2

Can somebody tell me how to prove that $f:[0,2π)→S^1$ given by $t↦⟨\cos t,\sin t⟩$, where $S^1$ is the unit circle in the plane, and $[0,2π)$ is the real interval, 1. is continuous at the point 0 (the Problem I am facing consists on the fact that if we consider an open interval $V$ at 0,for example $V=(0-d, 0+d)$, I can prove that there exists an open Interval $V'$ of the circle such that $f(V)$ is part of $V'$. But at the same time given the interval $[0,2π)$ we must have $V=(2π-d,0+d)$. In this case $f(V)$ is Not part of $V'$ rather then other way round. I think I am confusing concepts length and point. This is all about the critical point (0,0) of the circle,the other points are not concerned. And 2. that the inverse function of f is not continuous at the point (1,0) of the circle.

Lot of thanks for any comment.

Martin Sleziak
  • 53,687
ivo
  • 21

1 Answers1

3

You know that functions $\sin$ and $\cos$ are continuous. So the function $g\colon \mathbb R\to \mathbb R^2$ given by $g(x) = (\sin x, \cos x)$ is continuous. Your function is nothing else than a restriction of $g$.

To show that the inverse is not continuous take the sequence $p_n \in \mathbb S^1$ given by $p_n=f(2\pi-1/n)$. Notice that $p_n\to(1,0)$ but $f^{-1}(p_n) \to 2\pi \neq 0 = f^{-1}(1,0)$.

Emanuele Paolini
  • 21,447