The Heine–Borel theorem

Theorem

The Heine–Borel theorem

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Statement

Let \( K \subseteq \mathbb{R}^n \), where \( \mathbb{R}^n \) carries the standard Euclidean metric \( d(x,y) = \lVert x - y \rVert \) and the topology it induces. Then \( K \) is compact — every cover of \( K \) by open subsets of \( \mathbb{R}^n \) admits a finite subcover — if and only if \( K \) is closed in \( \mathbb{R}^n \) and bounded, i.e. \( \exists M \gt 0 \) with \( \lVert x \rVert \le M \) for all \( x \in K \). Both the finite dimension of the ambient space and the completeness of \( \mathbb{R} \) are essential: the equivalence fails in infinite-dimensional normed spaces and in incomplete metric spaces.

Why it matters

Compactness is the property that turns local information into global information: a condition verified near each point of a compact set can be enforced by finitely many verifications. But the covering definition is awkward to check directly. Heine–Borel replaces it, in \( \mathbb{R}^n \), by two conditions you can verify by inspection — closedness and boundedness. Almost every existence theorem of a first analysis course (extreme values, uniform continuity, attainment of distances) runs through this reduction.

The theorem is also the standard-bearer for a recurring theme: properties of \( \mathbb{R}^n \) that silently use finite dimensionality. Riesz's lemma shows that in any infinite-dimensional normed space the closed unit ball is not compact, so Heine–Borel is precisely a finite-dimensional phenomenon — the reason functional analysis needs weak topologies at all.

Hypotheses

Hypothesis.Ambient space \( \mathbb{R}^n \), Euclidean metric. In a general metric space, closed and bounded do not imply compact: in \( \ell^2 \), the set \( \{e_1, e_2, \dots\} \) of standard basis vectors is closed and bounded, yet \( \lVert e_i - e_j \rVert = \sqrt{2} \) for \( i \ne j \), so no subsequence converges and the balls \( B(e_k, \tfrac{1}{2}) \) form an open cover with no finite subcover.

Hypothesis.\( K \) closed. The half-open interval \( (0,1] \subseteq \mathbb{R} \) is bounded but not closed; the open cover \( \{ (\tfrac{1}{m}, 2) : m \in \mathbb{N} \} \) has no finite subcover, since any finite subfamily omits \( (0, \tfrac{1}{m_0}] \) for the largest index \( m_0 \) used.

Hypothesis.\( K \) bounded. The ray \( [0, \infty) \subseteq \mathbb{R} \) is closed but unbounded; the cover \( \{ (-1, m) : m \in \mathbb{N} \} \) has no finite subcover.

Hypothesis.Completeness of \( \mathbb{R} \) (used implicitly). The proof rests on the nested-interval property, a form of completeness. In \( \mathbb{Q} \) with the inherited metric, the set \( K = [0,2] \cap \mathbb{Q} \) is closed in \( \mathbb{Q} \) and bounded, yet not compact: the sets \( U_m = \{ q \in \mathbb{Q} : q^2 \lt 2 - \tfrac{1}{m} \} \cup \{ q \in \mathbb{Q} : q^2 \gt 2, \, q \gt 0 \} \) (each open in \( \mathbb{Q} \)) cover \( K \) once we adjoin \( \{q \lt \tfrac12\} \), but no finite subfamily covers, because rationals accumulate at the "hole" \( \sqrt{2} \notin \mathbb{Q} \).

Proof

We prove the two implications separately. Throughout, \( B(x,r) = \{ y \in \mathbb{R}^n : \lVert y - x \rVert \lt r \} \) denotes the open ball.

Part I: compact \( \Rightarrow \) closed and bounded.

Let \( K \) be compact. The family \( \{ B(0,m) : m \in \mathbb{N} \} \) is an open cover of \( K \), since every point of \( \mathbb{R}^n \) has finite norm.

Definition of open ball; the Archimedean property of \( \mathbb{R} \) guarantees some \( m \gt \lVert x \rVert \) for each \( x \). A

Compactness yields finitely many indices \( m_1 \lt \dots \lt m_k \) with \( K \subseteq B(0,m_1) \cup \dots \cup B(0,m_k) = B(0, m_k) \). Hence \( K \) is bounded by \( M = m_k \).

