Intermediate Value Theorem
Table of Contents
1. Definition
The Intermediate Value Theorem (IVT) states the following:
Let \( f: [a, b] \to \mathbb{R} \) be some function continuous on the interval \( [a, b] \), with \( f(a) = A \) and \( f(b) = B \). Let \( C \in \mathbb{R} \) be such that \( min\{A, \ B\} < C < max\{A, \ B\} \), then:
\[ \exists c \in [a, b] \ s.t. \ f(c) = C \]
And:
\[ S = \{x : x \in [a, b], f(z) \le C \ \forall z \in [a, x] \} \] \[ c = \sup\{S\} \]
2. Proof
Suppose WLOG that \( f(a) < f(b) \). Note since \( S \) is bounded above by \( b \) and nonempty, it must have a supremum by the completeness property of the reals, hence \( c \) is defined.
2.1. Case 1
Now suppose \( f(c) < C \implies C - f(c) = \epsilon > 0 \). Now since \( f \) is continuous on \( [a, b] \):
\[ \exists \delta > 0 \ s.t. \ x \in (c - \delta, c + \delta) \implies \ |f(c) - f(x)| < \epsilon \]
Now consider any :
\[ c^* \in (c, c + \delta) \]
Note since \( |f(c^*) - f(x)| < \epsilon \implies f(c^*) < f(c) + \epsilon = C \) there is no \( c^* \) such that \( f(c^*) > C \), hence all \( c^* \) are in \( S \), but this is a contradiction as their memebership in the set would mean
\[ c \ne sup\{S\} \]
So we can conclude that our assumption: \( f(c) < C \) is false.
2.2. Case 2
Alternatively, suppose \( f(c) > C \implies f(c) - C = \epsilon > 0 \). By continuity:
\[ \exists \delta > 0 \ s.t. \ x \in (c - \delta, c + \delta) \implies \ |f(c) - f(x)| < \epsilon \]
Now choose \( c' = c - \frac{\epsilon}{2} \). We prove the above supposition to be false by considering the membership of \( c' \) to \( S \). We know from the membership predicates of \( S \) that \( c \) cannot be in \( S \). Suppose:
\[ c' \not \in S \implies \forall x \ge c', \ x \not \in C \implies c' \text{ is an upper bound of } S \]
But that would contradict the definition of \( c \) being the least upper bound of \( S \), hence we must have \( c' \in S \).
\[ c' \in S \implies f(c') \le C \]
But we know from continuity: \( f(c) - f(c') \le \epsilon \)
\begin{align} & \implies f(c) - f(c') \le f(c) - C\\ & \implies C \le f(c')\\ & \implies c' \not \in S \end{align}Which again arrives us at a contradiction, hence the statement \( f(c) > C \) is false.
2.3. Synthesis
Given that the statements \( f(c) > C \) and \( f(c) < C \) are both false, we deduce that \( f(c) = C \) which proves the theorem.