Compactness

Proof:

A proof traversing all ten criteria requires a web of implications. We establish the central bridge (1) \(\Rightarrow\) (6) first, then close the remaining connections cluster by cluster, and state the forward direction of (9) without proof. That bridge exposes the deep link between the covering and sequential viewpoints.

Core implication: (1) \(\Rightarrow\) (6).
Suppose every open cover of \(X\) admits a finite subcover. Let \(\{x_n\}\) be a sequence in \(X\). Assume for contradiction that \(\{x_n\}\) has no convergent subsequence. Then for every \(p \in X\) there exists \(r_p \gt 0\) such that the ball \(\mathcal{B}[p; r_p)\) contains \(x_n\) for only finitely many indices \(n\). Otherwise the balls of radius \(1/k\) around \(p\) would each contain infinitely many terms, and we could inductively select indices \(n_1 \lt n_2 \lt \cdots\) with \(x_{n_k} \in \mathcal{B}[p; 1/k)\), producing a subsequence that converges to \(p\) and contradicts our assumption.

The family \(\{\mathcal{B}[p; r_p) : p \in X\}\) is an open cover of \(X\), since open balls are open subsets, so by (1) a finite subcollection \(\mathcal{B}[p_1; r_{p_1}), \ldots, \mathcal{B}[p_m; r_{p_m})\) covers \(X\). Then \[ \mathbb{N} = \bigcup_{i=1}^{m} \{\, n : x_n \in \mathcal{B}[p_i; r_{p_i}) \,\}, \] a finite union of finite sets. Yet \(\mathbb{N}\) is infinite, a contradiction. Hence a convergent subsequence exists.

Sequential \(\leftrightarrow\) covering cluster: (6) \(\Longleftrightarrow\) (7) \(\Longleftrightarrow\) (1).
For (6) \(\Rightarrow\) (7): a Cauchy sequence with a convergent subsequence converges to the same limit, so every Cauchy sequence in \(X\) converges and \(X\) is complete. Suppose \(X\) were not totally bounded. Then some \(\varepsilon \gt 0\) admits no finite cover of \(X\) by balls of radius \(\varepsilon\), so picking any \(y_1 \in X\) and then each \(y_{n+1}\) outside \(\mathcal{B}[y_1; \varepsilon) \cup \cdots \cup \mathcal{B}[y_n; \varepsilon)\) produces a sequence \(\{y_n\}\) with pairwise distances \(\ge \varepsilon\). Such a sequence admits no convergent subsequence, since any convergent subsequence is Cauchy and would force \(d(y_{n_k}, y_{n_l}) \lt \varepsilon\) for sufficiently large \(k \ne l\), contradicting the pairwise \(\ge \varepsilon\) condition. The existence of \(\{y_n\}\) therefore contradicts (6).

For (7) \(\Rightarrow\) (6): given any sequence, total boundedness provides, at each scale \(1/k\), a ball of radius \(1/k\) containing infinitely many terms. Iterating this inside successive subsequences and extracting the diagonal produces a subsequence in which any two terms beyond index \(k\) lie in a common ball of radius \(1/k\), hence within distance \(2/k\) of each other. That subsequence is Cauchy, and completeness delivers its limit.

For (6) \(\Rightarrow\) (1): let \(\mathcal{U}\) be an open cover of \(X\) and let \(\delta \gt 0\) be a Lebesgue number for it, which (6) supplies on its own. Total boundedness is already in hand from (6) \(\Rightarrow\) (7), so finitely many balls of radius \(\delta/3\) cover \(X\). Each of them is non-empty and has diameter \(\le 2\delta/3 \lt \delta\), hence lies in a single member of \(\mathcal{U}\), and those finitely many members form a finite subcover.

Covering-style criteria: (1) \(\Longleftrightarrow\) (2) \(\Longleftrightarrow\) (3).
(1) \(\Rightarrow\) (2) because ball covers are a special case of open covers, so the finite-subcover property transfers directly. (2) \(\Rightarrow\) (1): given any open cover \(\mathcal{U}\), for each \(x \in X\) pick \(U_x \in \mathcal{U}\) with \(x \in U_x\) and, since \(U_x\) is open, a ball \(\mathcal{B}[x; r_x) \subseteq U_x\). The family \(\{\mathcal{B}[x; r_x)\}\) is a ball cover, so (2) yields a finite subcollection \(\mathcal{B}[x_1; r_{x_1}), \ldots, \mathcal{B}[x_m; r_{x_m})\) covering \(X\), and the corresponding \(\{U_{x_1}, \ldots, U_{x_m}\}\) is a finite subcover of \(\mathcal{U}\).

