MAT158H1/Unit 05
Unit 05 · Analysis 1A

The Three Hard Theorems

Intermediate Value, Boundedness, and Extreme Value — the theorems that make calculus work. Each looks obvious, and each genuinely needs the completeness of $\mathbb{R}$ from Unit 2. This unit is where the least upper bound axiom earns its keep.

By the end you can
Why it matters & where it comes from
Why it exists

"A continuous function that changes sign has a root" and "a continuous function on a closed interval attains a maximum" feel obvious — but over $\mathbb{Q}$ both are false, and proving them over $\mathbb{R}$ is the whole point of building the reals carefully.

In the real world

The IVT guarantees that root-finding by bisection converges; the EVT guarantees that optimization problems actually have a solution to find. Both are silent assumptions in countless algorithms.

Story & open questions

Bolzano proved the IVT rigorously in 1817, decades ahead of his time and largely unread. The companion Bolzano–Weierstrass theorem (every bounded sequence has a convergent subsequence) carries both their names.

Why $\mathbb{Q}$ breaks all three

Take $f(x)=x^2-2$ on rational $[0,2]$. It is continuous, negative at $0$, positive at $2$ — and has no rational root: the graph slips through the hole at $\sqrt2$. Or take $g(x)=\dfrac{1}{x^2-2}$ on the same rational interval: continuous everywhere it is defined, yet unbounded. Completeness is not decoration; it is the load-bearing wall.

The Intermediate Value Theorem

Theorem IVT

If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$, then $f(c)=y$ for some $c\in[a,b]$.

Proof (for $f(a)<0<f(b)$, $y=0$; the general case applies this to $f-y$)
Let $A=\{x\in[a,b]:f(x)<0\}$. It contains $a$ and is bounded above by $b$, so $c=\sup A$ exists by completeness.
If $f(c)>0$: sign preservation (Unit 4) gives $\delta>0$ with $f>0$ on $(c-\delta,c+\delta)$. Then no point of $A$ exceeds $c-\delta$, so $c-\delta$ is an upper bound smaller than $c$ — contradiction.
If $f(c)<0$: then $c<b$ (as $f(b)>0$), and $f<0$ on some $(c-\delta,c+\delta)\subseteq[a,b]$, putting points of $A$ above $c$ — contradicting "upper bound." So $f(c)=0$. ∎
Example Roots exist

Every $c>0$ has an $n$th root: $f(x)=x^n-c$ is continuous, $f(0)=-c<0$, and $f(1+c)\ge1+nc-c>0$ by Bernoulli's inequality (Unit 1), so the IVT gives a root in $(0,1+c)$.

Boundedness, then Extreme Value

Theorem Boundedness

A continuous function on $[a,b]$ is bounded there.

Proof idea
Let $A=\{x\in[a,b]:f\text{ is bounded on }[a,x]\}$; it contains $a$ and is bounded above by $b$, so $\alpha=\sup A$ exists.
Continuity at $\alpha$ bounds $f$ on a small interval around $\alpha$ (take $\varepsilon=1$). Splicing this onto $[a,x]$ for some $x\in A$ close to $\alpha$ shows: $\alpha\in A$, and if $\alpha<b$ then points beyond $\alpha$ would also lie in $A$ — impossible. So $\alpha=b$ and $f$ is bounded on all of $[a,b]$. ∎
Theorem Extreme Value (EVT)

A continuous function on $[a,b]$ attains a maximum and a minimum value.

Proof
By boundedness, $M=\sup\{f(x):x\in[a,b]\}$ exists. Suppose $f(x)<M$ for every $x$.
Then $g=\dfrac{1}{M-f}$ is continuous on $[a,b]$, hence bounded: $g\le K$ for some $K>0$. That rearranges to $f\le M-\tfrac1K$, making $M-\tfrac1K$ a smaller upper bound than the least one — contradiction. So the max is attained; apply this to $-f$ for the min. ∎
Both hypotheses are essential $1/x$ on $(0,1]$ is continuous but unbounded (interval not closed at $0$); $x$ on $(0,1)$ is bounded but attains neither max nor min; $\mathbb{R}$ itself is closed but not bounded, and $f(x)=x$ attains nothing. And a single jump discontinuity breaks the IVT.
Common pitfalls

• The interval must be closed and bounded for boundedness and the EVT.
• The IVT gives existence of $c$, never its value or uniqueness.
• "Attains" means some $c$ has $f(c)=M$ — a supremum alone is not a maximum.
• Continuity is needed on the whole interval; one bad point can break all three theorems.

