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

  • Thread starter Bashyboy
  • Start date
  • #1
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##?
 

Answers and Replies

  • #2
35,499
11,946
You lost something between the third and second last expression.
 
  • #3
1,421
5
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
35,499
11,946
Should work.
 

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

  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
46K
Replies
4
Views
1K
Replies
7
Views
1K
M
Replies
2
Views
636
M
Replies
19
Views
27K
Replies
8
Views
1K
  • Last Post
Replies
6
Views
6K
Replies
3
Views
881
  • Last Post
Replies
4
Views
62K
Top