Theorem

The Heine–Borel theorem

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Statement

Let \( n \ge 1 \) and let \( K \subseteq \mathbb{R}^n \), the ambient space carrying the Euclidean norm \( \lVert x \rVert = \left( \sum_{i=1}^{n} x_i^2 \right)^{1/2} \) and the metric \( d(x,y) = \lVert x - y \rVert \). Then the following three conditions are equivalent: (i) \( K \) is closed in \( \mathbb{R}^n \) and bounded, meaning \( K \subseteq \{ x \in \mathbb{R}^n : \lVert x \rVert \le M \} \) for some \( M \lt \infty \); (ii) \( K \) is compact, meaning every family \( \mathcal{U} \) of open subsets of \( \mathbb{R}^n \) with \( K \subseteq \bigcup_{U \in \mathcal{U}} U \) has a finite subfamily \( U_1, \dots, U_N \) with \( K \subseteq U_1 \cup \cdots \cup U_N \); (iii) \( K \) is sequentially compact, meaning every sequence \( (x_k) \) in \( K \) has a subsequence converging to a point of \( K \). The standing hypotheses are that the ambient space is \( \mathbb{R}^n \) for a finite \( n \) — a complete, finite-dimensional normed space — and that “bounded” is measured by that norm and not by some other metric inducing the same topology.

Why it matters

Compactness is the hypothesis under which local information becomes global information, and Heine–Borel is the theorem that makes it checkable. “Every open cover has a finite subcover” quantifies over all covers and is impossible to verify directly; “closed and bounded” is two elementary tests. Every existence theorem of multivariable analysis runs through this translation: a continuous real function on a closed bounded set attains its extreme values, a continuous function on such a set is uniformly continuous, a continuous bijection from such a set onto its image is a homeomorphism, and a decreasing family of such sets has a common point. In optimisation the theorem is the licence to say “a maximiser exists” before any algorithm is written.

It also draws a hard line. The equivalence is false in a general metric space, and false in every infinite-dimensional normed space; the closed unit ball is compact precisely when the dimension is finite. Heine–Borel is therefore not a soft topological fact but a statement about \( \mathbb{R}^n \) purchased with two specific structural inputs, completeness and finite dimension. Recognising which one is missing is what tells you whether a limiting argument in a function space needs Arzelà–Ascoli, weak-\( * \) compactness, or nothing that exists at all. For linear algebra the payoff is immediate: because all norms on a finite-dimensional space are equivalent, the theorem transfers verbatim to any \( n \)-dimensional real or complex vector space, which is why matrix groups such as \( \mathrm{O}(n) \) and \( \mathrm{U}(n) \) are compact and why continuous functions of a matrix — determinant, trace, spectral radius, condition number — attain extrema on them.

Hypotheses

The ambient space is complete.Completeness of \( \mathbb{R} \) is used exactly once, in the nested-box step. Work inside \( \mathbb{Q} \) instead and the theorem is false: \( K = \{ q \in \mathbb{Q} : q^2 \le 2 \} \) is closed in \( \mathbb{Q} \) (it is the preimage of a closed set under a continuous map) and bounded by \( 2 \), but the open cover \( U_k = \{ q \in \mathbb{Q} : q^2 \lt 2 - 1/k \} \cup \{ q \in \mathbb{Q} : q \lt -1 \} \) for \( k \ge 1 \) has no finite subcover, since the truncations \( 1.4,\ 1.41,\ 1.414, \dots \) of \( \sqrt{2} \) escape every \( U_k \) eventually.

The dimension \( n \) is finite.Drop it and only the trivial half survives. In the Hilbert space \( \ell^2 \) the closed unit ball \( \overline{B} = \{ x : \lVert x \rVert_2 \le 1 \} \) is closed and bounded, yet the orthonormal vectors \( e_k \) satisfy \( \lVert e_j - e_k \rVert_2 = \sqrt{2} \) for \( j \ne k \), so no subsequence is Cauchy and \( \overline{B} \) is not sequentially compact. By Riesz’s lemma, compactness of the closed unit ball characterises finite dimension.

“Bounded” is boundedness in the norm.Boundedness is a metric notion, not a topological one. Give \( \mathbb{R} \) the metric \( \rho(x,y) = \frac{|x-y|}{1 + |x-y|} \), which induces the usual topology. Then \( \mathbb{R} \) is closed in itself and has \( \rho \)-diameter \( 1 \), so it is “closed and bounded”, but the cover \( \{ (-k, k) \}_{k \in \mathbb{N}} \) has no finite subcover. Rescaling the metric destroys the theorem while changing no open set.

Closed in the ambient space, not merely closed in itself.Every metric space is closed as a subset of itself, so “\( K \) is closed” is only meaningful relative to \( \mathbb{R}^n \). The set \( (0,1] \) is closed in the metric space \( (0,2) \) and bounded, but it is not compact; what fails is closedness in \( \mathbb{R} \), where the limit point \( 0 \) is missing. Compactness, unlike closedness, is intrinsic: it does not depend on the ambient space, which is why it is the more robust hypothesis.

Proof

We prove (i) \( \Rightarrow \) (ii) by bisection (Steps 1–8), then (ii) \( \Rightarrow \) (i) (Steps 9–10), then close the cycle through sequential compactness (Steps 11–12). Throughout, a closed box means a product \( Q = \prod_{i=1}^{n} [a_i, b_i] \) with all sides of equal length \( s(Q) \); its diameter is \( \operatorname{diam} Q = s(Q)\sqrt{n} \), since the greatest distance between two points of a cube is its main diagonal.

\[ K \ \text{bounded} \ \implies \ K \subseteq Q_0 := [-M, M]^n, \qquad s(Q_0) = 2M, \quad \operatorname{diam} Q_0 = 2M\sqrt{n}. \]

If \( \lVert x \rVert \le M \) then \( |x_i| \le \lVert x \rVert \le M \) for each coordinate, so \( x \) lies in the cube. The problem is reduced to two claims: \( Q_0 \) is compact, and a closed subset of a compact set is compact. A

Suppose, for contradiction, that some open cover \( \mathcal{U} \) of \( Q_0 \) admits no finite subcover. Bisect each of the \( n \) edges to write \[ Q_0 = \bigcup_{j=1}^{2^n} Q^{(j)}, \qquad s\!\left( Q^{(j)} \right) = \tfrac{1}{2} s(Q_0), \] the union of the \( 2^n \) closed congruent sub-boxes.

Each edge \( [a_i, b_i] \) splits as \( [a_i, m_i] \cup [m_i, b_i] \) with \( m_i = \frac{a_i + b_i}{2} \); taking one half per coordinate and forming the product gives \( 2^n \) closed boxes whose union is \( Q_0 \). They overlap only on shared faces, which is harmless. A

At least one \( Q^{(j)} \) admits no finite subcover from \( \mathcal{U} \).