(1) \(\Longleftrightarrow\) (3) is De Morgan duality. A family of open sets covers \(X\) if and only if the complementary family of closed sets has empty intersection, and a finite subcover corresponds to a finite subfamily with empty intersection, which is exactly the negation of the finite intersection property.

Nested-set criteria: (3) \(\Rightarrow\) (4) \(\Rightarrow\) (5).
(3) \(\Rightarrow\) (4) \(\Rightarrow\) (5) by specialization. A nest (totally ordered family under \(\subseteq\)) of non-empty closed sets has the finite intersection property (any finite subfamily has a minimum element among its members, which is non-empty), and any nested sequence \(F_1 \supseteq F_2 \supseteq \cdots\) is a countable nest. The converse directions, restoring (3) from (4) or (5), are obtained indirectly, by passing through (6).

The cleanest closure is sequential. Given a sequence \(\{x_n\}\), the nested closed sets \(F_n = \overline{\{x_k : k \ge n\}}\) have non-empty intersection by any of (3), (4), (5). If \(x \in \bigcap_n F_n\), every ball about \(x\) meets \(\{x_k : k \ge n\}\) for every \(n\), so we may choose indices \(n_1 \lt n_2 \lt \cdots\) with \(x_{n_k} \in \mathcal{B}[x; 1/k)\), and \(\{x_{n_k}\}\) converges to \(x\). Combined with (6) \(\Rightarrow\) (1) above, this closes the equivalence loop.

Geometric and functional characterizations: (1) \(\Longleftrightarrow\) (8), (1) \(\Longleftrightarrow\) (10), (9).
(1) \(\Rightarrow\) (8): compactness implies boundedness, since for any fixed \(x_0 \in X\) the cover \(\{\mathcal{B}[x_0; n) : n \in \mathbb{N}\}\) admits a finite subcover, placing \(X\) inside a single ball. For the nearest-point property: for any metric superspace \((Y, e)\) of \(X\) and any \(y \in Y\), the function \(e(\cdot, y): X \to \mathbb{R}\) is continuous (Lipschitz with constant 1). The Extreme Value Theorem, proved below as a consequence of (1), then shows that this function attains its minimum on the compact \(X\), and any minimizer is a nearest point.

(8) \(\Rightarrow\) (6): suppose \(X\) is bounded and has the nearest-point property, and suppose for contradiction that some sequence \(\{x_n\}\) in \(X\) has no convergent subsequence. Put \(D = \operatorname{diam}(X)\), which is finite because \(X\) lies in some ball \(\mathcal{B}[a; R)\) and then \(D \le 2R\). Adjoin a new point \(z \notin X\), and define \(e\) on \(X \cup \{z\}\) by \(e = d\) on \(X \times X\), \(e(z, z) = 0\), and \(e(x, z) = e(z, x)\) for \(x \in X\), where \[ e(z, x) = \inf_{n \in \mathbb{N}} \Bigl( d(x, x_n) + D + 1 + \frac{1}{n} \Bigr) \quad (x \in X). \] Every term of the infimum exceeds \(D + 1\), so \(e(z, x) \ge D + 1 \gt 0\). Taking the infimum over \(n\) in \(d(x, x_n) \le d(x, y) + d(y, x_n)\) gives \(e(z, x) \le d(x, y) + e(z, y)\) for \(x, y \in X\). In the other direction, \(d(x, y) \le d(x, x_n) + d(x_n, x_m) + d(x_m, y) \le d(x, x_n) + D + d(x_m, y)\) for all \(n, m\), and the right side is smaller than the sum of the \(n\)-th and \(m\)-th terms defining \(e(z, x)\) and \(e(z, y)\), so \(d(x, y) \le e(z, x) + e(z, y)\). Hence \(e\) is a metric and \(X \cup \{z\}\) is a metric superspace of \(X\). Since \(e(z, x_m) \le D + 1 + 1/m\) for every \(m\), we get \(\operatorname{dist}(z, X) = D + 1\). A nearest point \(x\) would satisfy \(e(z, x) = D + 1\), that is, \(\inf_n \bigl(d(x, x_n) + 1/n\bigr) = 0\). Each term is at least \(1/n\), so a term below \(1/m\) forces \(n \gt m\). We may therefore choose \(n_1\) with term below \(1\), and inductively \(n_{k+1}\) with term below \(\min\{1/(k+1), 1/n_k\}\), which forces \(n_{k+1} \gt n_k\) and gives \(d(x, x_{n_k}) + 1/n_k \lt 1/k\) for every \(k\). Then \(\{x_{n_k}\}\) converges to \(x\), which is excluded. So \(z\) has no nearest point in \(X\), contradicting (8). Together with (6) \(\Rightarrow\) (1) above, this closes (8) \(\Rightarrow\) (1).

