Path Connectedness

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  • Опубліковано 22 гру 2024

КОМЕНТАРІ • 8

  • @drewduncan5774
    @drewduncan5774 5 років тому +11

    "The proof that the Topologist's Sine Curve is not path connected is left as an exercise to the viewer."

    • @DanielChanMaths
      @DanielChanMaths  5 років тому +4

      Yes. But here's a hint which gives the rough argument. Let f(x) = sin(1/x). Suppose (x(t),y(t)) is the path joining (X,f(X)) to (0,Y) so x(0) = X, y(0) = f(X), x(1) = 0 and y(1) = Y. As t goes to 1, x(t) goes to 0. But essentially, y(t) = f(x(t)) then has no limit so doesn't converge to Y. There are some technical details that need to be sorted out, but that's the basic idea.

    • @topiado2073
      @topiado2073 5 років тому +1

      Waoo sir, sir m an Indian girl thank you

  • @filippogovi2340
    @filippogovi2340 11 місяців тому

    Hello,
    When you prove that the path from x to z is transitive, can you really speak of a composition? I mean, as defined f:[0,1]->X will map either to x or y, what you would have to do is show that either [0,1] is homeomorphic to the set connecting f(0)=x to f(1)=y, or you could define a piecewise function g:[0,2]-->y such that g(x)={(f(x), if 0\leq x < 1) and (f'(x-1), if 1 \leq x < 2), where f:[0,1] --> X such taht f'(0)=x and f'(1)=y. Then since [0,2] is an interval of R, we can very easily prove it is connected. Am I mistaken?
    I apologise if this is completely wrong.
    Cheers!

  • @laxmividyaclasses9770
    @laxmividyaclasses9770 4 роки тому

    I am from India. Your explanation is very easy to understand. Thanks for this video.

  • @aminu.j
    @aminu.j 2 роки тому

    Great work

  • @drewduncan5774
    @drewduncan5774 5 років тому +1

    The wait is over

  • @mathbooster977
    @mathbooster977 4 роки тому

    Wow...