Contrapositive: if every \( Q^{(j)} \) were covered by finitely many members \( \mathcal{F}_j \subseteq \mathcal{U} \), then \( \mathcal{F}_1 \cup \cdots \cup \mathcal{F}_{2^n} \) would be a finite subfamily covering \( \bigcup_j Q^{(j)} = Q_0 \), contradicting Step 2. This is the pigeonhole of the argument: a finite union of “good” pieces is good, so a bad box has a bad child. B

Recursively define \( Q_{k+1} \) to be the first (in a fixed enumeration of the \( 2^n \) children) sub-box of \( Q_k \) with no finite subcover. Then \[ Q_0 \supseteq Q_1 \supseteq Q_2 \supseteq \cdots, \qquad \operatorname{diam} Q_k = \frac{2M\sqrt{n}}{2^{k}}. \]

Step 3 guarantees the recursion never stalls; fixing the enumeration makes the choice canonical, so no appeal to the axiom of choice is needed. The diameter formula follows by induction, each bisection halving every edge and hence the diagonal. B

\[ \bigcap_{k=0}^{\infty} Q_k = \{ x^{*} \} \ \ne \ \varnothing. \]

Write \( Q_k = \prod_{i=1}^{n} \left[ a_i^{(k)}, b_i^{(k)} \right] \). For each fixed \( i \) the intervals \( \left[ a_i^{(k)}, b_i^{(k)} \right] \) are nested, so \( \left( a_i^{(k)} \right)_k \) increases and is bounded above by every \( b_i^{(j)} \); completeness of \( \mathbb{R} \) supplies \( x_i^{*} = \sup_k a_i^{(k)} \) with \( a_i^{(k)} \le x_i^{*} \le b_i^{(k)} \) for all \( k \). Doing this in each of the \( n \) coordinates gives a point of every \( Q_k \); uniqueness follows because two such points would differ by at most \( \operatorname{diam} Q_k \to 0 \). This is the only step that uses completeness, and the only one that uses \( n \lt \infty \). C

\[ x^{*} \in Q_0 \subseteq \bigcup_{U \in \mathcal{U}} U \ \implies \ \exists\, U^{*} \in \mathcal{U}, \ \exists\, \varepsilon \gt 0 : \ B\!\left( x^{*}, \varepsilon \right) := \left\{ y : \lVert y - x^{*} \rVert \lt \varepsilon \right\} \subseteq U^{*}. \]

\( \mathcal{U} \) covers \( Q_0 \), so some member contains \( x^{*} \); that member is open, so it contains an open ball about each of its points. A

Choose \( k \) with \( \dfrac{2M\sqrt{n}}{2^{k}} \lt \varepsilon \). Since \( x^{*} \in Q_k \), every \( y \in Q_k \) satisfies \( \lVert y - x^{*} \rVert \le \operatorname{diam} Q_k \lt \varepsilon \), hence \[ Q_k \subseteq B\!\left( x^{*}, \varepsilon \right) \subseteq U^{*}. \]

Such a \( k \) exists by the Archimedean property, using \( 2^{k} \ge k + 1 \). But then \( \{ U^{*} \} \) is a subcover of \( Q_k \) with one element, contradicting the defining property of \( Q_k \). The contradiction refutes Step 2, so \( Q_0 \) is compact. B

If \( K \subseteq Q_0 \) is closed in \( \mathbb{R}^n \) and \( \mathcal{V} \) is an open cover of \( K \), then \[ \mathcal{V} \cup \left\{ \mathbb{R}^n \setminus K \right\} \ \text{is an open cover of } Q_0, \] so it has a finite subcover; deleting \( \mathbb{R}^n \setminus K \) from it leaves finitely many members of \( \mathcal{V} \) that still cover \( K \).

\( \mathbb{R}^n \setminus K \) is open because \( K \) is closed, and it contains every point of \( Q_0 \) outside \( K \), so the enlarged family covers \( Q_0 \). Deleting it can only lose points of \( \mathbb{R}^n \setminus K \), and \( K \) contains none. Hence (i) \( \Rightarrow \) (ii). B

Conversely, let \( K \) be compact. From \( K \subseteq \bigcup_{m=1}^{\infty} B(0, m) \) and a finite subcover \( B(0, m_1), \dots, B(0, m_N) \), \[ K \subseteq B\!\left( 0, \max_j m_j \right) \ \implies \ K \ \text{is bounded}. \]

The open balls \( B(0,m) \) exhaust \( \mathbb{R}^n \) because every point has finite norm (Archimedean property), and they are nested increasing, so a finite subfamily is contained in its largest member. A

Fix \( p \notin K \). The sets \( V_m = \left\{ y : \lVert y - p \rVert \gt 1/m \right\} \), \( m \ge 1 \), are open and cover \( K \); a finite subcover gives \( K \subseteq V_{m_0} \) for some \( m_0 \), i.e. \[ B\!\left( p, 1/m_0 \right) \cap K = \varnothing \ \implies \ \mathbb{R}^n \setminus K \ \text{is open} \ \implies \ K \ \text{is closed}. \]

Each \( V_m \) is the preimage of the open ray \( (1/m, \infty) \) under the continuous map \( y \mapsto \lVert y - p \rVert \). They cover \( K \) because \( \lVert y - p \rVert \gt 0 \) for \( y \in K \) and the reciprocals \( 1/m \) reach below any positive number. The \( V_m \) increase with \( m \), so a finite subcover collapses to one. Hence (ii) \( \Rightarrow \) (i). B

(i) \( \Rightarrow \) (iii). Let \( (x_k) \subseteq K \) with \( K \) closed and bounded. Bolzano–Weierstrass in \( \mathbb{R}^n \) — extract in coordinate \( 1 \), then a further subsequence in coordinate \( 2 \), through all \( n \) coordinates — gives \( x_{k_j} \to x \in \mathbb{R}^n \), and \[ x_{k_j} \in K \ \text{for all } j, \ \ K \ \text{closed} \ \implies \ x \in K. \]

Boundedness of \( K \) bounds each coordinate sequence by \( M \), so Bolzano–Weierstrass applies; the successive extractions are finite in number precisely because \( n \lt \infty \), and each preserves the convergence already achieved. A closed set contains all its limit points, which is what places \( x \) in \( K \). B

(iii) \( \Rightarrow \) (i). If \( K \) is unbounded pick \( x_k \in K \) with \( \lVert x_k \rVert \gt k \); if \( K \) is not closed pick \( p \in \overline{K} \setminus K \) and \( x_k \in K \) with \( \lVert x_k - p \rVert \lt 1/k \). In the first case every subsequence has \( \lVert x_{k_j} \rVert \gt k_j \ge j \to \infty \) and cannot converge; in the second every subsequence converges to \( p \notin K \). Either way (iii) fails. \[ \text{(i)} \iff \text{(ii)} \iff \text{(iii)}. \qquad \blacksquare \]

Contrapositive of (iii) in both directions. Note the asymmetry: unboundedness kills convergence outright, while non-closedness produces convergence to the wrong place — which is why sequential compactness demands the limit lie in \( K \). B

Result

\[ K \subseteq \mathbb{R}^n \ \ (n \lt \infty): \qquad K \ \text{compact} \ \iff \ K \ \text{closed and bounded} \ \iff \ K \ \text{sequentially compact}. \]

Reading. In \( \mathbb{R}^n \), and only there among the familiar spaces, the infinitary condition “every open cover reduces to a finite one” is decided by two finitary tests you can apply by inspection: does the set contain its boundary, and does it fit inside a ball. Closedness stops points from leaking out at the edge; boundedness stops them from escaping to infinity. Remove either and a cover with no finite subcover can be written down explicitly.

