- #1

Bashyboy

- 1,421

- 5

## Homework Statement

Let ##f : [0, 2 \pi) \rightarrow S^1##, where ##S^1## denotes the unit circle in plane, and ##f(x) := (\cos x, \sin x)##. I am trying to prove various properties of this map, the first of which is showing continuity.

## Homework Equations

A function ##f : M \rightarrow N## between the metric spaces ##M## and ##N## is continuous at some point ##p \in M## if and only if for every sequence ##(p_n)## in ##M## converging ##p##, the sequence ##(f(p_n))## converges to ##f(p)## in ##N##.

The sequence ##(p_n)## is said to converge to ##p## if and only if for every ##\epsilon > 0##, there exists a number number ##N_\epsilon## such that whenever ##n \ge N_\epsilon##, we have ##d(p,p_n) < \epsilon##.

## The Attempt at a Solution

My goal is show that continuity of ##f## reduces (somehow) to a question of whether trig function is a continuous map from between the real numbers, which we are given, although I am unsure whether this is possible. However, I encountered a simple problem along the way. Taking ##(p_n)## to be some arbitrary sequence that converges to some point ##p \in [0,2 \pi)##,

$$d(f(p),f(p_n)) = \sqrt{(\cos p - \cos p_n)^2 + (\sin p - \sin p_n)^2}$$

$$= \sqrt{\cos^2 p - 2 \cos p \cos p_n + \cos^2 p_n + \sin^2 p - 2 \sin p \sin p_n + \sin^2 p_n}$$

$$= \sqrt{-2(\cos p \cos p_n + \sin p \sin p_n)}$$

$$= \sqrt{-2 \cos(p_n-p)}$$

What am I to do with this negative sign, and the fact that ##\cos x## oscillates between ##-1## and ##1##?