Definition of compactness applied to the cover in Step 1; the balls are nested, so the union is the largest one. A

To show \( K \) is closed, fix \( x \in \mathbb{R}^n \setminus K \). For each \( y \in K \) set \( r_y = \tfrac{1}{2}\lVert x - y \rVert \gt 0 \). Then \( \{ B(y, r_y) : y \in K \} \) is an open cover of \( K \).

Each \( y \in K \) satisfies \( y \ne x \), so \( r_y \gt 0 \) and \( y \in B(y, r_y) \). This is the key idea: use compactness to separate \( x \) from \( K \) uniformly. B

Extract a finite subcover \( B(y_1, r_{y_1}), \dots, B(y_p, r_{y_p}) \) and set \( r = \min_{1 \le i \le p} r_{y_i} \gt 0 \). Then \( B(x, r) \cap B(y_i, r_{y_i}) = \emptyset \) for every \( i \).

If \( z \) lay in both balls, the triangle inequality would give \( \lVert x - y_i \rVert \le \lVert x - z \rVert + \lVert z - y_i \rVert \lt r + r_{y_i} \le 2 r_{y_i} = \lVert x - y_i \rVert \), a contradiction. A minimum over a finite set of positive numbers is positive — this is where finiteness of the subcover is indispensable. C

Hence \( B(x,r) \cap K = \emptyset \), so \( \mathbb{R}^n \setminus K \) is open, i.e. \( K \) is closed.

\( K \) is contained in the union of the \( B(y_i, r_{y_i}) \), each disjoint from \( B(x,r) \); \( x \) was an arbitrary point of the complement. A

Part II: closed and bounded \( \Rightarrow \) compact. We first prove that every closed box \( Q = [a_1,b_1] \times \dots \times [a_n,b_n] \) is compact (Steps 6–9), then deduce the general case (Steps 10–11).

Suppose, for contradiction, that some open cover \( \mathcal{U} \) of the box \( Q_0 = Q \) admits no finite subcover. Bisect each of the \( n \) edges of \( Q_0 \), partitioning it into \( 2^n \) closed subboxes. At least one subbox, call it \( Q_1 \), is not covered by any finite subfamily of \( \mathcal{U} \).

If every one of the \( 2^n \) subboxes had a finite subcover, the union of those finitely many finite subfamilies would be a finite subcover of \( Q_0 \) — contradiction. This bisection is the key idea of the proof. B

Iterating, we obtain nested closed boxes \( Q_0 \supseteq Q_1 \supseteq Q_2 \supseteq \dots \), none finitely coverable by \( \mathcal{U} \), with \[ \operatorname{diam}(Q_k) = 2^{-k} \operatorname{diam}(Q_0) \to 0 . \]

Each bisection halves every edge length, hence halves the diameter \( \operatorname{diam}(Q) = \big( \sum_{i=1}^n (b_i - a_i)^2 \big)^{1/2} \). The construction uses dependent choice over a countable sequence of finite sets, which is unobjectionable. A

\[ \bigcap_{k=0}^{\infty} Q_k = \{ x^* \} \quad \text{for a single point } x^* \in Q_0 . \]

Write \( Q_k = \prod_{i=1}^n [a_i^{(k)}, b_i^{(k)}] \). For each fixed coordinate \( i \), the intervals \( [a_i^{(k)}, b_i^{(k)}] \) are nested with lengths \( \to 0 \); the sequence \( (a_i^{(k)})_k \) is increasing and bounded above by \( b_i^{(0)} \), so by the monotone convergence theorem (equivalently, the least-upper-bound axiom of \( \mathbb{R} \)) it converges to \( x_i^* = \sup_k a_i^{(k)} \), and \( a_i^{(k)} \le x_i^* \le b_i^{(k)} \) for all \( k \). This is Cantor's nested interval lemma, applied coordinatewise; it is exactly where the completeness of \( \mathbb{R} \) enters, and the theorem fails over \( \mathbb{Q} \) for this reason. Uniqueness: any two points of the intersection are at distance \( \le \operatorname{diam}(Q_k) \to 0 \), hence equal. C

Since \( x^* \in Q_0 \), some \( U \in \mathcal{U} \) contains \( x^* \); as \( U \) is open, \( B(x^*, \varepsilon) \subseteq U \) for some \( \varepsilon \gt 0 \). Choose \( k \) with \( \operatorname{diam}(Q_k) \lt \varepsilon \). Since \( x^* \in Q_k \), every \( y \in Q_k \) satisfies \( \lVert y - x^* \rVert \le \operatorname{diam}(Q_k) \lt \varepsilon \), so \[ Q_k \subseteq B(x^*, \varepsilon) \subseteq U . \] Thus \( Q_k \) is covered by the single set \( U \) — a finite subcover — contradicting Step 6. Hence every closed box is compact.