Scope. Valid in \( \mathbb{R}^n \) and \( \mathbb{C}^n \) for every finite \( n \), and hence in every finite-dimensional normed space over \( \mathbb{R} \) or \( \mathbb{C} \), with any norm, since all such norms are equivalent. It fails in \( \mathbb{Q}^n \) (no completeness), in every infinite-dimensional normed space (no finite dimension), and in general metric spaces, where the correct replacement is: compact \( \iff \) complete and totally bounded. Boundedness must be read in the norm; remetrising by a bounded equivalent metric preserves the topology and destroys the theorem.

Corollaries & converses

Extreme value theorem. A continuous \( f : K \to \mathbb{R} \) on a closed bounded \( K \subseteq \mathbb{R}^n \) attains a maximum and a minimum: \( f(K) \) is compact in \( \mathbb{R} \), hence closed and bounded, hence contains \( \sup f(K) \) and \( \inf f(K) \).
Heine–Cantor. A continuous function on a closed bounded \( K \subseteq \mathbb{R}^n \) is uniformly continuous; the pointwise \( \delta(\varepsilon, x) \) can be replaced by a single \( \delta(\varepsilon) \) by taking a finite subcover of the balls \( B(x, \delta(\varepsilon/2, x)/2) \).
Continuous images. If \( f \) is continuous and \( K \) is closed and bounded then \( f(K) \) is closed and bounded. Note the hypothesis is needed on both counts: \( f(x) = \arctan x \) maps the closed unbounded set \( \mathbb{R} \) onto the non-closed \( (-\pi/2, \pi/2) \).
Cantor intersection. A decreasing chain \( K_1 \supseteq K_2 \supseteq \cdots \) of nonempty closed bounded sets has \( \bigcap_j K_j \ne \varnothing \), and if \( \operatorname{diam} K_j \to 0 \) the intersection is a single point.
Finite-dimensional normed spaces. If \( \dim V = n \lt \infty \) over \( \mathbb{R} \), every two norms on \( V \) are equivalent, so “closed and bounded \( \iff \) compact” holds in \( V \) with any norm. Consequently \( \mathrm{O}(n) \), \( \mathrm{SO}(n) \), \( \mathrm{U}(n) \) and the set of \( n \times n \) stochastic matrices are compact, while \( \mathrm{GL}(n, \mathbb{R}) \) and \( \mathrm{SL}(n, \mathbb{R}) \) are not (the first is open, the second closed but unbounded).
Riesz: the sharp converse. The closed unit ball of a normed space is compact if and only if the space is finite-dimensional. So the theorem does not merely happen to hold in \( \mathbb{R}^n \) — it holds nowhere else among normed spaces.
Converse of “compact \( \Rightarrow \) closed and bounded” in general: FALSE. In any infinite metric space with the discrete metric \( d(x,y) = 1 \) for \( x \ne y \), every subset is closed and bounded, and no infinite subset is compact: the cover by singleton balls \( B(x, 1/2) = \{x\} \) has no finite subcover.
Metric-space replacement. A metric space is compact if and only if it is complete and totally bounded. Heine–Borel is the special case in which boundedness upgrades to total boundedness for free — and it does so exactly because \( \mathbb{R}^n \) is finite-dimensional.

Fails without

Closedness dropped: \( K = (0,1] \subseteq \mathbb{R} \) is bounded but not closed. The open cover \( U_k = \left( \tfrac{1}{k}, 2 \right) \), \( k \ge 1 \), satisfies \( \bigcup_k U_k = (0,2) \supseteq K \), yet any finite subfamily lies inside the largest \( U_N = \left( \tfrac{1}{N}, 2 \right) \), which misses the point \( \tfrac{1}{N+1} \in K \). For \( N = 20 \) the escaping point is \( 1/21 \approx 0.047619 \).
Boundedness dropped: \( K = \mathbb{R} \) is closed but unbounded. The cover \( V_k = (-k, k) \), \( k \ge 1 \), has \( \bigcup_k V_k = \mathbb{R} \), yet a finite subfamily lies in \( V_N \) and misses \( N \). Equivalently \( f(x) = x \) is continuous on \( \mathbb{R} \) with no maximum, so the extreme value theorem fails with it.
Completeness dropped (work in \( \mathbb{Q} \)): \( K = \left\{ q \in \mathbb{Q} : 0 \le q, \ q^2 \le 2 \right\} \) is closed in \( \mathbb{Q} \) and bounded by \( 2 \). The relatively open sets \( W_k = \left\{ q \in \mathbb{Q} : q^2 \lt 2 - \tfrac{1}{k} \right\} \) cover \( K \), because no rational has \( q^2 = 2 \); but a finite subfamily lies in \( W_N \), which excludes any rational \( q \) with \( 2 - \tfrac{1}{N} \le q^2 \le 2 \), and such rationals exist for every \( N \) (take a decimal truncation of \( \sqrt{2} \) with enough digits: for \( N = 100 \), \( q = 1.41421 \) has \( q^2 = 1.9999899\ldots \ge 1.99 \)). The bisection of Steps 2–4 produces a nested sequence of boxes whose intersection, taken in \( \mathbb{Q} \), is empty: Step 5 is what fails.
Finite dimension dropped: in \( \ell^2 \), the closed unit ball \( \overline{B} \) is closed and bounded, and \( \ell^2 \) is complete. The standard vectors \( e_k \) satisfy \( \lVert e_j - e_k \rVert_2 = \sqrt{2} \approx 1.4142 \) for \( j \ne k \), so \( E = \{ e_k : k \ge 1 \} \) has no limit point and is closed. Then \[ \left\{ \ell^2 \setminus E \right\} \cup \left\{ B\!\left( e_k, \tfrac{1}{2} \right) : k \ge 1 \right\} \] is an open cover of \( \overline{B} \) in which the first set contains no \( e_j \) and each ball contains exactly one, so no finite subfamily covers the infinite set \( E \subseteq \overline{B} \). The same separation shows \( \overline{B} \) is not sequentially compact.
Norm-boundedness weakened to metric boundedness: put \( \rho(x,y) = \frac{|x-y|}{1+|x-y|} \) on \( \mathbb{R} \). This metric induces the standard topology and gives \( \mathbb{R} \) diameter \( 1 \), so \( \mathbb{R} \) is closed and \( \rho \)-bounded — and still not compact, by the cover \( (-k,k) \) above. Boundedness is not a topological invariant, so “closed and bounded” is not a topological criterion.

