The Euclidean Characterization of Compactness
The previous tutorial established two important restrictions on compact subsets of \(\mathbb{R}^n\): they are closed and bounded. The converse is the central point of this tutorial. In Euclidean space, these two geometric conditions are not merely necessary; together, they guarantee compactness. This result is the Heine–Borel Theorem.
The proof uses the sequential description of compactness established in “Compactness in \(\mathbb{R}^n\).” It will be enough to show that every sequence in a closed bounded set has a subsequence converging to a point of that set. Boundedness gives a convergent subsequence in the ambient space, while closedness ensures that its limit remains in the set.
Both conditions matter. Boundedness controls how far points can travel, but does not prevent a sequence from approaching a point that is missing from the set. Closedness prevents such missing limit points, but does not prevent points from escaping arbitrarily far. The theorem says that in finite-dimensional Euclidean space, these two controls together are exactly what compactness requires.
A Subsequence Fact for Bounded Sets
We first isolate the sequence argument needed for the converse direction of Heine–Borel. The real Bolzano–Weierstrass Theorem says that every bounded sequence of real numbers has a convergent subsequence. Applying it successively to the finitely many coordinate sequences yields the corresponding result in \(\mathbb{R}^n\).
Proof. Let \((x^{(k)})_{k\geq1}\) be a bounded sequence in \(\mathbb{R}^n\), where \(x^{(k)}=(x_1^{(k)},\ldots,x_n^{(k)})\). There is a number \(M\geq0\) such that \(\|x^{(k)}\|_2\leq M\) for every \(k\). For each coordinate \(i\),
Thus each coordinate sequence is bounded. Apply the real Bolzano–Weierstrass Theorem to the first coordinate to obtain a subsequence along which that coordinate converges. From this subsequence, select a further subsequence along which the second coordinate converges. Continue in this way through coordinate \(n\). There are only finitely many coordinates, so the final subsequence has convergent coordinate sequences in every coordinate. Write their limits as \(a_1,\ldots,a_n\), and set \(a=(a_1,\ldots,a_n)\). By the Coordinate Criterion for Convergence, the final subsequence converges in \(\mathbb{R}^n\) to \(a\). \(\square\)
The finiteness of \(n\) is essential to this argument: after finitely many selections, one subsequence has all the required coordinate limits. The result gives a limit in \(\mathbb{R}^n\), not necessarily in the set containing the original sequence. That distinction is why closedness is needed in Heine–Borel.
Proof of the Heine–Borel Theorem
Proof. First suppose \(K\subseteq\mathbb{R}^n\) is compact. The Euclidean space \(\mathbb{R}^n\) is Hausdorff, so the theorem “Compact Subsets of Hausdorff Spaces Are Closed” implies that \(K\) is closed in \(\mathbb{R}^n\). The previous tutorial also proved that every compact subset of \(\mathbb{R}^n\) is bounded. Hence \(K\) is closed and bounded.
Conversely, suppose \(K\) is closed and bounded. If \(K\) is empty, it is compact by the definition of compactness. Assume now that \(K\) is nonempty, and take any sequence \((x^{(k)})_{k\geq1}\) in \(K\). Since \(K\) is bounded, this sequence is bounded in \(\mathbb{R}^n\). The Bolzano–Weierstrass Theorem in \(\mathbb{R}^n\), proved above, gives a subsequence \((x^{(k_j)})_{j\geq1}\) converging to some \(x\in\mathbb{R}^n\).
We check that \(x\in K\). If \(x\notin K\), then \(x\) belongs to the open set \(\mathbb{R}^n\setminus K\), because \(K\) is closed. There is therefore an open Euclidean ball centered at \(x\) that is contained in \(\mathbb{R}^n\setminus K\). Convergence of \(x^{(k_j)}\) to \(x\) implies that all sufficiently late terms of this subsequence lie in that ball. But every term \(x^{(k_j)}\) belongs to \(K\), which is disjoint from the ball. This is a contradiction. Therefore \(x\in K\).
We have shown that every sequence in \(K\) has a subsequence converging to a point of \(K\). Thus \(K\) is sequentially compact. By the theorem “Compactness and Sequential Compactness in \(\mathbb{R}^n\),” \(K\) is compact. This completes both directions. \(\square\)
The proof separates the roles of the hypotheses. Boundedness supplies a convergent subsequence in the ambient space; closedness keeps the subsequence’s limit inside the set. The final step converts this sequential property to compactness using the equivalence for subsets of \(\mathbb{R}^n\) established in the previous tutorial.
Worked Applications
Worked Example: A Closed Annulus in \(\mathbb{R}^n\)
Let \(n\geq1\), and consider the annulus \[ A=\{x\in\mathbb{R}^n:1\leq\|x\|_2\leq3\}. \] It is bounded because every \(x\in A\) satisfies \(\|x\|_2\leq3\).
To check closedness, let \((x^{(k)})\) be a sequence in \(A\) converging to \(x\in\mathbb{R}^n\). The reverse triangle inequality gives \[ \big|\|x^{(k)}\|_2-\|x\|_2\big|\leq\|x^{(k)}-x\|_2\longrightarrow0. \] Hence \(\|x^{(k)}\|_2\to\|x\|_2\). Since \(1\leq\|x^{(k)}\|_2\leq3\) for every \(k\), taking limits gives \(1\leq\|x\|_2\leq3\), so \(x\in A\). Thus \(A\) is closed. By Heine–Borel, \(A\) is compact.
