# Showing a Function From $[0, 2 \pi)$ to $S^1$ is cont.

1. Mar 2, 2017

### Bashyboy

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 $d(p,p_n) < \epsilon$.

3. 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 $1$?

2. Mar 2, 2017

### Staff: Mentor

You lost something between the third and second last expression.

3. Mar 2, 2017

### Bashyboy

Ah! Shoot! Here the correct expression is $\sqrt{2-2 \cos(p_n - p)}$. So, since $2 - 2 \cos(p_n-p) \ge 0$, we can write $d(f(p_n),f(p))^2 = |2 - 2 \cos(p_n-p)|$. Thus, the problem reduces to showing that for every $\epsilon > 0$, there exists a natural number $N$ such that

$$|2 - 2 \cos(p_n-p)| < \epsilon^2$$ for all $n \ge N$.

Let $g(x) = 2 \cos x$. Since the cosine function is continuous at every $x$, this last statement is merely an assertion that $g(x)$ is continuous at $x=0$ using the sequence $(p_n-p)$, which converges to zero. That is, we have that for every $\epsilon^2 > 0$, there exists an $N' \in \mathbb{N}$ such that $|2 \cos(p_n-p) - 2 \cos 0| < \epsilon^2$ or $|2 \cos (p_n -p) - 2| < \epsilon^2$ for every $n \ge N'$, which we know holds because $g$ is continuous.

Does this appear correct? It appears so, since each step is an "if and only if."

4. Mar 2, 2017

Should work.