Common errors

“Compact means closed and bounded.” That is the conclusion of a theorem about \( \mathbb{R}^n \), not the definition. Taking it as a definition makes the closed unit ball of \( C[0,1] \) “compact” and every extreme-value argument in a function space invalid.
“\( K \) is closed, since it is closed in itself.” Closedness is relative. \( (0,1] \) is closed in \( (0,2) \) and in itself; it is not closed in \( \mathbb{R} \), and that is the closedness the theorem asks for. Always name the ambient space.
Covering with closed sets. The finite-subcover property is about open covers. \( [0,1] = \bigcup_{x \in [0,1]} \{x\} \) is a cover by closed sets with no finite subcover, and it says nothing at all.
“The cover has a finite subcover of \( N \) sets, with \( N \) depending only on \( K \).” The number \( N \) depends on the cover. For \( [0,1] \) and the cover by intervals of length \( 1/m \) one needs at least \( m \) of them; there is no uniform bound.
Extracting infinitely many subsequences. The coordinatewise proof of Step 11 works because there are \( n \) coordinates and \( n \) is finite. Repeating it “for every coordinate” in \( \ell^2 \) needs a diagonal argument, and the diagonal subsequence then converges coordinatewise but not in norm — which is exactly why \( \overline{B} \subseteq \ell^2 \) is not compact.
Confusing bounded with totally bounded. In \( \mathbb{R}^n \) they coincide; in general they do not. A set is totally bounded when it is covered by finitely many \( \varepsilon \)-balls for every \( \varepsilon \gt 0 \); the set \( \{ e_k \} \subseteq \ell^2 \) is bounded but needs infinitely many balls of radius \( \tfrac{1}{2} \).
“A continuous function on a bounded set is bounded.” \( f(x) = 1/x \) on \( (0,1] \). Closedness is not optional decoration; it is half the hypothesis.

Discussion

The name records a tangled history. Eduard Heine’s 1872 paper on uniform continuity of continuous functions on a closed interval contains the covering idea in embryo, though a similar argument had appeared earlier in Dirichlet’s lectures; Émile Borel’s 1895 thesis proved that a countable open cover of a closed bounded interval has a finite subcover; and the extension to arbitrary covers is due to Lebesgue, with Schoenflies and Young among those who circulated the general form soon after. What we now regard as the definition of compactness — finite subcovers, full stop — only settled in the 1920s, once general topology had separated the notion from the interval it was invented on. The historical order is worth remembering: the theorem came first, and the definition of compactness was reverse-engineered from it.

Structurally, the proof above has one load-bearing beam and a lot of scaffolding. Steps 6–10 are pure bookkeeping — set algebra and the definition of an open set, valid in any metric space — and Steps 1–4 are the geometry of subdividing a box, which needs the coordinate structure of \( \mathbb{R}^n \) but no limit process at all. Step 5, the nested-box argument, is the entire mathematical content, and it consumes both completeness (the supremum exists) and finite dimension (there are only \( n \) coordinates to control). The completeness and finite-dimension failure regimes listed above are failures of Step 5 and of nothing else; the remetrisation failure is different in kind, striking at Step 1, where it is boundedness in the norm that puts \( K \) inside a box in the first place. The same skeleton with “bisect the box” replaced by “refine the finite \( \varepsilon \)-net” proves the general metric criterion, compact \( \iff \) complete and totally bounded; \( \mathbb{R}^n \) is special only in that bounded sets are automatically totally bounded there, because a cube of side \( s \) is covered by \( \lceil s\sqrt{n}/\varepsilon \rceil^{\,n} \) cubes of diameter \( \varepsilon \) — a count that is finite precisely because the exponent \( n \) is.

The theorem has two natural generalisations pointing in opposite directions. Upward, Tychonoff’s theorem says an arbitrary product of compact spaces is compact in the product topology, which recovers compactness of \( [-M,M]^n \) instantly — but Tychonoff in full generality is equivalent to the axiom of choice, whereas the bisection proof above uses none, so the elementary argument is genuinely stronger information about \( \mathbb{R}^n \). Downward, in infinite dimensions one weakens the topology until balls become compact again: the Banach–Alaoglu theorem makes the closed unit ball of a dual space weak-\( * \) compact, and Arzelà–Ascoli replaces “bounded” by “bounded and equicontinuous” for subsets of \( C(K) \). Both are the working analyst’s substitute for Heine–Borel where Heine–Borel has failed.

Logically, the covering form is weaker than the sequential form. Over the base theory \( \mathsf{RCA}_0 \) of reverse mathematics, the Heine–Borel covering property for \( [0,1] \) is equivalent to weak König’s lemma \( \mathsf{WKL}_0 \), while the Bolzano–Weierstrass property is equivalent to arithmetical comprehension \( \mathsf{ACA}_0 \), a strictly stronger system. Classically the two are interchangeable in metric spaces; proof-theoretically they are not, and the bisection proof above quietly crosses that gap by using a supremum. Constructively the situation is sharper still: in Bishop-style mathematics the covering theorem for \( [0,1] \) is not provable, and the standard repair is to work with formal or point-free covers, where a localic version of Heine–Borel is constructively valid — compactness survives, but the points do not.

Common misconceptions. Heine–Borel is not a definition, not a topological criterion (boundedness is metric), and not a theorem about metric spaces in general. It does not assert that a compact set is finite, or that the finite subcover has a size depending only on the set. And it does not say that closed bounded sets are “small”: the Cantor set is compact and uncountable, of Lebesgue measure zero, while \( [0,1]^{10} \) is compact and of measure one.

