Topological Connectedness & Compactness

Proof:

A singleton is connected trivially. For a subset \(J\) with more than one point, the equivalence "connected \(\iff\) interval" depends only on the order structure of \(\mathbb{R}\), its Dedekind completeness, and the open sets these generate. These are the same whether \(\mathbb{R}\) is viewed as a topological space or as a metric space. The argument is carried out in full for the metric setting, and applies verbatim here.

In outline, suppose first that \(J\) is not an interval. Points \(a, b \in J\) and \(c \notin J\) with \(a \lt c \lt b\) yield the separation \(J = \bigl(J \cap (-\infty, c)\bigr) \sqcup \bigl(J \cap (c, \infty)\bigr)\), so \(J\) is disconnected. Conversely, suppose \(J\) is an interval and \(J = U \sqcup V\) is a hypothetical separation with \(a \in U\), \(b \in V\), \(a \lt b\). Examining \(c = \sup\{\, t \in [a,b] : [a,t] \subseteq U \,\}\) rules this out, since \(c\) can lie in neither \(U\) nor \(V\) without contradicting either the openness of that set or the defining supremum.

Proof:

(1) View \(F\) as a surjection onto \(F(X)\). If \(F(X) = A \sqcup B\) were a separation by sets open in \(F(X)\), then \(F^{-1}(A)\) and \(F^{-1}(B)\) would be open by continuity, disjoint, and nonempty by surjectivity, with union \(X\). That is a separation of \(X\), contradicting its connectedness.

(3) Let \(\{C_\alpha\}\) be connected subspaces sharing a point \(p\), and suppose \(\bigcup_\alpha C_\alpha = U \sqcup V\) is a separation with \(p \in U\). For each \(\alpha\), the sets \(U \cap C_\alpha\) and \(V \cap C_\alpha\) are open in \(C_\alpha\) and partition it. Since \(C_\alpha\) is connected and \(p \in U \cap C_\alpha \neq \emptyset\), we must have \(V \cap C_\alpha = \emptyset\), that is, \(C_\alpha \subseteq U\). Then \(V = \emptyset\), a contradiction.

(2) The union of all connected subsets containing a point \(x\) is connected by (3) and is by construction maximal, hence the component of \(x\). Any connected subset meeting a component is absorbed into it by (3).

(4) The components are nonempty and partition \(X\) by (2)-(3). For closedness, we first note that the closure of a connected subset is connected. Suppose \(C\) is connected and \(\overline{C} = A \sqcup B\) were a separation into sets open in \(\overline{C}\). Then \(C\), being connected, would lie entirely in one part, say \(C \subseteq A\). But then any point \(x \in B\) would have \(B\) as a neighborhood (in \(\overline{C}\)) disjoint from \(C\), contradicting the characterization of the closure, by which every neighborhood of a point of \(\overline{C}\) meets \(C\). Hence \(\overline{C}\) is connected, and by maximality each component equals its own closure and is therefore closed.

(5) A clopen \(S\) meets each component \(C\) in a subset that is clopen in the connected space \(C\), hence equal to \(\emptyset\) or to \(C\). Thus \(S\) is the union of the components it contains.

(6) For two factors, assume both are nonempty, since otherwise the product is empty and therefore connected, and fix a base point \((x_0, y_0) \in X \times Y\). For each \(x \in X\) the "cross" \(T_x = (\{x\} \times Y) \cup (X \times \{y_0\})\) is connected. It is the union of \(\{x\} \times Y \cong Y\) and \(X \times \{y_0\} \cong X\), which share the point \((x, y_0)\), so (3) applies. Every \(T_x\) contains the common slice \(X \times \{y_0\}\), hence \(\bigcup_{x} T_x\) is connected by (3) again, and this union is all of \(X \times Y\). Finite products follow by induction.

(7) is a special case of (1), since a quotient map is a continuous surjection.

Proof:

For (1), each clause is proved by exhibiting paths rather than ruling out separations. A continuous image carries a path \(f\) from \(p\) to \(q\) to the path \(F \circ f\) from \(F(p)\) to \(F(q)\). For a union of path-connected sets sharing a point \(r\), any two points are joined by concatenating a path into \(r\) with a path out of \(r\). If \(g, h : I \to X\) satisfy \(g(1) = h(0)\), their concatenation \(g \ast h\), equal to \(g(2t)\) on \([0, \tfrac12]\) and \(h(2t - 1)\) on \([\tfrac12, 1]\), is continuous by the gluing lemma, since it is continuous on each of the two closed subintervals and the two definitions agree at \(t = \tfrac12\), where \(g(1) = h(0)\). For a finite product, two points \((x_1, y_1)\) and \((x_2, y_2)\) of \(X \times Y\) are joined by first moving in the \(X\)-factor along a path from \(x_1\) to \(x_2\) at height \(y_1\), then in the \(Y\)-factor at base \(x_2\). These assemble into a path in the product. A quotient \(q : X \to X/{\sim}\) is a continuous surjection, so \(q \circ f\) joins the images of the endpoints of any path \(f\).

