The Timpson Lyceum · Real Analysis (2020)

Lecture 10 (Real Analysis 2020)

Overview

This lecture proves two of the central theorems about continuous functions on closed intervals: the Extreme Value Theorem and the Intermediate Value Theorem. Both match the intuition that a continuous function drawn without lifting your pen must "fill in" all its intermediate values and reach its highest and lowest points — but that intuition is not the definition, so each statement requires proof from the sequential definition of continuity. Along the way we establish a boundedness lemma, recall the Sequential Compactness Theorem, and prove the Nested Interval Theorem (the tool needed for the IVT proof). We close with the corollary that a continuous function maps intervals to intervals.

Throughout, continuity means: if $x_n \to x$ in the domain, then $f(x_n) \to f(x)$. Every "obvious" picture-fact below is earned from that single definition.

Key Concepts

Setup and notation. For $f : D \to \mathbb{R}$, the set $D$ is the domain and the image is $f(D) = \{\, y : \exists\, x \in D,\ f(x) = y \,\}.$ A function attains a maximum (minimum) if its image $f(D)$ has a largest (smallest) element. The point $x_1$ where $f(x_1)$ is the max is the maximizer; similarly the minimizer.

Why a closed interval matters. On open intervals extrema can fail to exist:

• $f(x) = 2x$ on $(0,1)$ has image $(0,2)$ — no max, no min (no smallest positive number).
• $f(x) = \tfrac{1}{x}$ on $(0,1)$ has image $(1,\infty)$ — unbounded above (since $f(\tfrac1n) = n$), so no max.

Boundedness Lemma. If $f : [a,b] \to \mathbb{R}$ is continuous, then $f$ is bounded. (Proof by contradiction; uses Sequential Compactness — see Worked Example.)

Extreme Value Theorem. If $f : [a,b] \to \mathbb{R}$ is continuous, then $f$ attains a maximum and a minimum value on $[a,b]$.

• Proof idea (max case): By the lemma $f([a,b])$ is bounded above, so by completeness it has a supremum $C$. Build $x_n \in [a,b]$ with $C - \tfrac1n < f(x_n) \le C$, so $f(x_n) \to C$. A convergent subsequence $x_{n_k} \to x_c$ gives, by continuity, $f(x_c) = C$. Since the supremum is attained, it is the maximum.

Sequential Compactness Theorem (recalled). A sequence inside a closed interval $[a,b]$ is bounded, hence (by Bolzano–Weierstrass) has a convergent subsequence, and its limit again lies in $[a,b]$.

Nested intervals. A sequence of closed intervals $I_n = [a_n, b_n]$ is nested if $I_{n+1} \subseteq I_n$ for all $n$, equivalently $a_n \le a_{n+1} \le b_{n+1} \le b_n.$

Nested Interval Theorem. If $\{I_n\}$ are nested and $\lim_{n\to\infty}(b_n - a_n) = 0$, then there exists exactly one point $x$ with $x \in I_n$ for all $n$, and $\lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n = x.$

• Proof idea: $(a_n)$ is increasing and bounded above (every $b_n$ is an upper bound), so it converges to its supremum $x$. Since $b_n - a_n \to 0$, $(b_n) \to x$ too. As $x$ is the sup of $(a_n)$ and inf of $(b_n)$, it is the unique number sandwiched in every interval.

Intermediate Value Theorem. If $f : [a,b] \to \mathbb{R}$ is continuous and $C$ is any value between $f(a)$ and $f(b)$, then there exists $x_0 \in (a,b)$ with $f(x_0) = C$.

• Proof idea (bisection, assuming $f(a) \le f(b)$): Start with $[a_1,b_1] = [a,b]$. At each step take the midpoint $m_n$; keep the half whose endpoint values still straddle $C$ — if $f(m_n) \le C$ set $a_{n+1} = m_n$, else $b_{n+1} = m_n$. This keeps $f(a_n) \le C \le f(b_n)$ and halves the length, so $b_n - a_n = (b-a)(\tfrac12)^{n-1} \to 0$. By the Nested Interval Theorem there is a unique $x_0$ with $a_n, b_n \to x_0$. Continuity gives $f(a_n) \to f(x_0)$ and $f(b_n) \to f(x_0)$; since $f(a_n) \le C \le f(b_n)$, both limits equal $C$, so $f(x_0) = C$.

Applications of the IVT.

• Square roots exist: every $C > 0$ has a positive square root.
• Roots of high-degree polynomials: there is no quadratic-formula analogue for degree $\ge 5$, but the IVT still certifies that a solution exists. E.g. $x^5 + x + 1 = 0$ has a root in $(-2, 0)$ since $f(-2) = -33 < 0 < 1 = f(0)$.

Convex sets and the image of an interval. A set $D \subseteq \mathbb{R}$ is convex if for any two points in $D$ all points between them are also in $D$; on $\mathbb{R}$ this is exactly "being an interval." Corollary of the IVT: if $I$ is an interval and $f : I \to \mathbb{R}$ is continuous, then $f(I)$ is an interval — continuous images of intervals are intervals.

Worked Example

Boundedness Lemma — $f$ continuous on $[a,b]$ is bounded.

Suppose, for contradiction, $f$ is not bounded above. Then for every $n \in \mathbb{N}$ there is a point $x_n \in [a,b]$ whose value exceeds $n$. This defines a sequence whose images are forced upward; a convergent subsequence then forces its image to be bounded — a contradiction.

A symmetric argument rules out unboundedness below, so $f$ is bounded.

IVT application — every $C > 0$ has a positive square root.

Define a continuous function on a closed interval whose values bracket $C$, then invoke the IVT.

So $x_0 = \sqrt{C}$ exists and is positive.

Summary

• On a closed interval, a continuous function is bounded and attains its maximum and minimum (Extreme Value Theorem); on open intervals these can fail.
• The proofs run through sequences: build a sequence whose images approach the target value, extract a convergent subsequence via Sequential Compactness, then apply the sequential definition of continuity.
• The Nested Interval Theorem — nested closed intervals with lengths shrinking to $0$ pin down exactly one common point — is the engine behind the IVT.
• The Intermediate Value Theorem says a continuous function on $[a,b]$ hits every value between $f(a)$ and $f(b)$; it is proved by repeated bisection to construct the right nested intervals.
• The IVT proves existence of square roots and of roots of polynomials (e.g. degree-5 equations) even when no solving formula exists.
• As a corollary, continuous functions map intervals to intervals; on $\mathbb{R}$, convex sets are exactly intervals.