Definition of openness; definition of diameter; contradiction with the construction, which guaranteed no \( Q_k \) has a finite subcover. B

Now let \( K \) be closed and bounded, say \( \lVert x \rVert \le M \) for all \( x \in K \), so \( K \subseteq Q := [-M, M]^n \). Let \( \mathcal{V} \) be any open cover of \( K \). Then \[ \mathcal{V}' = \mathcal{V} \cup \{ \mathbb{R}^n \setminus K \} \] is an open cover of \( Q \).

\( \mathbb{R}^n \setminus K \) is open because \( K \) is closed — this is exactly where the closedness hypothesis is used. Every point of \( Q \) lies either in \( K \) (covered by \( \mathcal{V} \)) or in the complement. B

By Steps 6–9, \( Q \) is compact, so \( \mathcal{V}' \) has a finite subcover \( \{ V_1, \dots, V_q \} \), possibly including \( \mathbb{R}^n \setminus K \). Discarding \( \mathbb{R}^n \setminus K \) if present leaves finitely many members of \( \mathcal{V} \) that still cover \( K \), since \( \mathbb{R}^n \setminus K \) contains no point of \( K \). Hence \( K \) is compact. \( \blacksquare \)

This is the standard lemma "a closed subset of a compact set is compact", proved inline rather than cited. A

Result

\[ K \subseteq \mathbb{R}^n \text{ is compact} \iff K \text{ is closed and bounded.} \]

Reading. In Euclidean space, and only there among the spaces you meet early, the abstract covering property collapses to two checkable conditions: the set contains all its limit points, and it fits inside some ball. Whenever you can verify those two facts, you may deploy the full power of compactness — finite subcovers, attained extrema, uniform estimates.

Scope. Valid in \( \mathbb{R}^n \) for every finite \( n \) (and in any finite-dimensional normed space over \( \mathbb{R} \) or \( \mathbb{C} \), since all norms there are equivalent). The implication compact \( \Rightarrow \) closed and bounded holds in every metric space; the reverse implication fails in every infinite-dimensional normed space (Riesz) and in incomplete spaces such as \( \mathbb{Q} \).

Corollaries & converses

Extreme value theorem. A continuous \( f : K \to \mathbb{R} \) on compact \( K \subseteq \mathbb{R}^n \) is bounded and attains \( \sup_K f \) and \( \inf_K f \) (continuous images of compact sets are compact, and a compact subset of \( \mathbb{R} \) is closed and bounded, hence contains its supremum).
Bolzano–Weierstrass. Every bounded sequence in \( \mathbb{R}^n \) has a convergent subsequence: its terms lie in a closed box, which is compact, and in metric spaces compactness is equivalent to sequential compactness.
Heine–Cantor. A continuous function on a closed bounded subset of \( \mathbb{R}^n \) is uniformly continuous.
Cantor intersection property. A decreasing sequence of nonempty closed bounded subsets of \( \mathbb{R}^n \) has nonempty intersection (Problem 5).
Converses. The statement is itself a biconditional, so both directions hold in \( \mathbb{R}^n \). In a general metric space only "compact \( \Rightarrow \) closed and bounded" survives; the converse fails, e.g. for the closed unit ball of \( \ell^2 \), or for \( [0,2] \cap \mathbb{Q} \) inside \( \mathbb{Q} \). Indeed Riesz's lemma shows the closed unit ball of a normed space is compact iff the space is finite-dimensional.