For (2), suppose \(X\) is path-connected but \(X = U \sqcup V\) is a separation. Pick \(p \in U\), \(q \in V\), and a path \(f : I \to X\) from \(p\) to \(q\). Then \(f^{-1}(U)\) and \(f^{-1}(V)\) are open in \(I\) by continuity, disjoint, and nonempty, containing \(0\) and \(1\) respectively, with union \(I\). That is a separation of \(I\), contradicting that \(I\) is connected as an interval.

Proof:

(1) Let \(C\) be a component and \(x \in C\). By hypothesis \(x\) has a path-connected open neighborhood \(U\). Since \(U\) is path-connected it is connected, and meeting \(C\) it is contained in \(C\). Thus \(C\) is a neighborhood of each of its points, hence open.

(2) Each path component is connected, so contained in a component. Conversely, fix \(x\) and let \(P\) be its path component. The set of points reachable from \(x\) by a path is open, since each such point has a path-connected open neighborhood all of whose points are reachable, and its complement within the component is open by the same argument. Since the component is connected, \(P\) exhausts it.

(3) Immediate from (2). A connected space is a single component, hence a single path component, hence path-connected. The converse is the previous proposition.

(4) Let \(S \subseteq X\) be open. The members of the ambient basis that are contained in \(S\) form a basis for \(S\), and each of them is path-connected, so \(S\) is locally path-connected.

Proof:

(1) This is the topological form of the statement that the continuous image of a compact metric space is compact. Given an open cover of \(F(X)\), its preimages cover \(X\), a finite subcollection covers \(X\) by compactness, and the corresponding images cover \(F(X)\).

(2) Applying (1) with \(Y = \mathbb{R}\), the image \(f(X)\) is a compact subset of \(\mathbb{R}\), hence closed and bounded by Heine-Borel. A closed bounded subset of \(\mathbb{R}\) contains its supremum and infimum, which are the attained maximum and minimum. This is the extreme value theorem.

(3) An open cover of a finite union \(K_1 \cup \cdots \cup K_m\) covers each \(K_j\). Extracting a finite subcover for each and taking the union of these finitely many finite families gives a finite subcover of the whole.

(4) For each \(x \in K\), the Hausdorff condition together with compactness of \(L\) yields disjoint open sets \(U_x \ni x\) and \(W_x \supseteq L\), by the point-compact-set form of the argument below. The sets \(\{U_x\}_{x \in K}\) cover \(K\), a finite subfamily \(U_{x_1}, \ldots, U_{x_k}\) covers \(K\), and \(U = \bigcup_i U_{x_i}\), \(V = \bigcap_i W_{x_i}\) are disjoint open sets separating \(K\) and \(L\).

(5) If \(K\) is closed in the compact space \(X\) and \(\{U_\alpha\}\) is an open cover of \(K\) by sets open in \(X\), then \(\{U_\alpha\} \cup \{X \setminus K\}\) is an open cover of \(X\). A finite subcover, with \(X \setminus K\) discarded, covers \(K\). Clause (6) is the separate theorem below.

(7) In a metric space a compact set is covered by the balls of radius \(1\) about its points, and a finite subcover bounds it.

(8) and (9) for products and quotients are the topological counterparts of the corresponding clauses for connectedness. The finite product case is the special case of Tychonoff's theorem for finitely many factors, and the quotient case follows from (1), since a quotient map is a continuous surjection.

Proof:

Fix a countable basis \(\{D_m\}_{m \in \mathbb{N}}\) for the topology of \(X\), and let \(\mathcal{U}\) be an open cover of \(X\). Let \(N \subseteq \mathbb{N}\) be the set of indices \(m\) for which \(D_m\) is contained in at least one member of \(\mathcal{U}\), and for each \(m \in N\) select one such member, calling it \(U_m\). The family \(\{U_m\}_{m \in N}\) is a subfamily of \(\mathcal{U}\) indexed by a subset of \(\mathbb{N}\), hence countable.