Worked examples

Example 1. In \( \mathbb{R}^2 \) let \( E = \left\{ (x,y) : \tfrac{x^2}{4} + y^2 \le 1 \right\} \) (a closed elliptical disc) and \( A = \left\{ (x,y) : 0 \lt x^2 + y^2 \le 1 \right\} \) (the punctured closed unit disc). Decide which is compact. For the non-compact one, exhibit an explicit open cover with no finite subcover and name a point that escapes any \( 50 \) of its members.

\[ E = g^{-1}\big( (-\infty, 1] \big), \qquad g(x,y) = \frac{x^2}{4} + y^2 . \]

\( g \) is a polynomial, hence continuous, and \( (-\infty, 1] \) is closed in \( \mathbb{R} \); the preimage of a closed set under a continuous map is closed. So \( E \) is closed in \( \mathbb{R}^2 \). A

\[ (x,y) \in E \implies x^2 \le 4, \ y^2 \le 1 \implies \lVert (x,y) \rVert = \sqrt{x^2 + y^2} \le \sqrt{4 + 1} = \sqrt{5} \approx 2.2361 . \]

From \( \frac{x^2}{4} + y^2 \le 1 \) each summand is at most \( 1 \). The bound \( \sqrt{5} \) is not sharp — the true maximum norm on \( E \) is \( 2 \), at \( (\pm 2, 0) \) — but any finite bound suffices for boundedness. A

\[ E \ \text{closed and bounded in } \mathbb{R}^2 \ \implies \ E \ \text{is compact}. \]

Heine–Borel, with \( n = 2 \) and \( M = \sqrt{5} \). A

\( A \) is bounded (\( \lVert p \rVert \le 1 \) on \( A \)) but not closed: \[ p_k = \left( \tfrac{1}{k}, 0 \right) \in A \ \ \text{for all } k \ge 1, \qquad p_k \to (0,0) \notin A . \]

\( (0,0) \) is a limit point of \( A \) not belonging to \( A \), so \( A \) fails to contain all its limit points. By Heine–Borel, \( A \) is therefore not compact. A

An explicit cover with no finite subcover: \[ U_k = \left\{ p \in \mathbb{R}^2 : \tfrac{1}{k} \lt \lVert p \rVert \lt 2 \right\}, \quad k \ge 1, \qquad \bigcup_{k \ge 1} U_k = \left\{ p : 0 \lt \lVert p \rVert \lt 2 \right\} \supseteq A . \]

Each \( U_k \) is open, being the intersection of the preimages of the open rays \( (1/k, \infty) \) and \( (-\infty, 2) \) under the continuous map \( p \mapsto \lVert p \rVert \). Every \( p \in A \) has \( 0 \lt \lVert p \rVert \le 1 \lt 2 \), so \( p \in U_k \) once \( 1/k \lt \lVert p \rVert \). B

The family is increasing, \( U_1 \subseteq U_2 \subseteq \cdots \), so any finite subfamily \( U_{k_1}, \dots, U_{k_r} \) has union \( U_N \) with \( N = \max_j k_j \). Taking \( N = 50 \), \[ q = \left( \tfrac{1}{51}, 0 \right) \approx (0.019608,\ 0), \qquad \lVert q \rVert = \tfrac{1}{51} \approx 0.019608 \lt \tfrac{1}{50} = 0.02, \] so \( q \in A \) but \( q \notin U_1 \cup \cdots \cup U_{50} \).

\( 1/k \) decreases in \( k \), so \( U_k \) grows with \( k \) and the union of finitely many is the largest. A point of \( A \) with norm below \( 1/N \) lies outside \( U_N \); such points exist for every \( N \) because \( A \) reaches arbitrarily close to the missing origin. B

Consequence for the extreme value theorem: \( f(p) = \lVert p \rVert^{-2} \) is continuous on \( A \) but \[ f(q) = 51^2 = 2601, \qquad \sup_{p \in A} f(p) = +\infty . \]

A continuous function on a compact set is bounded and attains its bounds; \( f \) does neither on \( A \). On \( E \), by contrast, every continuous function is bounded and attains its extremes. A

\[ E \ \text{compact} \ \left( \lVert p \rVert \le \sqrt{5} \approx 2.2361, \ \text{closed} \right); \qquad A \ \text{not compact}: \ \left\{ \tfrac{1}{k} \lt \lVert p \rVert \lt 2 \right\}_{k \ge 1} \ \text{has no finite subcover}. \]

Reading. Both sets are bounded and both are “almost” the same shape; the single missing point of \( A \) is enough to destroy compactness, and the destruction is witnessed by one concrete cover. Removing one point from a compact set never leaves a compact set unless the point was isolated.

Scope. The same recipe — sublevel set of a continuous function, plus a norm bound — certifies compactness for any set defined by finitely many non-strict polynomial inequalities and a bound; strict inequalities generally break closedness.

Example 2. A thin circular plate of radius \( 2\ \mathrm{m} \) occupies \( D = \left\{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 \le 4\ \mathrm{m}^2 \right\} \). Its steady temperature field is \( T(x,y) = 293\ \mathrm{K} + \left( 12\ \mathrm{K\,m^{-1}} \right) x - \left( 9\ \mathrm{K\,m^{-1}} \right) y \). Prove that the hottest and coldest points exist, and find them and the temperatures there.

\[ D = h^{-1}\big( (-\infty, 4] \big), \qquad h(x,y) = x^2 + y^2 \ \ \left[ \mathrm{m}^2 \right]. \]

\( h \) is continuous and \( (-\infty, 4] \) is closed, so \( D \) is closed in \( \mathbb{R}^2 \). Physically: the plate includes its rim. A

\[ (x,y) \in D \implies \lVert (x,y) \rVert = \sqrt{x^2+y^2} \le 2\ \mathrm{m} \quad \implies \quad D \ \text{is bounded}. \]

Immediate from the defining inequality. With Step 1 and Heine–Borel (\( n = 2 \), \( M = 2\ \mathrm{m} \)), \( D \) is compact. A

\( T \) is continuous on \( D \), so \( T(D) \subseteq \mathbb{R} \) is compact, hence closed and bounded, hence \[ \exists\, p_{\max}, p_{\min} \in D : \quad T(p_{\max}) = \max_D T, \qquad T(p_{\min}) = \min_D T . \]

Continuous images of compact sets are compact (a cover of the image pulls back to a cover of the domain); a nonempty compact subset of \( \mathbb{R} \) contains its supremum and infimum because it is closed and bounded. This is the extreme value theorem, and it is the only place existence is established — before any calculus is done. B

Write \( T(x,y) = T_0 + \langle c, (x,y) \rangle \) with \( T_0 = 293\ \mathrm{K} \) and \( c = (12, -9)\ \mathrm{K\,m^{-1}} \). Cauchy–Schwarz gives \[ \left| T(x,y) - T_0 \right| \le \lVert c \rVert \, \lVert (x,y) \rVert \le \sqrt{12^2 + 9^2} \cdot 2 = \sqrt{225} \cdot 2 = 15 \cdot 2 = 30\ \mathrm{K}. \]

Cauchy–Schwarz in \( \mathbb{R}^2 \), then the norm bound from Step 2. Symbols first: the bound is \( \lVert c \rVert R \) with \( R \) the plate radius; only then the numbers \( \lVert c \rVert = 15\ \mathrm{K\,m^{-1}} \), \( R = 2\ \mathrm{m} \). A

Equality in Cauchy–Schwarz holds exactly when \( (x,y) \) is a non-negative multiple of \( \pm c \) of length \( R \): \[ p_{\max} = \frac{R}{\lVert c \rVert} \, c = \frac{2\ \mathrm{m}}{15\ \mathrm{K\,m^{-1}}} \left( 12, -9 \right) \mathrm{K\,m^{-1}} = (1.6,\ -1.2)\ \mathrm{m}, \qquad p_{\min} = -p_{\max} = (-1.6,\ 1.2)\ \mathrm{m}. \]