Fails without

Drop closedness: \( (0,1) \subseteq \mathbb{R} \) is bounded but the open cover \( \{ (\tfrac{1}{m}, 1) : m \ge 2 \} \) has no finite subcover: any finite subfamily has a largest \( m_0 \) and misses \( (0, \tfrac{1}{m_0}] \cap (0,1) \).
Drop boundedness: \( \mathbb{Z} \subseteq \mathbb{R} \) is closed, but \( \{ (k - \tfrac12, k + \tfrac12) : k \in \mathbb{Z} \} \) covers it with each interval containing exactly one integer, so no finite subfamily covers infinitely many integers.
Drop finite dimension: in \( \ell^2 \), the closed unit ball is closed and bounded, yet the basis vectors satisfy \( \lVert e_i - e_j \rVert = \sqrt{2} \) for \( i \ne j \), so \( (e_k) \) has no Cauchy — hence no convergent — subsequence, and the ball is not (sequentially) compact.
Drop completeness of the ambient field: in \( \mathbb{Q} \), the set \( [0,2] \cap \mathbb{Q} \) is closed in \( \mathbb{Q} \) and bounded, but the sequence of rational truncations of \( \sqrt{2} \) has no limit in \( \mathbb{Q} \), so the set is not compact.

Common errors

Exporting the theorem beyond \( \mathbb{R}^n \). Writing "closed and bounded, hence compact" for a set of functions in \( C[0,1] \) or a subset of \( \ell^p \). In infinite dimensions you need Arzelà–Ascoli or weak-* compactness instead.
Closed relative to what? \( [0,2] \cap \mathbb{Q} \) is closed in \( \mathbb{Q} \) but not in \( \mathbb{R} \). Compactness of \( K \) is intrinsic, but "closed" in Heine–Borel means closed in \( \mathbb{R}^n \).
Finite subcover of a cover you chose. To prove compactness from the definition you must handle an arbitrary open cover; exhibiting one particular cover with a finite subcover proves nothing. Conversely, to disprove compactness a single bad cover suffices.
"Bounded" confused with "finite" or with "has a maximum". \( (0,1) \) is bounded and has no maximum; boundedness alone never yields attained extrema — closedness is doing that work.
Assuming the infimum of infinitely many radii is positive. Step 4 takes a minimum over a finite subcover; over an infinite family, \( \inf r_y \) can be \( 0 \), and the argument collapses. This is precisely why compactness, not mere coveredness, is needed.

Discussion

The theorem's name records a slow crystallisation rather than a single discovery. Dirichlet (in 1852 lectures, published 1904) and Heine (1872) used covering arguments implicitly to prove uniform continuity on closed intervals; Borel (1895) isolated and proved the covering statement for countable covers of \( [a,b] \); Lebesgue and Schoenflies (circa 1900) removed countability and popularised the modern form. The bisection proof given above descends from Bolzano's 1817 argument for the intermediate value theorem — the same divide-and-conquer engine drives Bolzano–Weierstrass.

Structurally, Heine–Borel splits into two independent ingredients. "Closed subset of a compact set is compact" (Steps 10–11) is pure topology and holds everywhere. "The closed box is compact" (Steps 6–9) is where geometry enters: it needs both completeness (Step 8, the nested intervals close up on a genuine point) and total boundedness (a box splits into finitely many pieces of half the diameter). In fact the correct generalisation to arbitrary metric spaces is exactly: \( K \) is compact iff \( K \) is complete and totally bounded. Heine–Borel is the special case in which completeness is inherited from \( \mathbb{R}^n \) (closed subsets of complete spaces are complete) and total boundedness collapses to boundedness because a bounded set in \( \mathbb{R}^n \) sits inside a box that can be chopped into finitely many small boxes — a manoeuvre that consumes finite dimensionality.

Two deeper vantage points. First, foundations: over the base theory \( \mathsf{RCA}_0 \) of reverse mathematics, the Heine–Borel covering lemma for \( [0,1] \) is equivalent to weak König's lemma (\( \mathsf{WKL}_0 \)) — the statement that every infinite binary tree has an infinite path — whereas Bolzano–Weierstrass is strictly stronger, equivalent to \( \mathsf{ACA}_0 \). So the two "compactness" facts that look interchangeable in a first course have genuinely different logical strength. Second, topology: Tychonoff's theorem gives compactness of \( [-M,M]^n \) as a product of compact intervals (for finite products no choice principle is needed), and Alaoglu's theorem restores a Heine–Borel-type statement in infinite dimensions by trading the norm topology for the weak-* topology — the closed unit ball of a dual space is weak-* compact, which is why weak topologies are the natural habitat of infinite-dimensional analysis.

