{ 0 \} [/tex]). We can devide A into two open subsets, namely the two rays (without their starting point). Because they are disjoint, A can not be connected.
  5. Aug 9, 2005 #4
    It is incorrect:

    [tex]S \subseteq \mathbb{R}^n[/tex] is connected [tex]\leftrightharpoons \exists A, B \subset \mathbb{R}^n : A \cup B = S , A \cap \overline{B} = \varnothing , B \cap \overline{A} = \varnothing[/tex]
  6. Aug 9, 2005 #5

    matt grime

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    it is unmathematical and meaningless in the english language to say something "has no distance between it"

    look at the definitions anyway (whatever those may be for you).
  7. Aug 10, 2005 #6
    Well, yes, that is if you do not require A and B to be open in S. Both statements are equivalent though.
    (Recall that A open in S ([tex]\subset \mathbb{R}^n[/tex]) means that there exist an open [tex]U \subset \mathbb{R}^n[/tex] such that [tex]U \cap S = A[/tex].)
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