Topological and Metric Spaces

Derives Proposition 6.10. Let \(\set{U_i}\) be an open cover of \(f(K)\). The sets \(f^{-1}(U_i)\) are open (Definition 6.6) and cover \(K\), so finitely many \(f^{-1}(U_{i_1}),\dots,f^{-1}(U_{i_n})\) cover \(K\); then \(U_{i_1},\dots,U_{i_n}\) cover \(f(K)\).

Derives Theorem 6.11. Let \(\set{U_i}\) be an open cover of \([a,b]\) and define

\(S\) is nonempty (\(a\) lies in some \(U_i\)) and bounded above by \(b\), so \(c = \sup S\) exists by the completeness of \(\R\) (Axiom 7.1, proven in Appendix A.7). Some \(U_{i_0}\) contains \(c\), and with it an interval \((c-\varepsilon, c+\varepsilon)\). Since \(c - \varepsilon\) is not an upper bound of \(S\), some \(x \in S\) exceeds it; the finite subcover of \([a,x]\) together with \(U_{i_0}\) covers \([a, \min(c+\varepsilon/2, b)]\). If \(c < b\) this puts points beyond \(c\) into \(S\), contradicting \(c = \sup S\); hence \(c = b\), and the same finite family covers \([a,b]\).

Derives Theorem 6.12. Necessity. Let \(K\) be compact. The balls \(B_{n}(\vect{0})\), \(n \in \N\), form an open cover of \(K\); a finite subcover is contained in the largest of its members, so \(K\) is bounded. For closedness, take \(\vect{y} \notin K\) and set \(r_{\vect{x}} = \abs{\vect{x} - \vect{y}}/2 > 0\) for each \(\vect{x} \in K\). The balls \(B_{r_{\vect{x}}}(\vect{x})\) cover \(K\), so finitely many of them, centred at \(\vect{x}_{1}, \dots, \vect{x}_{p}\), already do; abbreviate \(r_{j} = r_{\vect{x}_{j}}\) and put \(\rho = \min_{j \le p} r_{j} > 0\). No point \(\vect{z}\) can lie in both \(B_{\rho}(\vect{y})\) and \(B_{r_{j}}(\vect{x}_{j})\), since that would give

Hence \(B_{\rho}(\vect{y})\) misses \(K\), and the complement of \(K\) is open.

Sufficiency. Consider first the closed box \(Q_{0} = [-R, R]^{N}\), and suppose some open cover \(\set{U_i}\) of it admitted no finite subcover. Halving every edge splits \(Q_{0}\) into \(2^{N}\) closed boxes whose union is \(Q_{0}\); were each of them covered by finitely many \(U_{i}\), so would \(Q_{0}\) be, so at least one of them — call it \(Q_{1}\) — is not. Iterating produces nested closed boxes \(Q_{0} \supseteq Q_{1} \supseteq \cdots\), none of them finitely covered, the \(k\)-th of edge \(2R/2^{k}\) and hence of diameter \(\sqrt{N}\,2R/2^{k}\). In each coordinate the left endpoints of the \(Q_{k}\) form a nondecreasing bounded sequence and the right endpoints a nonincreasing one, the two differing by \(2R/2^{k}\); by completeness (Axiom 7.1) each converges, and to the same number, which therefore lies in every one of the nested coordinate intervals. The point \(\vect{x}\) with those \(N\) coordinates thus belongs to every \(Q_{k}\). Some \(U_{i_{0}}\) contains \(\vect{x}\), and with it a ball \(B_{\varepsilon}(\vect{x})\); choosing \(k\) with \(\sqrt{N}\,2R/2^{k} < \varepsilon\) gives \(Q_{k} \subseteq B_{\varepsilon}(\vect{x}) \subseteq U_{i_{0}}\), a subcover of \(Q_{k}\) by a single set — contradicting the choice of \(Q_{k}\). So \(Q_{0}\) is compact.

Let now \(K\) be closed and bounded, so that \(K \subseteq Q_{0}\) for some \(R\), and let \(\set{U_i}\) be an open cover of \(K\). Since \(K\) is closed, \(\R^{N} \setminus K\) is open (Definition 6.3), so \(\set{U_i}\) together with \(\R^{N} \setminus K\) is an open cover of \(Q_{0}\). Extract a finite subcover and discard \(\R^{N} \setminus K\) from it: what remains is a finite subfamily of \(\set{U_i}\), and it still covers \(K\), because the discarded set contains no point of \(K\).

Derives Theorem 6.14. Suppose \(I = U \cup V\) with \(U, V\) open in \(I\), disjoint and nonempty; pick \(u \in U\), \(v \in V\), say \(u < v\). Let \(c = \sup\set{x \in U \mid x < v}\), which exists by completeness and satisfies \(u \le c \le v\), so \(c \in I\). If \(c \in U\), openness of \(U\) puts an interval around \(c\) inside \(U\); since \(c < v\) (as \(v \in V\)), points of \(U\) above \(c\) would contradict the definition of \(c\). If \(c \in V\), openness of \(V\) puts \((c - \varepsilon, c] \subseteq V\), making \(c - \varepsilon\) an upper bound of the set defining \(c\) — again a contradiction. So \(c\) lies in neither, contradicting \(I = U \cup V\).