The lower bound excludes points near the origin, but it does not interfere with compactness: both boundary spheres are included. The upper bound supplies boundedness, and including the boundary ensures closedness.
Worked Example: A Bounded Set That Is Not Compact
In \(\mathbb{R}^n\), let \[ S=\{t e_1:0\leq t<1\}, \] where \(e_1=(1,0,\ldots,0)\). Every \(x=t e_1\in S\) satisfies \(\|x\|_2=t<1\), so \(S\) is bounded. However, \(S\) is not closed: for each integer \(k\geq2\), the point \[ x^{(k)}=\left(1-\frac{1}{k}\right)e_1 \] belongs to \(S\), since \(0\leq1-1/k<1\), and \[ \left\|x^{(k)}-e_1\right\|_2=\frac{1}{k}\longrightarrow0. \] Thus \(x^{(k)}\to e_1\), but \(e_1\notin S\), because its scalar coefficient is \(1\), which is excluded by the definition of \(S\). The set is not closed, so Heine–Borel shows that it is not compact.
This example also illustrates the sequential obstruction directly: the displayed sequence has no subsequence converging to a point of \(S\). Every subsequence still converges in \(\mathbb{R}^n\) to \(e_1\), and limits in the Hausdorff space \(\mathbb{R}^n\) are unique.
Worked Example: A Closed Unbounded Set That Is Not Compact
Consider the ray \[ R=\{t e_1:t\geq0\}\subseteq\mathbb{R}^n. \] It is unbounded because \(\|k e_1\|_2=k\) for every positive integer \(k\), and these norms have no finite upper bound.
The set is closed. Indeed, suppose \(t_j e_1\in R\) and \(t_j e_1\to x=(x_1,\ldots,x_n)\). Coordinatewise convergence gives \(t_j\to x_1\) and \(0\to x_i\) for \(i=2,\ldots,n\). Since every \(t_j\geq0\), its limit satisfies \(x_1\geq0\), and the other coordinates satisfy \(x_i=0\). Therefore \(x=x_1e_1\in R\). Heine–Borel now shows that \(R\) is not compact, because it is not bounded.
For \(n=1\), the conditions on coordinates \(i=2,\ldots,n\) are absent, and the same argument says that the closed ray \([0,\infty)\) is not compact. The example shows why closedness alone cannot replace boundedness.
Worked Example: A Compact Region Defined by Inequalities
Let \[ E=\{(x,y)\in\mathbb{R}^2:x^2+4y^2\leq4,\ y\geq x\}. \] We verify the two Heine–Borel conditions directly. If \((x,y)\in E\), then \(x^2\leq4\) and \(4y^2\leq4\), so \(|x|\leq2\) and \(|y|\leq1\). Thus \(E\) is bounded.
Now let \((x_k,y_k)\in E\) converge to \((x,y)\). The coordinate limits and the limit laws for sums and products give \[ x_k^2+4y_k^2\longrightarrow x^2+4y^2 \quad\text{and}\quad y_k-x_k\longrightarrow y-x. \] Since \(x_k^2+4y_k^2\leq4\) and \(y_k-x_k\geq0\) for every \(k\), taking limits gives \(x^2+4y^2\leq4\) and \(y-x\geq0\). Hence \((x,y)\in E\), so \(E\) is closed. By Heine–Borel, \(E\) is compact.
This method is useful when a set is described by inequalities: establish a uniform bound, then check that limits preserve the inequalities. It avoids having to construct finite subcovers directly.
Why Both Conditions Matter
Heine–Borel turns the open-cover definition of compactness into a practical test for subsets of finite-dimensional Euclidean space. To prove a set compact, one can often check that it is bounded and that it contains the limits of its convergent sequences. To prove it is not compact, it is enough to show that at least one of those conditions fails.
The characterization is specific to the setting at hand. In a general metric space, a set can be closed and bounded without being compact. The proof here relies on the Bolzano–Weierstrass property in finite-dimensional Euclidean space and on the equivalence between compactness and sequential compactness for subsets of \(\mathbb{R}^n\). Do not apply the closed-and-bounded test automatically outside \(\mathbb{R}^n\).
A final common pitfall is to prove only boundedness or only closedness and conclude compactness. The half-open segment above is bounded but not closed; the closed ray is closed but unbounded. Heine–Borel requires both conditions. The next topic, connectedness in \(\mathbb{R}^n\), studies a different kind of structure: whether a set can be separated into disjoint relatively open pieces.
Check Your Understanding
Use Heine–Borel and the sequence arguments in this tutorial to answer each question.
- State both directions of the Heine–Borel Theorem for subsets of \(\mathbb{R}^n\).
- Why does a bounded sequence in \(\mathbb{R}^n\) have a subsequence converging in every coordinate?
- Where is closedness used when proving that a closed bounded set is compact?
- Give an example of a bounded set that is not compact, and identify the failed condition.
- Why does the closedness of a set not by itself imply compactness in \(\mathbb{R}^n\)?
- Why should the closed-and-bounded test not be assumed for arbitrary metric spaces?