Common misconceptions. "Compact means closed and bounded" is a theorem about \( \mathbb{R}^n \), not a definition of compactness; treating it as a definition makes the \( \ell^2 \) unit ball look compact when it is not. Also, compactness of \( K \) does not depend on the ambient space ( it is intrinsic to \( K \) with its subspace topology), whereas closedness does — the interval \( (0,1) \) is closed in itself but is not compact.

Worked examples

Example 1. The unit sphere \( S^{n-1} = \{ x \in \mathbb{R}^n : \lVert x \rVert = 1 \} \) is compact, and every continuous \( f : S^{n-1} \to \mathbb{R} \) attains a minimum.

\( S^{n-1} \) is bounded: every \( x \in S^{n-1} \) satisfies \( \lVert x \rVert = 1 \le 1 \), so \( S^{n-1} \subseteq \overline{B}(0,1) \).

Immediate from the defining equation. A

\( S^{n-1} \) is closed: the norm \( \lVert \cdot \rVert : \mathbb{R}^n \to \mathbb{R} \) is continuous (indeed \( \big| \lVert x \rVert - \lVert y \rVert \big| \le \lVert x - y \rVert \) by the reverse triangle inequality), and \( S^{n-1} = \lVert \cdot \rVert^{-1}(\{1\}) \) is the preimage of the closed set \( \{1\} \) under a continuous map.

Preimages of closed sets under continuous maps are closed. B

By Heine–Borel, \( S^{n-1} \) is compact. Given continuous \( f \), the image \( f(S^{n-1}) \subseteq \mathbb{R} \) is compact, hence (Heine–Borel again, in \( \mathbb{R}^1 \)) closed and bounded; being bounded below, \( m = \inf f(S^{n-1}) \) is finite, and being closed, \( m \in f(S^{n-1}) \), i.e. \( m = f(x_0) \) for some \( x_0 \in S^{n-1} \).

Continuous images of compact sets are compact (extract a finite subcover of the pullback cover); the infimum of a nonempty bounded-below set is a limit point of the set, so a closed set contains it. B

\[ S^{n-1} \text{ is compact; every continuous } f : S^{n-1} \to \mathbb{R} \text{ attains its infimum.} \]

Reading. This is the engine behind the equivalence of all norms on \( \mathbb{R}^n \): one bounds any norm below by its minimum on the Euclidean sphere, which Heine–Borel guarantees is attained and positive.

Scope. Any level set \( \{ g = c \} \) of a continuous \( g \) that happens to be bounded is compact by the same two moves.

Example 2. A direct covering argument: every continuous \( f : [a,b] \to \mathbb{R} \) is bounded. (Here we use the finite-subcover property itself, not a corollary.)

For each \( x \in [a,b] \), continuity at \( x \) with \( \varepsilon = 1 \) gives \( \delta_x \gt 0 \) such that \( |f(y) - f(x)| \lt 1 \) for all \( y \in [a,b] \) with \( |y - x| \lt \delta_x \). Set \( U_x = (x - \delta_x, \, x + \delta_x) \).

The \( \varepsilon \)–\( \delta \) definition of continuity, one point at a time; note \( \delta_x \) varies with \( x \) and may shrink to \( 0 \) — no uniformity is claimed. A

\( \{ U_x : x \in [a,b] \} \) is an open cover of \( [a,b] \), which is closed and bounded, hence compact by Heine–Borel. Extract a finite subcover \( U_{x_1}, \dots, U_{x_k} \).

Each \( x \in U_x \), so the family covers; Heine–Borel supplies the finite subcover — this is the crux of the argument. B

Set \( M = 1 + \max_{1 \le i \le k} |f(x_i)| \). Any \( y \in [a,b] \) lies in some \( U_{x_i} \), whence \[ |f(y)| \le |f(y) - f(x_i)| + |f(x_i)| \lt 1 + |f(x_i)| \le M . \]

Triangle inequality plus Step 1's local estimate; the maximum exists because the subcover is finite — an infinite supremum of the \( |f(x_i)| \) could be infinite. B

\[ f \text{ continuous on } [a,b] \implies \sup_{y \in [a,b]} |f(y)| \le M \lt \infty . \]

Reading. Local boundedness (trivial from continuity) is upgraded to global boundedness by trading an infinite family of local certificates for finitely many — exactly what a finite subcover does.

