is the map $g:T\to [0,2\pi),\; e^{ir}\to r$ continuous?

Question

$T=\{z\in\mathbb{C}: |z|=1\}$. Is the map $g:T\to [0,2\pi),\; e^{is}\to s$ continuous?

Our teacher said, that $g$ is continuous on $T\setminus \{1\}$, what I don't understand. I tried to find a sequence $(a_n)\subseteq T$ such that $a_n\to 1$ but $g(a_n)\nrightarrow g(1)$. First of all, $g(1)=0$, because $e^{is}=1$ if $s=2n\pi$, $n\in\mathbb{Z}$. But I'm stuck.

Why is $g$ continuous in $T\setminus \{1\}$, and why is $g$ discontinuous in 1?

oh great, thanks! $a_n\to 1$ but $g(a_n)\to 2\pi\neq g(1)$. How can I argue that g is continuous on $T\setminus {1}$? — taglap, Aug 20 '15 at 15:12
There are many ways. Which is the most appropriate depends on what theory is available at this point. — Daniel Fischer, Aug 20 '15 at 15:18
In a neighbourhood of each point, you can write $g$ as a simple function of $\arcsin y$ or $\arccos x$. The maps $(x+iy) \mapsto x$ and $(x+iy) \mapsto y$ are continuous, so are $\arcsin$ and $\arccos$, hence the composition. — Daniel Fischer, Aug 20 '15 at 15:23

score 1 · Answer 1 · answered Aug 20 '15 at 15:08

1

$g$ can't be continuous on $T$ because $T$ is compact, and $[0,2\pi)$ is not.

answered Aug 20 '15 at 15:08

Tim Raczkowski

20,046

thanks. this is helpful, now I know how to see that g can't be continuous on $T$. – taglap Aug 20 '15 at 15:14

is the map $g:T\to [0,2\pi),\; e^{ir}\to r$ continuous?

1 Answers1