The equality case of Cauchy–Schwarz. Check that these points lie in \( D \): \( 1.6^2 + 1.2^2 = 2.56 + 1.44 = 4.00 \), so \( \lVert p_{\max} \rVert = 2\ \mathrm{m} \) exactly — both extremes sit on the rim, as they must for a non-constant affine field. B

\[ T_{\max} = 293 + 30 = 323\ \mathrm{K}, \qquad T_{\min} = 293 - 30 = 263\ \mathrm{K}, \qquad \Delta T = 60\ \mathrm{K}. \]

Substituting: \( T(1.6, -1.2) = 293 + 12(1.6) - 9(-1.2) = 293 + 19.2 + 10.8 = 323 \), and \( T(-1.6, 1.2) = 293 - 19.2 - 10.8 = 263 \). In Celsius, \( 49.85\,{}^{\circ}\mathrm{C} \) and \( -10.15\,{}^{\circ}\mathrm{C} \). A

If the rim is excluded, \( D^{\circ} = \left\{ x^2 + y^2 \lt 4 \right\} \) is bounded but not closed, and \[ \sup_{D^{\circ}} T = 323\ \mathrm{K} \ \ \text{is not attained}; \qquad T\!\left( (1 - \tfrac{1}{k}) p_{\max} \right) = 323 - \tfrac{30}{k}\ \mathrm{K} \to 323\ \mathrm{K}. \]

The supremum is approached along the radius but never reached, because every candidate maximiser lies on the deleted boundary. For \( k = 100 \) the value is \( 322.7\ \mathrm{K} \). This is exactly the failure regime “closedness dropped”. B

\[ T_{\max} = 323\ \mathrm{K} \ \text{at} \ (1.6,\ -1.2)\ \mathrm{m}, \qquad T_{\min} = 263\ \mathrm{K} \ \text{at} \ (-1.6,\ 1.2)\ \mathrm{m}, \qquad T_{\max} - T_{\min} = 2 \lVert c \rVert R = 60\ \mathrm{K}. \]

Reading. Compactness of the plate is what licenses the phrase “the hottest point”; Cauchy–Schwarz then locates it. The two steps are independent: existence comes from the geometry of the domain, the location from the algebra of the field.

Scope. The extremes are attained for any continuous \( T \) on \( D \), however wild; the explicit formula \( T_{\max} = T_0 + \lVert c \rVert R \) requires \( T \) affine. Removing the rim, or letting the plate extend to infinity, removes the guarantee.

Problems

Classify each subset of \( \mathbb{R} \) or \( \mathbb{R}^2 \) as compact or not, giving an explicit open cover with no finite subcover in the non-compact cases: (a) \( S = \left\{ \tfrac{1}{k} : k \in \mathbb{N},\ k \ge 1 \right\} \); (b) \( S \cup \{0\} \); (c) \( \mathbb{Z} \); (d) \( [0,1] \cap \mathbb{Q} \) as a subset of \( \mathbb{R} \); (e) \( H = \left\{ (x,y) \in \mathbb{R}^2 : xy = 1 \right\} \).
Solution

(a) Not compact. \( S \) is bounded by \( 1 \) but not closed: \( \tfrac{1}{k} \to 0 \notin S \). Cover: \( U_k = \left( \tfrac{1}{k+1}, 2 \right) \) for \( k \ge 1 \). Then \( \bigcup_k U_k = (0,2) \supseteq S \). The family increases, so a finite subfamily lies in \( U_N \), which excludes \( \tfrac{1}{N+1} \in S \) (since \( \tfrac{1}{N+1} \) is not \( \gt \tfrac{1}{N+1} \)). For \( N = 7 \) the escaping point is \( 0.125 \).

(b) Compact. Bounded by \( 1 \); closed because the only limit point of \( S \) is \( 0 \), which has been added. Heine–Borel applies. Directly: any open cover has a member \( U_0 \ni 0 \), which contains \( (-\delta, \delta) \) for some \( \delta \gt 0 \), hence contains all \( \tfrac{1}{k} \) with \( k \gt 1/\delta \); only the finitely many points \( 1, \tfrac{1}{2}, \dots, \tfrac{1}{\lceil 1/\delta \rceil} \) remain, and one member of the cover each finishes the job. With \( \delta = 0.1 \) that is \( U_0 \) plus at most \( 10 \) further sets.

(c) Not compact. \( \mathbb{Z} \) is closed in \( \mathbb{R} \) (its complement is the union of the open intervals \( (m, m+1) \)) but unbounded. Cover: \( V_k = (-k, k) \), \( k \ge 1 \); a finite subfamily lies in \( V_N \) and misses \( N \in \mathbb{Z} \). Alternatively the singleton cover \( \left( m - \tfrac{1}{2},\ m + \tfrac{1}{2} \right) \), \( m \in \mathbb{Z} \), is an open cover in which each set contains exactly one integer, so no finite subfamily can cover the infinite set \( \mathbb{Z} \).

(d) Not compact. Bounded by \( 1 \), but not closed in \( \mathbb{R} \): \( \tfrac{1}{\sqrt{2}} \approx 0.7071 \) is a limit of rationals in \( [0,1] \) but is irrational. Cover: with \( \alpha = \tfrac{1}{\sqrt{2}} \), put \( W_k = \left( -1,\ \alpha - \tfrac{1}{k} \right) \cup \left( \alpha + \tfrac{1}{k},\ 2 \right) \), \( k \ge 1 \). Since \( \alpha \notin \mathbb{Q} \), every rational in \( [0,1] \) lies in some \( W_k \). A finite subfamily lies in \( W_N \), which omits every rational in \( \left[ \alpha - \tfrac{1}{N},\ \alpha + \tfrac{1}{N} \right] \), and rationals exist there by density. For \( N = 1000 \), \( q = 0.7071 \) is omitted, as \( |0.7071 - \alpha| \approx 6.8 \times 10^{-6} \lt 10^{-3} \).

(e) Not compact. \( H \) is closed, being \( \varphi^{-1}(\{1\}) \) for the continuous \( \varphi(x,y) = xy \), but unbounded: \( \left( t, \tfrac{1}{t} \right) \in H \) has norm \( \sqrt{t^2 + t^{-2}} \to \infty \) as \( t \to \infty \). At \( t = 10 \) the norm is already \( \sqrt{100.01} \approx 10.0005 \). The cover \( V_k = \left\{ p : \lVert p \rVert \lt k \right\} \) has no finite subcover.
Show that the unit sphere \( S^2 = \left\{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \right\} \) is compact, and use this to prove that \( f(x,y,z) = x + 2y + 3z \) attains a maximum on \( S^2 \). Find the maximum and the maximiser numerically.
Solution

Compactness. Let \( h(x,y,z) = x^2+y^2+z^2 \), a polynomial and hence continuous. Then \( S^2 = h^{-1}(\{1\}) \) and \( \{1\} \) is closed in \( \mathbb{R} \), so \( S^2 \) is closed in \( \mathbb{R}^3 \). It is bounded, since \( \lVert p \rVert = 1 \) for every \( p \in S^2 \). By Heine–Borel with \( n = 3 \) and \( M = 1 \), \( S^2 \) is compact.