Scope. The identical argument works for continuous real-valued functions on any compact subset of \( \mathbb{R}^n \), and shows why it fails on \( (0,1] \): consider \( f(y) = \tfrac{1}{y} \), where no finite subfamily of the \( U_x \) reaches the left end.

Problems

Show directly from the theorem that \( K = [0,1] \cup [2,3] \) and the Cantor set \( C \) are compact.
Solution

\( [0,1] \cup [2,3] \) is a finite union of closed sets, hence closed, and is contained in \( [0,3] \), hence bounded; Heine–Borel gives compactness. For the Cantor set: \( C = \bigcap_{k=0}^{\infty} C_k \), where \( C_0 = [0,1] \) and each \( C_k \) is a finite union of \( 2^k \) closed intervals of length \( 3^{-k} \). Each \( C_k \) is closed (finite union of closed intervals), so the intersection \( C \) is closed (arbitrary intersections of closed sets are closed). Also \( C \subseteq [0,1] \), so \( C \) is bounded. By Heine–Borel, \( C \) is compact. Note that compactness follows without any analysis of what points \( C \) actually contains — the theorem converts two structural facts into compactness.
Let \( K \subseteq \mathbb{R}^n \) be compact and \( F \subseteq \mathbb{R}^n \) closed. Prove \( K \cap F \) is compact. Then give an example showing the intersection of two closed sets need not be compact.
Solution

By Heine–Borel, \( K \) is closed and bounded. Then \( K \cap F \) is closed, being an intersection of two closed sets, and bounded, being a subset of the bounded set \( K \). Heine–Borel (converse direction) gives compactness. For the example: \( F_1 = \mathbb{R} \times \{0\} \) and \( F_2 = \{ (x, 0) : x \ge 0 \} \) are both closed in \( \mathbb{R}^2 \), and \( F_1 \cap F_2 = [0,\infty) \times \{0\} \) is closed but unbounded, hence not compact — compactness of at least one factor is essential.
Let \( K \subseteq \mathbb{R}^n \) be nonempty and compact and let \( x \in \mathbb{R}^n \). Prove that the distance \( d(x, K) = \inf_{y \in K} \lVert x - y \rVert \) is attained: there exists \( y_0 \in K \) with \( \lVert x - y_0 \rVert = d(x,K) \). Show by example that attainment can fail if \( K \) is merely closed and we ask instead for two sets: exhibit disjoint closed sets \( A, B \subseteq \mathbb{R}^2 \) with \( \inf \{ \lVert a - b \rVert : a \in A, b \in B \} = 0 \).
Solution

The map \( g : K \to \mathbb{R} \), \( g(y) = \lVert x - y \rVert \), satisfies \( |g(y) - g(z)| \le \lVert y - z \rVert \) by the triangle inequality, so \( g \) is continuous (indeed 1-Lipschitz). \( K \) is compact, so by the extreme value theorem (Example 1, Step 3 — itself a consequence of Heine–Borel) \( g \) attains its infimum at some \( y_0 \in K \), giving \( \lVert x - y_0 \rVert = d(x, K) \).

For the failure with two closed sets: take \( A = \{ (t, 0) : t \in \mathbb{R} \} \) and \( B = \{ (t, e^{-t}) : t \in \mathbb{R} \} \) (the graph of a continuous function is closed, and \( e^{-t} \gt 0 \) means \( A \cap B = \emptyset \)). Then \( \lVert (t,0) - (t, e^{-t}) \rVert = e^{-t} \to 0 \) as \( t \to \infty \), so the infimum of distances is \( 0 \) but is never attained. Both sets are closed but neither is bounded; if either were compact, the distance between disjoint closed sets, one compact, would be attained and positive.
Prove, using only the covering definition, that \( K = \{0\} \cup \{ \tfrac{1}{m} : m \in \mathbb{N} \} \subseteq \mathbb{R} \) is compact, and that \( K \setminus \{0\} \) is not. Reconcile both facts with Heine–Borel.
Solution

\( K \) is compact: let \( \mathcal{U} \) be an open cover. Some \( U_0 \in \mathcal{U} \) contains \( 0 \); since \( U_0 \) is open, \( (-\varepsilon, \varepsilon) \subseteq U_0 \) for some \( \varepsilon \gt 0 \). By the Archimedean property, only finitely many points \( \tfrac{1}{m} \) satisfy \( \tfrac{1}{m} \ge \varepsilon \), namely those with \( m \le \tfrac{1}{\varepsilon} \). For each such point choose one covering set \( U_m \in \mathcal{U} \). Then \( U_0 \) together with these finitely many \( U_m \) covers \( K \).