(1) \(\Rightarrow\) (10) is the Extreme Value Theorem, proved below. (10) \(\Rightarrow\) (7): we exhibit a continuous function violating (10) whenever (7) fails. If \(X\) is not complete, then by the Cauchy criterion for completeness some Cauchy sequence \(\{x_n\}\) has no limit in \(X\). For each \(x \in X\), \(|d(x, x_n) - d(x, x_m)| \le d(x_n, x_m)\) makes \(\{d(x, x_n)\}\) a Cauchy sequence of real numbers, so, \(\mathbb{R}\) being complete, \(f(x) = \lim_n d(x, x_n)\) exists, and \(|f(x) - f(y)| \le d(x, y)\) makes \(f\) continuous. Each value \(f(x)\) is positive, since \(f(x) = 0\) would mean \(x_n \to x\). Yet \(f(x_m) \le \sup_{n \ge m} d(x_m, x_n) \to 0\) as \(m \to \infty\) because the sequence is Cauchy (in the \(\varepsilon\)-\(N\) form), so \(f\) does not attain its infimum \(0\). If instead \(X\) is not totally bounded, some \(r \gt 0\) admits no finite cover of \(X\) by balls of radius \(r\), and picking any \(y_1 \in X\) and then each \(y_{n+1}\) outside \(\mathcal{B}[y_1; r) \cup \cdots \cup \mathcal{B}[y_n; r)\) produces points with \(d(y_n, y_m) \ge r\) for \(n \ne m\). Put \[ f(x) = \sum_{n=1}^{\infty} n \max\Bigl\{0, 1 - \frac{4\, d(x, y_n)}{r}\Bigr\}. \] The \(n\)-th term vanishes unless \(d(x, y_n) \lt r/4\), and a ball \(\mathcal{B}[x; r/8)\) meets at most one of the sets \(\{w : d(w, y_n) \le r/4\}\), since meeting two of them would give \(d(y_n, y_m) \le r/4 + r/8 + r/8 + r/4 \lt r\). In particular at most one term is non-zero at any point, so the sum is finite, and on each such ball \(f\) coincides with a single continuous term (or with \(0\)), so \(f\) is continuous, while \(f(y_n) = n\) makes it unbounded. Either way (10) fails, and together with (7) \(\Rightarrow\) (6) \(\Rightarrow\) (1) above, this closes (10) \(\Rightarrow\) (1).

(9): the Hausdorff-Alexandroff Theorem states that every non-empty compact metric space is a continuous image of the Cantor set. This is a classical result in descriptive set theory. Its proof, which uses the separability of compact metric spaces and a binary encoding of a countable base, is beyond the scope of this page. The reverse direction (continuous images of the Cantor set are compact) follows from the Cantor set being compact (it is \(\bigcap_k C_k\), where \(C_k\) is the union of the \(2^k\) closed intervals left after \(k\) rounds of removing open middle thirds from \([0, 1]\), hence closed and bounded in \(\mathbb{R}\), and compact by the Heine-Borel theorem below) and from continuous images of compact sets being compact, established separately below.