It remains to check that this subfamily still covers \(X\). Let \(x \in X\). Since \(\mathcal{U}\) covers \(X\), some member \(U \in \mathcal{U}\) contains \(x\). The set \(U\) is open, so the basis property supplies an index \(m\) with \(x \in D_m \subseteq U\). That index lies in \(N\), witnessed by \(U\) itself, so \(U_m\) is defined and \(x \in D_m \subseteq U_m\). The point \(x\) was arbitrary, so \(\{U_m\}_{m \in N}\) covers \(X\).

Proof:

If \(X = \emptyset\) the conclusion holds vacuously with \(\delta = 1\), so assume \(X\) is nonempty. We first attach a radius to every point. Let \(x \in X\). Since \(\mathcal{U}\) covers \(X\), some member \(U_x \in \mathcal{U}\) contains \(x\). Being open, the set \(U_x\) contains no boundary point of itself, so \(x \notin \partial U_x\). The distance from \(x\) to \(U_x\) is \(0\), because \(x\) lies in \(U_x\), and hence the distance from \(x\) to \(X \setminus U_x\) is not \(0\). Choose \(r_x \gt 0\) such that every \(z \in X \setminus U_x\) satisfies \(d(x, z) \geq 2 r_x\). This is possible when that distance is a positive real number, and also when \(U_x = X\), since the condition is then vacuous. Every \(y \in X\) with \(d(x, y) \lt 2 r_x\) therefore lies in \(U_x\), that is, \(\mathcal{B}[x ; 2 r_x) \subseteq U_x\). The point \(x\) was arbitrary, so this construction runs at every point of \(X\).

The balls \(\mathcal{B}[x ; r_x)\), as \(x\) ranges over \(X\), cover \(X\), since each point lies in its own ball. By the ball cover criterion among the equivalent formulations of compactness, finitely many of them already cover \(X\), say \(X = \mathcal{B}[x_1 ; r_1) \cup \cdots \cup \mathcal{B}[x_n ; r_n)\), where \(r_i\) abbreviates \(r_{x_i}\). A cover of a nonempty set is itself nonempty, so \(n \geq 1\) and \(\delta = \min\{r_1, \ldots, r_n\}\) is a positive real number.

Let \(w \in X\) be arbitrary and choose \(i\) with \(w \in \mathcal{B}[x_i ; r_i)\). If \(y \in X\) satisfies \(d(w, y) \lt \delta\), the triangle inequality gives \(d(x_i, y) \leq d(x_i, w) + d(w, y) \lt r_i + \delta \leq 2 r_i\), so \(y \in \mathcal{B}[x_i ; 2 r_i) \subseteq U_{x_i}\). Hence \(\mathcal{B}[w ; \delta) \subseteq U_{x_i}\), which is the first assertion.

For the second assertion, let \(S \subseteq X\) be nonempty with \(\operatorname{diam}(S) \lt \delta\) and pick \(p \in S\). Every \(y \in S\) satisfies \(d(p, y) \leq \operatorname{diam}(S) \lt \delta\), so \(S \subseteq \mathcal{B}[p ; \delta)\), and the first assertion places that ball inside a single member of \(\mathcal{U}\).

Proof:

If \(K = \emptyset\), it is closed trivially, so assume \(K \neq \emptyset\). We show that \(X \setminus K\) is open. Let \(x \in X \setminus K\). For each \(y \in K\), the Hausdorff condition provides disjoint open sets \(U_y \ni x\) and \(V_y \ni y\). The collection \(\{V_y\}_{y \in K}\) covers \(K\), so by compactness there exist \(y_1, \ldots, y_n\) with \(K \subseteq V_{y_1} \cup \cdots \cup V_{y_n}\). Set \(U = U_{y_1} \cap \cdots \cap U_{y_n}\). Then \(U\) is a finite intersection of open sets containing \(x\), so \(U\) is an open neighborhood of \(x\). Moreover, \(U \subseteq U_{y_i}\) for each \(i\), so \(U \cap V_{y_i} \subseteq U_{y_i} \cap V_{y_i} = \emptyset\). Hence \(U\) is disjoint from \(V_{y_1} \cup \cdots \cup V_{y_n} \supseteq K\), giving \(U \subseteq X \setminus K\). Since \(x\) was arbitrary, \(X \setminus K\) is open.

