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

In summary, the goal of the given problem is to show that the trig function is a continuous map from between the real numbers, but 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##.
  • #1
Bashyboy
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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##?
 
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  • #2
You lost something between the third and second last expression.
 
  • #3
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
Should work.
 

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

1. What does it mean for a function to be continuous?

A function is continuous if there are no sudden jumps or breaks in its graph. This means that as the input values of the function change, the output values change in a smooth and continuous manner.

2. What is the significance of a function being continuous from ##[0, 2 \pi)## to ##S^1##?

This means that the function maps the interval ##[0, 2 \pi)## onto the unit circle, or the set of points that are equidistant from the center. This is important because it allows us to visualize the function in a more understandable way and make predictions about its behavior.

3. How can we show that a function is continuous from ##[0, 2 \pi)## to ##S^1##?

In order to show that a function is continuous, we must demonstrate that it satisfies the three conditions of continuity: 1) the limit of the function exists at the given point, 2) the function is defined at the given point, and 3) the limit and the function value are equal. We can do this by using the definition of continuity and analyzing the behavior of the function at different points in the interval.

4. What is the intuitive explanation for a function being continuous from ##[0, 2 \pi)## to ##S^1##?

An intuitive explanation for a function being continuous from ##[0, 2 \pi)## to ##S^1## is that it does not have any breaks or gaps in its graph. This means that as we move along the interval ##[0, 2 \pi)##, the corresponding points on the unit circle also move along in a smooth and unbroken manner, without any sudden jumps or changes.

5. Can a function be continuous from ##[0, 2 \pi)## to ##S^1## if it has a discontinuity at one point?

No, a function cannot be continuous if it has a discontinuity at any point within the given interval. This is because one of the conditions for continuity is that the limit and function value must be equal, and if there is a break or gap in the function at a certain point, the limit and function value will not be equal.

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