Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that a continuous function on an interval takes on every value between its endpoints. This intuitive result — continuous functions have "no jumps" — relies fundamentally on the completeness of $\mathbb{R}$ . The IVT guarantees roots of equations, fixed points, and is the basis for the bisection method in numerical analysis.

Statement

Theorem4.1Intermediate Value Theorem

Let $f : [a, b] \to \mathbb{R}$ be continuous. If $y$ is any value between $f(a)$ and $f(b)$ (i.e., $f(a) \leq y \leq f(b)$ or $f(b) \leq y \leq f(a)$ ), then there exists $c \in [a, b]$ such that $f(c) = y$ .

RemarkConnectedness

The IVT is equivalent to saying that the continuous image of a connected set (an interval) is connected. Since connected subsets of $\mathbb{R}$ are intervals, $f([a, b])$ is an interval, hence contains all values between its endpoints.

Proof

Without loss of generality, assume $f(a) < y < f(b)$ (the case $f(a) > y > f(b)$ is similar). Define

$S = \{x \in [a, b] \mid f(x) < y\}.$

Then $S \neq \varnothing$ (since $a \in S$ ) and $S$ is bounded above (by $b$ ). By completeness, $c = \sup S$ exists.

Claim: $f(c) = y$ .

We show $f(c) \not< y$ and $f(c) \not> y$ , leaving $f(c) = y$ .

Case 1: Suppose $f(c) < y$ . Since $f$ is continuous at $c$ , there exists $\delta > 0$ such that $|x - c| < \delta \implies |f(x) - f(c)| < (y - f(c))/2$ . In particular, for $x \in (c, c + \delta) \cap [a, b]$ ,

$f(x) < f(c) + \frac{y - f(c)}{2} < y.$

Thus $(c, c + \delta) \cap [a, b] \subseteq S$ , contradicting that $c = \sup S$ (since $c + \delta/2 > c$ is in $S$ ).

Case 2: Suppose $f(c) > y$ . By continuity, there exists $\delta > 0$ such that $|x - c| < \delta \implies |f(x) - f(c)| < (f(c) - y)/2$ . For $x \in (c - \delta, c) \cap [a, b]$ ,

$f(x) > f(c) - \frac{f(c) - y}{2} > y.$

Thus $f(x) > y$ for all $x \in (c - \delta, c)$ , so $c - \delta$ is an upper bound for $S$ , contradicting that $c = \sup S$ .

Thus $f(c) = y$ .

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Applications

ExampleExistence of roots

If $f(a) < 0 < f(b)$ , the IVT guarantees a root $c \in (a, b)$ with $f(c) = 0$ . For instance, $f(x) = x^3 + x - 1$ satisfies $f(0) = -1 < 0$ and $f(1) = 1 > 0$ , so $f$ has a root in $(0, 1)$ .

ExampleBrouwer fixed-point theorem (1D)

If $f : [0, 1] \to [0, 1]$ is continuous, then $f$ has a fixed point: there exists $x \in [0, 1]$ with $f(x) = x$ .

Proof: Consider $g(x) = f(x) - x$ . Then $g(0) = f(0) \geq 0$ and $g(1) = f(1) - 1 \leq 0$ . By IVT, there exists $c \in [0, 1]$ with $g(c) = 0$ , i.e., $f(c) = c$ .

ExampleBisection method

To approximate a root of $f(x) = 0$ where $f(a) f(b) < 0$ , repeatedly bisect the interval: if $f((a+b)/2) < 0$ , set $a' = (a+b)/2$ ; otherwise set $b' = (a+b)/2$ . The IVT guarantees each subinterval contains a root. After $n$ bisections, the root is localized to an interval of width $(b - a)/2^n$ .

ExampleNo discontinuous bijection [a, b] → R

There is no bijection $f : [0, 1] \to \mathbb{R}$ that is continuous. Proof: if $f$ were such a bijection, then by the IVT, $f([0, 1])$ would be an interval. But the only intervals mapping bijectively to $\mathbb{R}$ are unbounded intervals, contradicting $f([0, 1])$ being the continuous image of a compact set (which must be compact, hence closed and bounded).

IVT fails without completeness

ExampleIVT fails over Q

Define $f : [0, 2] \cap \mathbb{Q} \to \mathbb{Q}$ by $f(x) = x^2$ . Then $f(0) = 0$ and $f(2) = 4$ , so $0 < 2 < 4$ . However, there is no $c \in [0, 2] \cap \mathbb{Q}$ with $f(c) = 2$ (since $\sqrt{2} \notin \mathbb{Q}$ ). The IVT fails over $\mathbb{Q}$ because $\mathbb{Q}$ is not complete.

Summary

The Intermediate Value Theorem is fundamental in analysis:

Continuous functions on intervals take all intermediate values.
Proof uses completeness (existence of suprema).
Applications: root-finding, fixed points, bisection method.
Equivalent to connectedness of intervals.

See Extreme Value Theorem and Connectedness for related results.