The same construction yields more. The open set \(W = V_{y_1} \cup \cdots \cup V_{y_n} \supseteq K\) is disjoint from the open neighborhood \(U \ni x\), so a point and a disjoint compact set in a Hausdorff space can always be separated by disjoint open sets. This is the point-compact-set separation invoked in clause (4) above.

Proof:

We prove (1) directly, and the remaining three assertions then follow formally. Let \(K \subseteq X\) be closed. Since \(X\) is compact, the closed subset \(K\) is itself compact, by the clause of the compactness properties stating that a closed subset of a compact space is compact. The continuous image of a compact set is compact, so \(F(K)\) is compact in \(Y\). Finally, \(Y\) is Hausdorff, so the compact set \(F(K)\) is closed. Thus \(F\) carries closed sets to closed sets, which is to say that \(F\) is a closed map.

(2) Suppose \(F\) is also surjective. A quotient map is a continuous surjection for which \(V \subseteq Y\) is open precisely when \(F^{-1}(V)\) is open. Continuity gives the forward implication, so only the converse requires the hypothesis. Suppose \(F^{-1}(V)\) is open. Then its complement \(F^{-1}(Y \setminus V) = X \setminus F^{-1}(V)\) is closed, so by (1) its image \(F(X \setminus F^{-1}(V))\) is closed in \(Y\). Surjectivity gives \(F(X \setminus F^{-1}(V)) = Y \setminus V\), whence \(Y \setminus V\) is closed and \(V\) is open. Thus \(F\) is a quotient map.

(3) An injective map is a bijection onto its image \(F(X)\), and the corestriction \(F : X \to F(X)\) is a continuous bijection that is closed, since a closed map remains closed when the codomain is cut down to the image. By the characterization of homeomorphisms, a closed continuous bijection is a homeomorphism. Hence \(F : X \to F(X)\) is a homeomorphism, which is exactly the assertion that \(F\) is an embedding.

(4) A bijective map is simultaneously surjective and injective, so it is both a quotient map and an embedding. Equivalently, the corestriction in (3) is onto all of \(Y\), giving a homeomorphism \(F : X \to Y\).

Proof:

(1) Let \(K \subseteq Y\) be compact. Since \(Y\) is Hausdorff, \(K\) is closed, so \(F^{-1}(K)\) is closed in \(X\) by continuity. A closed subset of the compact space \(X\) is compact, so \(F^{-1}(K)\) is compact and \(F\) is proper.

(2) Let \(K \subseteq Y\) be compact. We show \(F^{-1}(K)\) is compact by extracting a finite subcover from an arbitrary open cover \(\mathcal{U}\) of \(F^{-1}(K)\). For each \(y \in K\), the fiber \(F^{-1}(\{y\}) \subseteq F^{-1}(K)\) is compact, so finitely many members of \(\mathcal{U}\) cover it. Let \(U_y\) be their union, an open set containing the fiber. The set \(V_y = Y \setminus F(X \setminus U_y)\) is open, because \(F\) is a closed map. It also contains \(y\), because the entire fiber over \(y\) lies in \(U_y\) and therefore no point of \(X \setminus U_y\) maps to \(y\). Moreover \(F^{-1}(V_y) \subseteq U_y\). The sets \(\{V_y\}_{y \in K}\) form an open cover of \(K\), so finitely many \(V_{y_1}, \ldots, V_{y_n}\) suffice. The corresponding \(U_{y_1}, \ldots, U_{y_n}\) cover \(F^{-1}(K)\), and each is a finite union of members of \(\mathcal{U}\). Thus \(F^{-1}(K)\) admits a finite subcover and is compact.

(3) Suppose \(F\) is an embedding with closed image \(F(X)\). Let \(K \subseteq Y\) be compact. Then \(K \cap F(X)\) is a closed subset of the compact set \(K\), hence compact, and it lies in \(F(X)\). Since \(F : X \to F(X)\) is a homeomorphism onto its image, \(F^{-1}(K) = F^{-1}(K \cap F(X))\) is the homeomorphic image of a compact set, hence compact.

(4) Suppose \(G : Y \to X\) is continuous with \(G \circ F = \operatorname{id}_X\). This identity forces \(F\) to be injective. Let \(K \subseteq Y\) be compact. For any \(x \in F^{-1}(K)\) we have \(x = G(F(x)) \in G(K)\), so \(F^{-1}(K) \subseteq G(K)\). The set \(G(K)\) is compact, being the continuous image of the compact set \(K\). Since \(Y\) is Hausdorff, \(K\) is closed, so \(F^{-1}(K)\) is closed in \(X\) by continuity. Being a closed subset of the compact set \(G(K)\), it is compact.