\( K \setminus \{0\} \) is not compact: the sets \( V_m = ( \tfrac{1}{m} - r_m, \, \tfrac{1}{m} + r_m ) \) with \( r_m = \tfrac{1}{2} \big( \tfrac{1}{m} - \tfrac{1}{m+1} \big) \) are open, pairwise disjoint, and each contains exactly one point of \( K \setminus \{0\} \); hence \( \{V_m\} \) is an open cover with no finite subcover (a finite subfamily covers only finitely many points).

Reconciliation: \( K \) is bounded (contained in \( [0,1] \)) and closed — its only limit point is \( 0 \), which it contains — so Heine–Borel predicts compactness. \( K \setminus \{0\} \) is still bounded but fails to contain its limit point \( 0 \), so it is not closed, and Heine–Borel correctly predicts non-compactness.
(Cantor intersection, and the failure of Heine–Borel in \( C[0,1] \).) (a) Let \( K_1 \supseteq K_2 \supseteq K_3 \supseteq \dots \) be nonempty compact subsets of \( \mathbb{R}^n \). Prove \( \bigcap_{k=1}^{\infty} K_k \ne \emptyset \). (b) In the space \( C[0,1] \) with the supremum norm \( \lVert f \rVert_\infty = \sup_{t \in [0,1]} |f(t)| \), show that the closed unit ball is closed and bounded but not compact.
Solution

(a) Suppose \( \bigcap_k K_k = \emptyset \). Set \( U_k = \mathbb{R}^n \setminus K_k \); each is open since \( K_k \) is compact, hence closed by Heine–Borel. For any \( x \in K_1 \), the assumption gives some \( k \) with \( x \notin K_k \), i.e. \( x \in U_k \); therefore \( \{U_k\}_{k \ge 2} \) is an open cover of \( K_1 \). Compactness of \( K_1 \) yields a finite subcover \( U_{k_1}, \dots, U_{k_p} \); since the \( K_k \) decrease, the \( U_k \) increase, so \( K_1 \subseteq U_{k^*} \) where \( k^* = \max_i k_i \). But then \( K_{k^*} \subseteq K_1 \subseteq \mathbb{R}^n \setminus K_{k^*} \), forcing \( K_{k^*} = \emptyset \) — contradicting nonemptiness. Hence the intersection is nonempty.

(b) The ball \( \overline{B} = \{ f : \lVert f \rVert_\infty \le 1 \} \) is bounded by definition and closed (if \( f_j \to f \) uniformly with \( \lVert f_j \rVert_\infty \le 1 \), then \( |f(t)| = \lim_j |f_j(t)| \le 1 \) for every \( t \)). To defeat compactness, consider \( f_m(t) = t^m \), all in \( \overline{B} \). For \( m \lt m' \), evaluate at \( t_0 = (1/2)^{1/m} \): \( f_m(t_0) = \tfrac12 \) while \( f_{m'}(t_0) = (1/2)^{m'/m} \lt \tfrac12 \), and more decisively \( \lVert f_m - f_{m'} \rVert_\infty \ge f_m(t_0) - f_{m'}(t_0) \to \tfrac12 - 0 \) along suitable subsequences; in fact one checks \( \sup_m \inf_{m' \gt m} \lVert f_m - f_{m'} \rVert_\infty \ge \tfrac14 \), so \( (f_m) \) has no uniformly convergent — i.e. no norm-convergent — subsequence. (Alternatively: the pointwise limit of \( t^m \) is the discontinuous function \( \mathbf{1}_{\{t = 1\}} \), and a uniform limit of continuous functions must be continuous, so no subsequence can converge uniformly.) Since compactness in metric spaces implies sequential compactness, \( \overline{B} \) is not compact. This exhibits concretely that Heine–Borel is a theorem about finite-dimensional spaces: \( C[0,1] \) is complete, so what fails is total boundedness, and the missing hypothesis is finite dimension (Riesz's lemma).