Section The Intermediate Value Theorem and a Fixed Point Theorem
In this section we present two important consequences of connectedness. The first consequence is the Intermediate Value Theorem. In calculus, the Intermediate Value Theorem tells us that if \(f\) is a continuous function on a closed interval \([a,b]\text{,}\) then \(f\) assumes all values between \(f(a)\) and \(f(b)\text{.}\) This result seems straightforward if one just draws a graph of a continuous function on a closed interval. But a picture is not a proof. We state and then prove a more general version of the Intermediate Value Theorem.
Theorem 18.14. The Intermediate Value Theorem.
Let \(X\) be a topological space and \(A\) a connected subset of \(X\text{.}\) If \(f : A \to \R\) is a continuous function, then for any \(a,b \in A\) and any \(y \in \R\) between \(f(a)\) and \(f(b)\text{,}\) there is a point \(x \in A\) such that \(f(x) = y\text{.}\)
Activity 18.10.
In this activity we prove the Intermediate Value Theorem. Let \(X\) be a topological space and \(A\) a connected subset of \(X\text{.}\) Assume that \(f : A \to \R\) is a continuous function, and let \(a,b \in A\text{.}\)
(a)
Explain why we can assume that \(a \neq b\text{.}\)
(b)
Explain what happens if \(y = f(a)\) or \(y=f(b)\text{.}\)
(c)
Now assume that \(f(a) \neq f(b)\text{.}\) Without loss of generality, assume that \(f(a) \lt f(b)\text{.}\) Why can we say that \(f(A)\) is an interval?
(d)
How does the fact that \(f(A)\) is an interval complete the proof?
Our second consequence of connectedness involves a fixed point theorem. Fixed point theorems are important in mathematics. A fixed point of a function \(f\) is an input \(c\) so that \(f(c) = c\text{.}\) There are many fixed point theorems — one of the most well-known is the Brouwer Fixed Point Theorem that shows that every continuous function from a closed ball \(B\) in \(\R^n\) to itself must have a fixed point. We can use the Intermediate Value Theorem to prove this result in \(\R\text{.}\)
Activity 18.11.
In this activity we prove the following theorem.
Theorem 18.15.
Let \(a,b \in \R\) with \(a \lt b\text{,}\) and let \(f : [a,b] \to [a,b]\) be a continuous function. Then there is a number \(c \in [a,b]\) such that \(f(c) = c\text{.}\)
So let \(a,b \in \R\) with \(a \lt b\text{,}\) and let \(f : [a,b] \to [a,b]\) be a continuous function.
(a)
Explain why we can assume that \(a \lt f(a)\) and \(f(b) \lt b\text{.}\)
(b)
Define \(g : [a,b] \to \R\) by \(g(x) = x-f(x)\text{.}\)
(i)
Why is \(g\) a continuous function?
(ii)
What can we say about \(g(a)\) and \(g(b)\text{?}\) Use the Intermediate Value Theorem to complete the proof.
(c)
Let \(f(x) = \frac{1}{6}x^3+\frac{1}{4}x\text{.}\)
(i)
Show that \(f\) maps the interval \([-1,2]\) into the interval \([-1,2]\text{.}\)
Hint.Use
Theorem 17.14 that a continuous function from a compact topological space to
\(\R\) assumes a maximum and minimum value.
(ii)
Find all values of \(c\) in \([-1,2]\) such that \(f(c) = c\text{.}\)