Existence of the maximum. \( f \) is linear, hence continuous; \( f(S^2) \) is a compact, therefore closed and bounded, subset of \( \mathbb{R} \), so it contains \( \sup f(S^2) \). Hence the maximum is attained at some \( p_{\max} \in S^2 \).

Value. Write \( f(p) = \langle c, p \rangle \) with \( c = (1,2,3) \). Cauchy–Schwarz gives \( f(p) \le \lVert c \rVert \lVert p \rVert = \lVert c \rVert \) for \( p \in S^2 \), with equality exactly when \( p = c / \lVert c \rVert \). Now \[ \lVert c \rVert = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \approx 3.741657, \] so \[ \max_{S^2} f = \sqrt{14} \approx 3.7417 \quad \text{at} \quad p_{\max} = \frac{(1,2,3)}{\sqrt{14}} \approx (0.267261,\ 0.534522,\ 0.801784). \] Check: \( 0.267261 + 2(0.534522) + 3(0.801784) = 0.267261 + 1.069045 + 2.405351 = 3.741657 \), and \( \lVert p_{\max} \rVert^2 = 0.071429 + 0.285714 + 0.642857 = 1.000000 \). The minimum is \( -\sqrt{14} \) at \( -p_{\max} \), by \( f(-p) = -f(p) \).

Remark. Nothing here used differentiability. Lagrange multipliers would find the same critical points, but they only produce candidates; it is compactness that certifies one of them is a genuine maximum.
Let \( K_1 \supseteq K_2 \supseteq K_3 \supseteq \cdots \) be nonempty closed bounded subsets of \( \mathbb{R}^n \). Prove \( \bigcap_{j \ge 1} K_j \ne \varnothing \). Then show by explicit counterexamples that both hypotheses are needed: one where the \( K_j \) are closed but unbounded, one where they are bounded but not closed.
Solution

Proof. Suppose \( \bigcap_j K_j = \varnothing \). Put \( G_j = \mathbb{R}^n \setminus K_j \), which is open since \( K_j \) is closed. Then \[ \bigcup_{j \ge 1} G_j = \mathbb{R}^n \setminus \bigcap_{j \ge 1} K_j = \mathbb{R}^n \supseteq K_1, \] so \( \{ G_j \} \) is an open cover of \( K_1 \). By Heine–Borel, \( K_1 \) is compact, so finitely many suffice: \( K_1 \subseteq G_{j_1} \cup \cdots \cup G_{j_r} \). The \( K_j \) decrease, so the \( G_j \) increase, and the union is \( G_N \) with \( N = \max_i j_i \). Hence \( K_1 \subseteq G_N = \mathbb{R}^n \setminus K_N \), i.e. \( K_1 \cap K_N = \varnothing \). But \( K_N \subseteq K_1 \) and \( K_N \ne \varnothing \), so \( K_N = K_1 \cap K_N = \varnothing \) — a contradiction. Therefore the intersection is nonempty.

Closed but unbounded. \( K_j = [j, \infty) \subseteq \mathbb{R} \). Each is nonempty and closed, and \( K_1 \supseteq K_2 \supseteq \cdots \). Yet \( \bigcap_j K_j = \varnothing \), since any \( x \in \mathbb{R} \) fails \( x \ge j \) once \( j \gt x \) (Archimedean property): for \( x = 10^{6} \), \( x \notin K_{10^{6}+1} \).

Bounded but not closed. \( K_j = \left( 0, \tfrac{1}{j} \right] \subseteq \mathbb{R} \). Each is nonempty, bounded by \( 1 \), and nested decreasing. Yet \( \bigcap_j K_j = \varnothing \): a common point \( x \) would satisfy \( 0 \lt x \le \tfrac{1}{j} \) for every \( j \), forcing \( x \le 0 \) in the limit. For \( x = 10^{-6} \), \( x \notin K_{10^{7}} = \left( 0, 10^{-7} \right] \).

Sharpening. If in addition \( \operatorname{diam} K_j \to 0 \), the intersection is a single point: any two points of the intersection lie in every \( K_j \), so their distance is at most \( \operatorname{diam} K_j \to 0 \).
Let \( C[0,1] \) carry the supremum norm \( \lVert f \rVert_{\infty} = \max_{t \in [0,1]} |f(t)| \), and let \( \overline{B} = \left\{ f : \lVert f \rVert_{\infty} \le 1 \right\} \). Show that \( \overline{B} \) is closed and bounded but not compact, by constructing an infinite family in \( \overline{B} \) whose members are pairwise at distance exactly \( 1 \).
Solution

Closed and bounded. \( \overline{B} \) is bounded by definition (diameter at most \( 2 \)). It is closed because the norm is continuous: if \( f_k \in \overline{B} \) and \( \lVert f_k - f \rVert_{\infty} \to 0 \) then \( \lVert f \rVert_{\infty} \le \lVert f - f_k \rVert_{\infty} + \lVert f_k \rVert_{\infty} \le \lVert f - f_k \rVert_{\infty} + 1 \to 1 \). (\( C[0,1] \) with this norm is complete, so \( \overline{B} \) is even a complete metric space — what it lacks is total boundedness.)

The family. For \( k \ge 1 \) define the spike \[ g_k(t) = \max\left\{ 0,\ 1 - 2k(k+1)\left| t - \tfrac{1}{k} \right| \right\}, \qquad t \in [0,1]. \] Each \( g_k \) is continuous (a maximum of two continuous functions), has \( g_k(1/k) = 1 \) and \( 0 \le g_k \le 1 \), so \( \lVert g_k \rVert_{\infty} = 1 \) and \( g_k \in \overline{B} \). Its support is the interval of centre \( \tfrac{1}{k} \) and radius \( r_k = \tfrac{1}{2k(k+1)} \).