Proof:

Suppose no such \(\delta\) exists. In particular \(\delta = 1/n\) fails for each \(n \in \mathbb{N}\), which produces a point \(x_n \in X\) for which \(\mathcal{B}[x_n; 1/n)\) is contained in no single member of \(\mathcal{U}\). By hypothesis the sequence \(\{x_n\}\) has a subsequence \(\{x_{n_k}\}\) converging to some \(x \in X\).

Since \(\mathcal{U}\) covers \(X\), there is \(U \in \mathcal{U}\) containing \(x\), and since \(U\) is open, a ball \(\mathcal{B}[x; \varepsilon) \subseteq U\). Since \(x_{n_k} \to x\) and \(n_k \ge k\), we may fix \(k\) with both \(d(x_{n_k}, x) \lt \varepsilon/2\) and \(1/n_k \lt \varepsilon/2\). For every \(y \in \mathcal{B}[x_{n_k}; 1/n_k)\) the triangle inequality then gives \[ d(y, x) \le d(y, x_{n_k}) + d(x_{n_k}, x) \lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon, \] so \(\mathcal{B}[x_{n_k}; 1/n_k) \subseteq \mathcal{B}[x; \varepsilon) \subseteq U\). That contradicts the choice of \(x_{n_k}\), and the first assertion follows.

For the second assertion, let \(A\) be a non-empty subset with \(\operatorname{diam}(A) \lt \delta\) and fix \(a \in A\). Every \(y \in A\) satisfies \(d(y, a) \le \operatorname{diam}(A) \lt \delta\), so \(A \subseteq \mathcal{B}[a; \delta)\), and that ball lies in a single member of \(\mathcal{U}\).

Proof Sketch:

(\(\Rightarrow\)) Compactness implies closed and bounded in any metric space, so we let \(K\) be a compact subset of a metric space \(X\) (for the theorem, \(X = \mathbb{R}^n\)). The case \(K = \emptyset\) is trivial in both directions, so we assume \(K \neq \emptyset\) throughout. Boundedness:
the cover \(\{\mathcal{B}[x; 1) : x \in K\}\) admits a finite subcover \(\mathcal{B}[x_1; 1), \ldots, \mathcal{B}[x_m; 1)\). Since the finite set \(\{x_1, \ldots, x_m\}\) has bounded diameter \(M := \max_{i,j} d(x_i, x_j)\), any \(y \in K\) lies in some \(\mathcal{B}[x_i; 1)\) and satisfies \(d(y, x_1) \le d(y, x_i) + d(x_i, x_1) \lt 1 + M\), placing \(K\) inside \(\mathcal{B}[x_1; 1 + M)\).

Closedness:
if \(p\) is a limit point of \(K\) not in \(K\), consider the sets \(X \setminus \mathcal{B}[p; 1/k]\) for \(k \in \mathbb{N}\). Each is open, since whenever \(d(y, p) \gt 1/k\) the triangle inequality gives \(\mathcal{B}[y; d(y, p) - 1/k) \subseteq X \setminus \mathcal{B}[p; 1/k]\). They cover \(K\) (every \(x \in K\) has positive distance to \(p\), so \(1/k \lt d(x, p)\) for some \(k\)), but no finite subcover exists. A finite subfamily, indexed by \(k_1, \ldots, k_m\), would leave \(K \cap \mathcal{B}[p; 1/\max_i k_i]\) uncovered, and that set is non-empty. Since \(p \notin K\), the limit-point assumption gives \(\operatorname{dist}(p, K) = 0\), so the open ball \(\mathcal{B}[p; 1/\max_i k_i)\) meets \(K\), and it lies inside the closed ball. This contradicts compactness of \(K\), so \(p \in K\).

(\(\Leftarrow\)) This direction is the distinctively Euclidean content. By the equivalence above, it suffices to show closed-and-bounded subsets of \(\mathbb{R}^n\) are complete and totally bounded (criterion (7)). Completeness:
\(\mathbb{R}^n\) is complete, and a closed subset of a complete metric space is complete (see complete subset theorem).

Total boundedness:
a bounded subset \(K \subseteq \mathbb{R}^n\) sits inside some cube \([-R, R]^n\). Given \(\varepsilon \gt 0\), partition this cube into finitely many sub-cubes of side length at most \(\varepsilon / (2\sqrt{n})\), each having diameter at most \(\varepsilon / 2 \lt \varepsilon\). From each sub-cube that meets \(K\) choose one point of \(K\). Any other point of \(K\) in the same sub-cube is within \(\varepsilon / 2\) of that chosen point, so the finitely many balls of \(K\) of radius \(\varepsilon\) about the chosen points cover \(K\).

