3.2 Compactness, extreme values and uniform continuity

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

Uniform continuity

Definition

One delta for the entire domain

A function $f:A\to\mathbb R$ is uniformly continuous on $A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in A$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. The chosen $\delta$ can depend on $\varepsilon$ but not on $x$ or $y$.

Theorem

Heine–Cantor theorem

Every continuous real-valued function on a compact set is uniformly continuous.

Proof

Contradiction using compactness

If uniform continuity failed, some $\varepsilon_0>0$ would permit, for every $n$, points $x_n,y_n\in K$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge\varepsilon_0$. Compactness supplies a subsequence $x_{n_k}\to x\in K$. Since the paired distances tend to zero, also $y_{n_k}\to x$. Continuity gives both $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$, contradicting the output separation.

Worked example

Pointwise but not uniform on an unbounded domain

The function $f(x)=x^2$ is continuous on $\mathbb R$ but not uniformly continuous there. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|=1/n\to0$, but $|f(y_n)-f(x_n)|=2+1/n^2\not\to0$. On any closed bounded interval, Heine–Cantor does give uniform continuity. It is also direct: if $|x|,|y|\le M$, then $|x^2-y^2|\le2M|x-y|$.

Method comparison

Use compactness and continuity to guarantee boundedness, attainment and one common continuity scale; check both closedness and boundedness in $\mathbb R$. Use a Lipschitz estimate $|f(x)-f(y)|\le C|x-y|$ to prove uniform continuity directly, even on noncompact domains. To disprove uniform continuity, construct pairs whose input distances approach zero while output differences stay above a fixed positive threshold.

Practice

1. Prove $f(x)=3x-2$ is uniformly continuous on $\mathbb R$.
2. Show $f(x)=1/x$ is not uniformly continuous on $(0,1)$.
3. Does every continuous function on $[0,1]$ attain its maximum and minimum? State the theorem used.
4. Is compactness necessary for $f(x)=0$ to be uniformly continuous on $\mathbb R$?

Full solutions

1. Given $\varepsilon>0$, choose $\delta=\varepsilon/3$. Then $|f(x)-f(y)|=3|x-y|<\varepsilon$ whenever $|x-y|<\delta$.
2. Take $x_n=1/n$ and $y_n=1/(2n)$ for $n\ge2$. Their distance tends to zero, but $|1/x_n-1/y_n|=n$, which does not tend to zero. Thus no single $\delta$ works for $\varepsilon=1$.
3. Yes. The interval is nonempty and compact and the function continuous; the extreme value theorem guarantees both extrema are attained.
4. No. The zero function is uniformly continuous on any domain; the compactness hypothesis is sufficient, not necessary.

Synthesis

Compactness allows sequences to yield limits inside the domain. Continuity then turns these limits into global attainment and a uniform input scale. The absence of compactness removes these guarantees, but does not force every conclusion to fail.