Practice

Original problems written for this guide in the style of Spivak's exercises. Write a full solution before opening the answer.

Practice 1 · A root

Show $x^3-x-1=0$ has a real root.

Answer
$f(1)=-1<0$ and $f(2)=5>0$; by the IVT a root lies in $(1,2)$.
Practice 2 · Fixed point

Show any continuous $f:[0,1]\to[0,1]$ has a fixed point: $f(c)=c$.

Solution
Let $g(x)=f(x)-x$; then $g(0)=f(0)\ge0$ and $g(1)=f(1)-1\le0$. If either is $0$ we are done; otherwise the IVT gives $g(c)=0$.
Practice 3 · Counterexample

Why does the EVT fail for $f(x)=x$ on $(0,1)$?

Answer
The interval is open: the sup $1$ and inf $0$ are limits of values but are never attained. No contradiction with the EVT, whose hypothesis requires a closed interval.
Practice 4 · Odd degree

Prove every polynomial of odd degree has a real root.

Solution
Divide by the leading coefficient, so $p(x)=x^n+\cdots$ with $n$ odd. For $|x|$ large the leading term dominates: $p(x)>0$ for large $x>0$ and $p(x)<0$ for large negative $x$ (odd power). The IVT on a large interval gives a root.
Practice 5 · Rational values

Suppose $f$ is continuous on $[0,1]$ and takes only rational values. Prove $f$ is constant.

Solution
If $f(x_0)\ne f(x_1)$, some irrational $y$ lies between them, and the IVT forces $f$ to attain $y$ — contradicting "only rational values."
Practice 6 · Equal values half apart

Let $f$ be continuous on $[0,1]$ with $f(0)=f(1)$. Prove there is $c\in[0,\tfrac12]$ with $f(c)=f(c+\tfrac12)$.

Solution
Let $g(x)=f(x+\tfrac12)-f(x)$ on $[0,\tfrac12]$. Then $g(0)+g(\tfrac12)=f(1)-f(0)=0$, so $g$ takes values of opposite sign (or a zero) at the endpoints; the IVT gives $g(c)=0$.
Practice 7 · Two roots

Show $x^4-2x-5=0$ has at least two real roots.

Solution
$p(0)=-5<0$, $p(2)=16-4-5=7>0$, and $p(-2)=16+4-5=15>0$. The IVT gives one root in $(-2,0)$ and another in $(0,2)$.
Practice 8 · Where EVT is used

Let $f$ be continuous on $[a,b]$ with $f(x)>0$ for all $x$. Prove there is $m>0$ with $f(x)\ge m$ for every $x$ — and show this can fail on $(a,b)$.

Solution
By the EVT, $f$ attains a minimum $m=f(c)>0$, which works. On an open interval it fails: $f(x)=x$ on $(0,1)$ is positive with infimum $0$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT158H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT158H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The spine is: completeness gives compactness behavior, compactness plus continuity gives extrema, and connectedness plus continuity gives intermediate values.

Study routine. For each theorem, memorize the hypotheses by finding examples that fail when one hypothesis is removed.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · IVT

Show that $\cos x=x$ has a solution in $(0,1)$.

Guide
Let $f(x)=\cos x-x$. It is continuous, $f(0)=1>0$, and $f(1)<0$, so IVT gives a root.
Extension 2 · EVT hypothesis

Give an example of a continuous function on an open interval that has no maximum.

Guide
The function $f(x)=x$ on $(0,1)$ is continuous but has no maximum because values approach $1$ without attaining it.
Extension 3 · Uniform continuity

Why is $f(x)=x^2$ uniformly continuous on $[0,10]$?

Guide
It is continuous on a closed bounded interval, so the Heine-Cantor theorem applies. Directly, $|x^2-y^2|\le20|x-y|$ there.
Extension 4 · Hypothesis failure

Why can IVT fail for a discontinuous function?

Guide
A jump function can take negative values on one side and positive values on the other while skipping $0$, so sign change alone is not enough.
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