Proof:

Let \(\mathcal{C} = \{V_i\}_{i \in I}\) be any open cover of \(f(K)\) in \(Y\). By continuity of \(f\), each preimage \(f^{-1}(V_i)\) is open in \(X\), and since \(\mathcal{C}\) covers \(f(K)\), the family \(\{f^{-1}(V_i)\}_{i \in I}\) covers \(K\). Compactness of \(K\) yields a finite subcollection \(f^{-1}(V_{i_1}), \ldots, f^{-1}(V_{i_n})\) covering \(K\). Applying \(f\) gives \(V_{i_1}, \ldots, V_{i_n}\) covering \(f(K)\). Thus every open cover of \(f(K)\) has a finite subcover, so \(f(K)\) is compact.

Proof:

By the Preservation of Compactness theorem, \(f(K) \subseteq \mathbb{R}\) is compact. By the Heine-Borel theorem, \(f(K)\) is closed and bounded in \(\mathbb{R}\). Since \(K \ne \emptyset\) we have \(f(K) \ne \emptyset\), so boundedness of \(f(K)\) means \(\sup f(K)\) and \(\inf f(K)\) exist and are finite.

Closedness of \(f(K)\) means these supremum and infimum belong to \(f(K)\) itself. Indeed, if \(\sup f(K) \notin f(K)\), then by the defining property of the supremum every \(\varepsilon\)-neighborhood contains some \(y \in f(K)\) with \(y \gt \sup f(K) - \varepsilon\), giving \(\operatorname{dist}(\sup f(K), f(K)) = 0\). Combined with \(\sup f(K) \in f(K)^c\) (which gives \(\operatorname{dist}(\sup f(K), f(K)^c) = 0\) trivially), this places \(\sup f(K)\) in \(\partial f(K) \subseteq \overline{f(K)} = f(K)\), a contradiction. Hence \(\sup f(K) \in f(K)\), and analogously \(\inf f(K) \in f(K)\).

Therefore there exist \(x_{\max}, x_{\min} \in K\) with \(f(x_{\max}) = \sup f(K)\) and \(f(x_{\min}) = \inf f(K)\), and for every \(x \in K\) we have \(f(x) \in f(K)\), so \(f(x_{\min}) \le f(x) \le f(x_{\max})\).

Proof Sketch:

Fix \(\varepsilon \gt 0\). By continuity, for each \(x \in K\) there exists \(\delta_x \gt 0\) such that \(f(\mathcal{B}[x; \delta_x)) \subseteq \mathcal{B}[f(x); \varepsilon/2)\). The family \(\{\mathcal{B}[x; \delta_x / 2) : x \in K\}\), whose radii are deliberately halved, is an open cover of \(K\), so by compactness it admits a finite subcover \(\{\mathcal{B}[x_1; \delta_{x_1}/2), \ldots, \mathcal{B}[x_n; \delta_{x_n}/2)\}\).

Set \(\delta = \min_{i} \delta_{x_i} / 2 \gt 0\). This choice ensures \(\delta + \delta_{x_i}/2 \le \delta_{x_i}\) for every \(i\), so that any point within distance \(\delta\) of a half-size ball lies inside the corresponding full-size ball where \(f\) is controlled. Concretely, for any \(x, y \in K\) with \(d(x, y) \lt \delta\), the point \(x\) lies in some \(\mathcal{B}[x_i; \delta_{x_i}/2)\), so \(d(x, x_i) \lt \delta_{x_i}/2\). Then \(d(y, x_i) \le d(y, x) + d(x, x_i) \lt \delta + \delta_{x_i}/2 \le \delta_{x_i}\), placing \(y\) in \(\mathcal{B}[x_i; \delta_{x_i})\). Both \(f(x)\) and \(f(y)\) therefore lie in \(\mathcal{B}[f(x_i); \varepsilon/2)\), so \(d(f(x), f(y)) \lt \varepsilon\). This \(\delta\) depends only on \(\varepsilon\), proving uniform continuity.