(5) Suppose \(F\) is proper and \(A = F^{-1}(F(A))\) is saturated. Let \(L \subseteq F(A)\) be compact in the subspace \(F(A)\). Then \(L\) is compact in \(Y\), so \(F^{-1}(L)\) is compact by properness of \(F\). Since \(L \subseteq F(A)\), every point of \(F^{-1}(L)\) maps into \(F(A)\), so \(F^{-1}(L) \subseteq F^{-1}(F(A)) = A\). Consequently \((F|_A)^{-1}(L) = F^{-1}(L) \cap A = F^{-1}(L)\), which is compact. Hence \(F|_A\) is proper.

Proof:

The implications \((3) \Rightarrow (2) \Rightarrow (1)\) are immediate. A precompact open set is a neighborhood whose closure is compact, and any neighborhood is contained in such a compact closure. The substance is \((1) \Rightarrow (3)\). Let \(p \in X\) and let \(U\) be any open neighborhood of \(p\). We seek a precompact open set \(W\) with \(p \in W \subseteq U\). By local compactness, \(p\) has a neighborhood \(V\) contained in some compact set \(C\). Since \(X\) is Hausdorff, \(C\) is closed, so \(\overline{V} \subseteq C\), and a closed subset of a compact set is compact. The set \(\overline{V} \setminus U\) is a closed subset of the compact \(\overline{V}\), hence compact, and it does not contain \(p\).

Since \(X\) is Hausdorff and the singleton \(\{p\}\) and the compact set \(\overline{V} \setminus U\) are disjoint compact subsets, the separation of disjoint compact sets provides disjoint open sets \(A \supseteq \{p\}\) and \(B \supseteq \overline{V} \setminus U\). Put \(W = A \cap V\). Then \(W\) is open, \(p \in W\), and \(W \subseteq V\) gives \(\overline{W} \subseteq \overline{V}\), so \(\overline{W}\) is compact. Moreover \(W \subseteq A \subseteq X \setminus B\), and \(X \setminus B\) is closed, so \(\overline{W} \subseteq X \setminus B \subseteq X \setminus (\overline{V} \setminus U)\). Combined with \(\overline{W} \subseteq \overline{V}\), this forces \(\overline{W} \subseteq U\). Thus \(W\) is a precompact open neighborhood of \(p\) contained in \(U\), and the collection of all such \(W\) is a basis of precompact open sets.

For the heredity statement, let \(S \subseteq X\) be open. Given \(p \in S\), condition (3) applied with the neighborhood \(S\) yields a precompact open \(W\) with \(p \in W\) and \(\overline{W} \subseteq S\). This \(W\) is a precompact neighborhood of \(p\) in \(S\), and \(S\) is Hausdorff as a subspace. If instead \(S\) is closed, take any \(p \in S\) and a precompact neighborhood \(N\) of \(p\) in \(X\). Then \(N \cap S\) is a neighborhood of \(p\) in \(S\). Its closure in \(S\) is \(\overline{N \cap S} \cap S\) (closure taken in \(X\)), which is contained in \(\overline{N} \cap S\). Since \(S\) is closed, \(\overline{N} \cap S\) is a closed subset of the compact set \(\overline{N}\), hence compact, and the closure of \(N \cap S\) in \(S\) is a closed subset of it, hence also compact. Thus \(N \cap S\) is a precompact neighborhood of \(p\) in \(S\). In either case \(S\) satisfies (2), so \(S\) is a locally compact Hausdorff space.

Proof:

Let \(K \subseteq X\) be closed. To show \(F(K)\) is closed in \(Y\), we show it contains all of its limit points. Let \(y\) be a limit point of \(F(K)\). Since \(Y\) is locally compact Hausdorff, we may let \(U\) be a precompact neighborhood of \(y\), so that \(\overline{U}\) is compact. Then \(y\) is also a limit point of \(F(K) \cap \overline{U}\), because every neighborhood of \(y\) contained in \(U\) meets \(F(K)\) in a point necessarily lying in \(U \subseteq \overline{U}\). Because \(F\) is proper, the preimage \(F^{-1}(\overline{U})\) is compact, so its closed subset \(K \cap F^{-1}(\overline{U})\) is compact. Its continuous image \(F\bigl(K \cap F^{-1}(\overline{U})\bigr) = F(K) \cap \overline{U}\) is therefore compact, and a compact subset of the Hausdorff space \(Y\) is closed. A closed set contains its limit points, so \(y \in F(K) \cap \overline{U} \subseteq F(K)\). Hence \(F(K)\) is closed.