Pairwise disjoint supports. For \( j \lt k \) the centres satisfy \[ \frac{1}{j} - \frac{1}{k} \ \ge \ \frac{1}{j} - \frac{1}{j+1} = \frac{1}{j(j+1)}, \] while the radii satisfy \[ r_j + r_k \le r_j + r_{j+1} = \frac{1}{2j(j+1)} + \frac{1}{2(j+1)(j+2)} \lt \frac{1}{2j(j+1)} + \frac{1}{2j(j+1)} = \frac{1}{j(j+1)} . \] The separation strictly exceeds the sum of the radii, so the supports are disjoint. Numerically, \( g_2 \) is supported on \( [0.5 - \tfrac{1}{12},\ 0.5 + \tfrac{1}{12}] = [0.41\overline{6},\ 0.58\overline{3}] \) and \( g_3 \) on \( [\tfrac13 - \tfrac{1}{24},\ \tfrac13 + \tfrac{1}{24}] = [0.291\overline{6},\ 0.375] \) — disjoint, as claimed.

Consequence. For \( j \ne k \), at \( t = 1/j \) we have \( g_j = 1 \) and \( g_k = 0 \), so \( \lVert g_j - g_k \rVert_{\infty} \ge 1 \); and \( \lVert g_j - g_k \rVert_{\infty} \le 1 \) since the supports are disjoint and both functions take values in \( [0,1] \). Hence \( \lVert g_j - g_k \rVert_{\infty} = 1 \) exactly. No subsequence of \( (g_k) \) is Cauchy, so none converges: \( \overline{B} \) is not sequentially compact, hence not compact. Equivalently, the open balls \( B\!\left( g_k, \tfrac{1}{2} \right) \) each contain exactly one \( g_j \), so together with \( C[0,1] \setminus \{ g_k : k \ge 1 \} \) — open, because the \( g_k \) form a closed discrete set — they give an open cover of \( \overline{B} \) with no finite subcover.

Moral. Closed \( + \) bounded \( + \) complete is still not enough. What fails is total boundedness: no finite family of balls of radius \( \tfrac{1}{2} \) covers \( \overline{B} \), because each such ball catches at most one spike.
Let \( V \) be a real vector space with \( \dim V = n \lt \infty \) and let \( \lVert \cdot \rVert \) be any norm on \( V \). Prove that a subset of \( V \) is compact if and only if it is closed and bounded. (Hint: fix a basis, transport the problem to \( \mathbb{R}^n \), and use Heine–Borel on the Euclidean unit sphere to compare the two norms.) Illustrate the comparison constants for \( V = \mathbb{R}^2 \) with \( \lVert \cdot \rVert = \lVert \cdot \rVert_1 \).
Solution

Setup. Let \( b_1, \dots, b_n \) be a basis of \( V \) and let \( \Phi : \mathbb{R}^n \to V \), \( \Phi(a) = \sum_{i=1}^{n} a_i b_i \), be the induced linear isomorphism. Write \( N(a) = \lVert \Phi(a) \rVert \) and let \( \lVert a \rVert_2 \) be the Euclidean norm on \( \mathbb{R}^n \). It suffices to prove that \( N \) and \( \lVert \cdot \rVert_2 \) are equivalent, i.e. \[ \exists\, c, C \gt 0 : \quad c \lVert a \rVert_2 \le N(a) \le C \lVert a \rVert_2 \quad \text{for all } a \in \mathbb{R}^n, \] because equivalent norms have the same open sets, the same closed sets, the same bounded sets and the same compact sets, and \( \Phi \) then carries Heine–Borel from \( \left( \mathbb{R}^n, \lVert \cdot \rVert_2 \right) \) to \( \left( V, \lVert \cdot \rVert \right) \).

Upper bound. By the triangle inequality and Cauchy–Schwarz, \[ N(a) = \left\lVert \sum_{i=1}^{n} a_i b_i \right\rVert \le \sum_{i=1}^{n} |a_i| \, \lVert b_i \rVert \le \left( \sum_{i=1}^{n} \lVert b_i \rVert^2 \right)^{1/2} \lVert a \rVert_2 =: C \lVert a \rVert_2 , \] with \( C \gt 0 \) since the \( b_i \) are nonzero.

\( N \) is continuous. The reverse triangle inequality gives \( |N(a) - N(a')| \le N(a - a') \le C \lVert a - a' \rVert_2 \), so \( N \) is Lipschitz, hence continuous, on \( \left( \mathbb{R}^n, \lVert \cdot \rVert_2 \right) \).

Lower bound. The Euclidean unit sphere \( S = \left\{ a \in \mathbb{R}^n : \lVert a \rVert_2 = 1 \right\} \) is closed (preimage of \( \{1\} \) under the continuous \( a \mapsto \lVert a \rVert_2 \)) and bounded, hence compact by Heine–Borel. The continuous function \( N \) therefore attains a minimum on \( S \): \( c := \min_{a \in S} N(a) = N(a^{*}) \) for some \( a^{*} \in S \). Since \( a^{*} \ne 0 \) and \( \Phi \) is injective, \( \Phi(a^{*}) \ne 0 \), so \( c = \lVert \Phi(a^{*}) \rVert \gt 0 \). For general \( a \ne 0 \), homogeneity gives \[ N(a) = \lVert a \rVert_2 \, N\!\left( \frac{a}{\lVert a \rVert_2} \right) \ge c \lVert a \rVert_2 , \] and the inequality is trivial at \( a = 0 \).

Conclusion. The two norms are equivalent, so \( \Phi \) is a homeomorphism taking bounded sets to bounded sets in both directions. Given \( K \subseteq V \) closed and bounded, \( \Phi^{-1}(K) \subseteq \mathbb{R}^n \) is closed and bounded, hence compact by Heine–Borel, hence \( K = \Phi\!\left( \Phi^{-1}(K) \right) \) is compact as a continuous image of a compact set. Conversely a compact \( K \subseteq V \) is closed and bounded by exactly the arguments of Steps 9 and 10 of the proof, which used only the norm. Note where finite dimension entered: the sphere \( S \) is compact only because \( n \lt \infty \), and this is the sole ingredient — the theorem is false verbatim in infinite dimensions, where inequivalent norms exist.

Illustration. For \( V = \mathbb{R}^2 \) with the standard basis and \( \lVert a \rVert_1 = |a_1| + |a_2| \): the upper bound above gives \( C = \left( \lVert b_1 \rVert_1^2 + \lVert b_2 \rVert_1^2 \right)^{1/2} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142 \), attained at \( a = \tfrac{1}{\sqrt{2}}(1,1) \), where \( \lVert a \rVert_1 = \sqrt{2} \) and \( \lVert a \rVert_2 = 1 \). The lower constant is \( c = \min_{\lVert a \rVert_2 = 1} \lVert a \rVert_1 = 1 \), attained at \( a = (1,0) \). So \( \lVert a \rVert_2 \le \lVert a \rVert_1 \le \sqrt{2} \, \lVert a \rVert_2 \), and both constants are sharp. In \( \mathbb{R}^n \) the same computation yields \( \lVert a \rVert_2 \le \lVert a \rVert_1 \le \sqrt{n} \, \lVert a \rVert_2 \); the constant \( \sqrt{n} \) degrading as \( n \to \infty \) is the quantitative shadow of the theorem’s failure in infinite dimensions.