Compactness

Proof Sketch (Spine):

A full proof traversing all ten criteria requires a web of implications. Rather than writing every one, we establish the central bridge (1) \(\Rightarrow\) (6) in detail and record the remaining connections as short structural remarks organized by cluster. 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 \(B(p, r_p)\) 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 \(x_{n_k} \in B(p, 1/k)\), producing a subsequence that converges to \(p\) and contradicts our assumption.

The family \(\{B(p, r_p) : p \in X\}\) is an open cover of \(X\), since open balls are open subsets, so by (1) a finite subcollection \(B(p_1, r_{p_1}), \ldots, B(p_m, r_{p_m})\) covers \(X\). Then \[ \mathbb{N} = \bigcup_{i=1}^{m} \{\, n : x_n \in B(p_i, r_{p_i}) \,\}, \] 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 \(\varepsilon \gt 0\) would allow constructing 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 \(B(x, r_x) \subseteq U_x\). The family \(\{B(x, r_x)\}\) is a ball cover, so (2) yields a finite subcollection \(B(x_1, r_{x_1}), \ldots, B(x_m, r_{x_m})\) 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 subtle. They rely either on Zorn's lemma or on an indirect route.

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), and any point in \(\bigcap_n F_n\) is an accumulation point of \(\{x_n\}\), so a convergent subsequence exists. Combined with (6) \(\Rightarrow\) (1) above, this closes the equivalence loop.

Geometric and functional characterizations: (7) \(\Longleftrightarrow\) (8), (1) \(\Longleftrightarrow\) (10), (9).
(1) \(\Rightarrow\) (8): compactness implies boundedness, since the cover \(\{B(x_0, n) : n \in \mathbb{N}\}\) admits a finite subcover, placing \(X\) inside a single ball. For the nearest-point property: for any superspace \(Y \supseteq X\) and any \(y \in Y\), the distance function \(d(\cdot, y): X \to \mathbb{R}\) 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. The converse (8) \(\Rightarrow\) (7) is a classical equivalence in metric-space theory. Bounded sets with the nearest-point property possess enough structure to force completeness and the existence of finite \(\varepsilon\)-nets. This direction is non-trivial and we do not reproduce it in this spine.

(1) \(\Rightarrow\) (10) is the Extreme Value Theorem, proved below. (10) \(\Rightarrow\) (1): from an open cover without a finite subcover, one constructs a continuous real-valued function on \(X\) that is unbounded (or attains no supremum), contradicting (10). The construction uses a Urysohn-type interpolation along a nested sequence of non-empty closed sets with empty intersection, which is non-trivial but standard.

(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 whose full proof uses the separability of compact metric spaces and a binary encoding of a countable base. The reverse direction (continuous images of the Cantor set are compact) follows from the Cantor set being compact 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 \(B(x_n, 1/n)\) 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 \(B(x, \varepsilon) \subseteq U\). 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 \(y \in B(x_{n_k}, 1/n_k)\) 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 \(B(x_{n_k}, 1/n_k) \subseteq B(x, \varepsilon) \subseteq U\). 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 \(A \subseteq B(a, \delta)\), and that ball lies in a single member of \(\mathcal{U}\).

Proof Sketch:

(\(\Rightarrow\)) Compactness implies closed and bounded in any metric space. The case \(K = \emptyset\) is trivial in both directions, so we assume \(K \neq \emptyset\) throughout. Boundedness:
the cover \(\{B(x, 1) : x \in K\}\) admits a finite subcover \(B(x_1, 1), \ldots, B(x_m, 1)\). 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 \(B(x_i, 1)\) and satisfies \(d(y, x_1) \le d(y, x_i) + d(x_i, x_1) \lt 1 + M\), placing \(K\) inside \(B(x_1, 1 + M)\).

Closedness:
if \(p\) is a limit point of \(K\) not in \(K\), the open sets \(\{X \setminus \overline{B}(p, 1/k) : k \in \mathbb{N}\}\) 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. Any finite subfamily would leave \(K \cap \overline{B}(p, 1/\max k_i)\) uncovered, yet every neighborhood of \(p\) meets \(K\) by the limit-point assumption. 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 \(\varepsilon / (2\sqrt{n})\), each having diameter \(\varepsilon / 2 \lt \varepsilon\). 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 \(f(B(x, \delta_x)) \subseteq B(f(x), \varepsilon/2)\). The family \(\{B(x, \delta_x / 2) : x \in K\}\), whose radii are deliberately halved, is an open cover of \(K\), so by compactness it admits a finite subcover \(\{B(x_1, \delta_{x_1}/2), \ldots, B(x_n, \delta_{x_n}/2)\}\).

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 \(B(x_i, \delta_{x_i}/2)\), 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 \(B(x_i, \delta_{x_i})\). Both \(f(x)\) and \(f(y)\) therefore lie in \(B(f(x_i), \varepsilon/2)\), so \(d(f(x), f(y)) \lt \varepsilon\). This \(\delta\) depends only on \(\varepsilon\), proving uniform continuity.