Derives Proposition 6.16. If \(X = U \cup V\) were a disconnection, join \(x \in U\) to \(y \in V\) by a path \(\gamma\). Then \(\gamma^{-1}(U)\) and \(\gamma^{-1}(V)\) disconnect \([0,1]\): both are open by continuity, disjoint, nonempty (\(0\) and \(1\)), and cover. This contradicts Theorem 6.14.

Derives Lemma 6.19. Existence. The function \(t \longmapsto \abs{\gamma(t)}\) is continuous and strictly positive on \([0,1]\), so it attains a minimum \(m > 0\) (Theorem 7.24); and \(\gamma\) is uniformly continuous, because its real and imaginary parts are (Theorem 7.25). Uniform continuity supplies a \(\delta > 0\) with \(\abs{\gamma(t) - \gamma(s)} < m\) whenever \(\abs{t - s} < \delta\); choose any partition \(0 = t_{0} < t_{1} < \dots < t_{M} = 1\) of mesh smaller than \(\delta\), so that \(\abs{\gamma(t) - \gamma(t_{j-1})} < m\) for every \(t \in [t_{j-1}, t_{j}]\) and every \(j\). On such an interval the quotient \(w(t) = \gamma(t)/\gamma(t_{j-1})\) satisfies

so \(w(t)\) lies in the open disc of radius \(1\) centred at \(1\), hence in the half-plane \(\Re w > 0\). On that half-plane the function

is continuous, takes values in \((-\pi/2, \pi/2)\), vanishes at \(w = 1\), and satisfies \(w = \abs{w}\,\ee^{\ii A(w)}\). Fix \(\theta(0)\) with \(\gamma(0) = \abs{\gamma(0)}\,\ee^{\ii\theta(0)}\) and define recursively

Each piece is continuous and, since \(A(1) = 0\), the pieces agree at the nodes, so \(\theta\) is continuous on \([0,1]\). Multiplying \(\gamma(t_{j-1}) = \abs{\gamma(t_{j-1})}\,\ee^{\ii\theta(t_{j-1})}\) by the polar form of the quotient gives Equation (6.3), one interval at a time.

Uniqueness up to \(2\pi\Z\). If \(\theta_{1}\) and \(\theta_{2}\) both satisfy Equation (6.3), then \(\ee^{\ii(\theta_{1} - \theta_{2})} = 1\), so \(\varphi = (\theta_{1} - \theta_{2})/2\pi\) is continuous with values in \(\Z\). Were it to take two distinct values, the intermediate value theorem (Theorem 7.23) would make it take some value strictly between two consecutive integers as well, which is not in \(\Z\); hence \(\varphi\) is constant.

Derives Lemma 6.22. The quotient \(w = \sigma/\gamma\) is a continuous loop, and Equation (6.6) says \(\abs{w(s) - 1} < 1\); in particular \(w\) never vanishes and takes its values in the half-plane \(\Re w > 0\), where the function \(A\) of Equation (6.4) provides the continuous argument \(s \longmapsto A(w(s))\). Hence \(n(w) = \bigl(A(w(1)) - A(w(0))\bigr)/2\pi = 0\), because \(w(1) = w(0)\). If \(\theta\) is a continuous argument for \(\gamma\), then \(\theta + A(w)\) is one for \(\sigma = w\,\gamma\), so \(n(\sigma) = n(\gamma) + n(w) = n(\gamma)\).

Derives Proposition 6.23. Path-connectedness is immediate: any two points of the punctured plane are joined by a circular arc followed by a radial segment, and neither meets the origin. Suppose then that \(H : [0,1]\times[0,1] \longrightarrow \C \setminus \set{0}\) were a contraction of \(\gamma\), that is, a continuous map with \(H(s,0) = \gamma(s)\), \(H(s,1) = 1\) and \(H(0,t) = H(1,t) = 1\). For each fixed \(t\) the map \(\gamma_{t}(s) = H(s,t)\) is then a loop in \(\C\setminus\set{0}\), and we may set \(n(t) = n(\gamma_{t})\).

