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 \(\mathbb{R}\) and the open sets it generates, which 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: if \(J\) is not an interval, points \(a, b \in J\) and \(c \notin J\) with \(a < c < b\) yield the separation \(J = \bigl(J \cap (-\infty, c)\bigr) \sqcup \bigl(J \cap (c, \infty)\bigr)\), so \(J\) is disconnected; conversely, if \(J\) is an interval, a hypothetical separation \(J = U \sqcup V\) with \(a \in U\), \(b \in V\), \(a < b\) is ruled out by examining \(c = \sup\{\, t \in [a,b] : [a,t] \subseteq U \,\}\), which can lie in neither \(U\) nor \(V\) without contradicting either the openness of that set or the defining supremum. \(\blacksquare\)

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 \(X\) — 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; say \(p \in U\). 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 \(V \cap C_\alpha = \emptyset\), i.e. \(C_\alpha \subseteq U\). 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, the closure of a connected set is connected, so 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 \(\emptyset\) or \(C\); thus \(S\) is the union of the components it contains. (6) For two factors, 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. \(\blacksquare\)

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 because 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 (\(0\) and \(1\) respectively), with union \(I\) — a separation of \(I\), contradicting that \(I\) is connected as an interval. \(\blacksquare\)

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 (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) An open subset inherits a basis of path-connected open sets from the ambient basis intersected with it. \(\blacksquare\)

Proof:

(1) The continuous image of a compact set 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 (by the point–compact-set form of the argument below) disjoint open sets \(U_x \ni x\) and \(W_x \supseteq L\); 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; 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. \(\blacksquare\)

Proof:

If \(K = \emptyset\), it is closed trivially; 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. \(\blacksquare\)

Proof:

We prove (1) directly; the remaining three assertions then follow formally. Let \(K \subseteq X\) be closed. Since \(X\) is compact, the closed subset \(K\) is itself compact — this is 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: \(F\) 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 (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 \(F : X \to Y\). \(\blacksquare\)

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; hence \(F^{-1}(K)\) 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, and it contains \(y\): no point of \(X \setminus U_y\) maps to \(y\), since the entire fiber over \(y\) lies in \(U_y\). 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. \(\blacksquare\)

Proof:

\((3) \Rightarrow (2) \Rightarrow (1)\) 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 contained in some compact set, so we may fix an open set \(V\) with \(p \in V\) and \(\overline{V}\) compact — \(\overline{V}\) is a closed subset of that compact set, hence 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 \(U\); 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 \(S\) whose closure in \(S\) equals \(\overline{N} \cap S\), a closed subset of the compact set \(\overline{N}\), hence compact. In either case \(S\) satisfies (2), so \(S\) is a locally compact Hausdorff space. \(\blacksquare\)

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 \(F(K)\), and — using that \(Y\) is locally compact Hausdorff — let \(U\) be a precompact neighborhood of \(y\), so that \(\overline{U}\) is compact. Then \(y\) is also a limit point of \(F(K) \cap \overline{U}\): 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. \(\blacksquare\)

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 \(A\) — contradicting that \(A\), being the whole space, is open in itself with nonempty interior. Therefore \(A\) has an isolated point. \(\blacksquare\)

Proof:

Let \(X\) be such a space. By the characterization above it has a basis of precompact open sets; being second-countable, it is covered by countably many of them, say \((U_i)_{i=1}^{\infty}\) with each \(\overline{U_i}\) 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 \(K_k \subseteq \operatorname{Int} K_{k+1}\); and 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. \(\blacksquare\)