Compactness and the Heine–Borel theorem

Try

Cover the interval [0,1] with open intervals, perhaps infinitely many. Compactness says some finite selection already covers every point. It does not say the original cover is finite, nor that one universal finite selection works for all covers. The cover is given first; the finite subcover is chosen afterward. This quantified property is stronger than having finite length or looking visually enclosed.

Predict

Predict why (0,1) fails compactness. The intervals (1/n,1), for n≥2, cover it: every positive x is greater than 1/n for some n. Any finite selection has a largest denominator N and misses sufficiently small positive points below 1/N. The omitted endpoint matters even though the set is bounded. In contrast, an unbounded closed set such as R fails because no finite collection of the intervals (−n,n) covers it.

Experiment

Compare the examples.

Set in R	Closed	Bounded	Compact
[0,1]	Yes	Yes	Yes
(0,1)	No	Yes	No
R	Yes	No	No
A finite set	Yes	Yes	Yes

For the finite-set row, prove compactness directly: choose one cover member for each point. For the open interval, write the uncovered point after a proposed finite subcollection is chosen. A finite diagram cannot display every cover member, so the proof must describe the whole family.

Understand

Prove a closed bounded interval is compact by contradiction. Suppose an open cover has no finite subcover. Bisect the interval; at least one closed half also has no finite subcover, or combining two finite subcovers would cover the original. Repeat to obtain nested bad intervals of shrinking length. Their common point p lies in some open cover member, which contains a neighborhood of p. Eventually the entire retained interval lies in that neighborhood, so that single member covers it—a contradiction.

A closed subset F of a compact set K is compact: add the open complement of F to any open cover of F, obtain a finite subcover of K, then remove the added member. Consequently every closed bounded subset of R is compact because it is closed inside some compact interval.

Apply

Conversely, a compact set is bounded by applying a finite subcover to the intervals (−n,n). It is also closed. For a point x outside a nonempty compact K, cover K by neighborhoods around each y∈K of radius |x−y|/3. A finite selection has a positive minimum radius. A sufficiently small neighborhood of x misses all those selected neighborhoods, hence misses K. Thus the complement is open. Together these arguments prove Heine–Borel on R.

Sequential compactness follows from closedness and boundedness: Bolzano–Weierstrass extracts a convergent subsequence, and closedness keeps its limit in the set. The converse equivalence is valid on R too, but it is a theorem rather than the definition used here.

Continue

Optional depth asks why closed and bounded is not a universal compactness criterion in arbitrary spaces. The equivalence proved here depends on real-line structure and does extend to finite-dimensional Euclidean spaces, but later metric-space theory requires more care.

Save an open-cover proof with the order of quantifiers explicit and a sequence extraction using the same set. These complementary views will power uniform continuity and existence of extrema. They do not provide a numerical algorithm for locating an extremum without further information about the function.