The function \(n\) is locally constant. Fix \(t_{0}\) and let \(m = \min_{s}\abs{H(s,t_{0})} > 0\), the minimum being attained by Theorem 7.24. For each \(s_{0} \in [0,1]\), continuity of \(H\) at \((s_{0}, t_{0})\) supplies a \(\delta(s_{0}) > 0\) such that \(\abs{H(s,t) - H(s_{0},t_{0})} < m/2\) whenever \(\abs{s - s_{0}} < \delta(s_{0})\) and \(\abs{t - t_{0}} < \delta(s_{0})\). The open intervals \((s_{0} - \delta(s_{0})/2,\ s_{0} + \delta(s_{0})/2)\) cover \([0,1]\), which is compact (Theorem 6.11), so finitely many of them suffice, say those centred at \(s_{1}, \dots, s_{k}\); put \(\delta = \min_{j}\delta(s_{j})/2 > 0\). Let \(\abs{t - t_{0}} < \delta\) and let \(s \in [0,1]\). Then \(\abs{s - s_{j}} < \delta(s_{j})/2\) for some \(j\), and \(\abs{t - t_{0}} < \delta \le \delta(s_{j})\); the defining property of \(\delta(s_{j})\) applies both at \(t\) and at \(t_{0}\), so \(H(s,t)\) and \(H(s,t_{0})\) both lie within \(m/2\) of \(H(s_{j},t_{0})\) and

Lemma 6.22 then gives \(n(t) = n(t_{0})\) for every \(t\) with \(\abs{t - t_{0}} < \delta\).

Conclusion. A locally constant function is continuous, so \(n : [0,1] \longrightarrow \Z\) is continuous and integer-valued, and the intermediate value theorem (Theorem 7.23) forces it to be constant, exactly as in Lemma 6.19. But \(n(0) = 1\), since \(\gamma_{0} = \gamma\), and \(n(1) = 0\), since \(\gamma_{1}\) is the constant loop (Example 6.21). No such \(H\) can exist.

Derives Proposition 6.28. (\(\Rightarrow\)) \(B_\varepsilon(f(x_0))\) is open, so \(f^{-1}(B_\varepsilon(f(x_0)))\) is open and contains \(x_0\), hence contains a ball \(B_\delta(x_0)\) — which is Equation (6.11). (\(\Leftarrow\)) Let \(U\) be open in \(X_2\) and \(x_0 \in f^{-1}(U)\). Some \(B_\varepsilon(f(x_0)) \subseteq U\); the \(\delta\) provided by Equation (6.11) gives \(B_\delta(x_0) \subseteq f^{-1}(U)\), so \(f^{-1}(U)\) is open.

Derives Lemma 6.30. If no such \(\delta\) existed, then for every \(n \in \N\) there would be a point \(x_{n} \in X\) with \(B_{1/n}(x_{n})\) contained in no member of the cover. By sequential compactness some subsequence \(x_{n_{k}} \longrightarrow x \in X\), and \(x\) lies in some \(U_{i_{0}}\), which contains a ball \(B_{\varepsilon}(x)\). Choose \(k\) so large that \(d(x_{n_{k}}, x) < \varepsilon/2\) and \(1/n_{k} < \varepsilon/2\). Every \(y \in B_{1/n_{k}}(x_{n_{k}})\) then satisfies

so \(B_{1/n_{k}}(x_{n_{k}}) \subseteq B_{\varepsilon}(x) \subseteq U_{i_{0}}\) — contradicting the choice of \(x_{n_{k}}\).

Derives Theorem 6.31. Compact \(\implies\) sequentially compact. Let \((x_{n})\) be a sequence with no convergent subsequence. Then no point \(x \in X\) is the limit of a subsequence, and consequently each \(x\) admits a radius \(r_{x} > 0\) such that \(B_{r_{x}}(x)\) contains \(x_{n}\) for only finitely many indices \(n\): otherwise every ball \(B_{1/k}(x)\) would contain terms of arbitrarily large index, and picking \(n_{1} < n_{2} < \cdots\) with \(x_{n_{k}} \in B_{1/k}(x)\) would produce a subsequence converging to \(x\). The balls \(B_{r_{x}}(x)\) form an open cover of \(X\), so finitely many of them cover \(X\); their union then contains \(x_{n}\) for only finitely many indices, although it is all of \(X\) and so contains every term of the sequence. Since the index set is infinite, this is absurd.

Sequentially compact \(\implies\) compact. Let \(\set{U_i}\) be an open cover of \(X\) and let \(\delta > 0\) be a Lebesgue number for it (Lemma 6.30). We claim first that finitely many balls of radius \(\delta\) cover \(X\). If not, choose \(x_{1} \in X\) and, having chosen \(x_{1}, \dots, x_{n}\), a point \(x_{n+1}\) outside \(B_{\delta}(x_{1}) \cup \dots \cup B_{\delta}(x_{n})\); the resulting sequence satisfies \(d(x_{m}, x_{n}) \ge \delta\) for all \(m \neq n\), so none of its subsequences is Cauchy and none converges — contradicting sequential compactness. So \(X = B_{\delta}(y_{1}) \cup \dots \cup B_{\delta}(y_{p})\) for suitable points \(y_{1}, \dots, y_{p}\), and by the defining property of \(\delta\) each \(B_{\delta}(y_{q})\) lies inside some \(U_{i_{q}}\). The finitely many sets \(U_{i_{1}}, \dots, U_{i_{p}}\) cover \(X\).