Proof sketch (locally compact Hausdorff case):

Let \(A = \bigcup_{n=1}^{\infty} A_n\) with each \(A_n\) nowhere dense. We show \(A\) has empty interior by showing that every nonempty open set \(W_0\) contains a point outside \(A\). Shrinking \(W_0\) using the basis of precompact open sets, we may assume \(\overline{W_0}\) is compact. Since \(A_1\) is nowhere dense, \(\overline{A_1}\) has empty interior, so \(W_0 \setminus \overline{A_1}\) is nonempty and open. Choose a precompact open \(W_1\) with \(\overline{W_1} \subseteq W_0 \setminus \overline{A_1}\). Continuing inductively produces precompact open sets \(W_n\) with \(\overline{W_n} \subseteq W_{n-1} \setminus \overline{A_n}\), so the closures form a decreasing sequence of nonempty compact sets \(\overline{W_1} \supseteq \overline{W_2} \supseteq \cdots\).

By compactness this nested sequence has nonempty intersection. Suppose it were empty. Then the open sets \(X \setminus \overline{W_n}\) would cover \(\overline{W_1}\), so finitely many would suffice, and since they increase with \(n\) the largest index \(N\) among them already gives \(\overline{W_1} \subseteq X \setminus \overline{W_N}\). But \(\overline{W_N} \subseteq \overline{W_1}\) then forces \(\overline{W_N} = \emptyset\), contradicting the construction. Any point \(x\) in the intersection lies in every \(W_{n-1}\) and avoids every \(\overline{A_n}\), hence \(x \in W_0\) and \(x \notin A\). The complete-metric case is identical with \(\overline{W_n}\) replaced by closed balls of radius \(\to 0\), the nonempty intersection coming from completeness rather than compactness.

Proof:

Let \(X\) be such a space and let \(A \subseteq X\) be a nonempty countable closed subset. Suppose, for contradiction, that \(A\) has no isolated points. Because \(A\) is closed in \(X\), it is itself a locally compact Hausdorff space or a complete metric space, so the Baire category theorem applies within \(A\). For each \(a \in A\), the singleton \(\{a\}\) is closed in \(A\) (since \(A\) is Hausdorff), and it has empty interior in \(A\) precisely because \(a\) is not isolated. Thus \(\{a\}\) is nowhere dense in \(A\). But \(A\) is the countable union \(\bigcup_{a \in A} \{a\}\), which by Baire must have empty interior in \(A\). This contradicts the fact that \(A\), being the whole space, is open in itself with nonempty interior. Therefore \(A\) has an isolated point.

Proof:

Let \(X\) be such a space. By the characterization above it has a basis of precompact open sets. Those sets form an open cover of \(X\), and since \(X\) is second-countable the cover admits a countable subcover. If \(X = \emptyset\), the constant sequence \(K_i = \emptyset\) is an exhaustion, so assume \(X\) is nonempty. The subcover is then nonempty as well, and its members can be listed as a sequence \((U_i)_{i=1}^{\infty}\), with repetitions if the subcover is finite. Each \(\overline{U_i}\) is compact. We build the exhaustion inductively. Set \(K_1 = \overline{U_1}\). Assume compact sets \(K_1, \ldots, K_k\) have been constructed with \(U_j \subseteq K_j\) for each \(j\) and \(K_{j-1} \subseteq \operatorname{int}(K_j)\) for \(j \geq 2\). Since \(K_k\) is compact and the \((U_i)\) cover it, finitely many suffice: \(K_k \subseteq U_1 \cup \cdots \cup U_{m_k}\) for some \(m_k\), which we may take with \(m_k \geq k+1\). Define \(K_{k+1} = \overline{U_1} \cup \cdots \cup \overline{U_{m_k}}\), a finite union of compact sets, hence compact. Then \(K_k \subseteq U_1 \cup \cdots \cup U_{m_k} \subseteq \operatorname{int}(K_{k+1})\) and \(U_{k+1} \subseteq K_{k+1}\). By induction this yields compact sets with \(K_k \subseteq \operatorname{int}(K_{k+1})\). Since \(U_i \subseteq K_i\) for every \(i\) while the \(U_i\) cover \(X\), we have \(X = \bigcup_i K_i\). This is the required exhaustion.