Chapter 4

Prove that \(A\) is connected iff there is no continuous map $$ f: A \underset{\text { onto }}{\longrightarrow}\\{0,1\\} .^{5} $$ [Hint: If there is such a map, Theorem 1 shows that \(A\) is disconnected. (Why?) Conversely, if \(A=P \cup Q(P, Q\) as in Definition 3), put \(f=0\) on \(P\) and \(f=1\) on \(Q\). Use again Theorem 1 to show that \(f\) so defined is continuous on \(A\).]

Prove that if \(f_{n} \rightarrow f\) (uniformly) on \(B\) and if \(C \subseteq B,\) then \(f_{n} \rightarrow f\) (uniformly) on \(C\) as well.

Prove, independently, the principle of nested intervals in \(E^{n},\) i.e., Theorem 5 with $$ F_{m}=\left[\bar{a}_{m}, \bar{b}_{m}\right] \subseteq E^{n}; $$ where $$ \bar{a}_{m}=\left(a_{m 1}, \ldots, a_{m n}\right) \text { and } \bar{b}_{m}=\left(b_{m 1}, \ldots, b_{m n}\right) $$ [Hint: As \(F_{m+1} \subseteq F_{m}, \bar{a}_{m+1}\) and \(\bar{b}_{m+1}\) are in \(F_{m} ;\) hence by properties of closed intervals, \(a_{m k} \leq a_{m+1, k} \leq b_{m+1, k} \leq b_{m k}, \quad k=1,2, \ldots, n\) Fixing \(k\), let \(A_{k}\) be the set of all \(a_{m k}, m=1,2, \ldots\) Show that \(A_{k}\) is bounded above by each \(b_{m k},\) so let \(p_{k}=\sup A_{k}\) in \(E^{1}\). Then $$ (\forall m) \quad a_{m k} \leq p_{k} \leq b_{m k} . \text { (Why?) } $$ Unfixing \(k\), obtain such inequalities for \(k=1,2, \ldots, n\). Let \(\bar{p}=\left(p_{1}, \ldots, p_{k}\right)\). Then $$ (\forall m) \quad \bar{p} \in\left[\bar{a}_{m}, \bar{b}_{m}\right], \text { i.e., } \bar{p} \in \bigcap F_{m}, \text { as required. } $$ Note that the theorem fails for nonclosed intervals, even in \(E^{1} ;\) e.g., take \(F_{m}=\) \((0,1 / m]\) and show that \(\bigcap_{m} F_{m}=\emptyset\).]

Let \(B \subseteq A \subseteq(S, \rho)\). Prove that \(B\) is connected in \(S\) iff it is connected in \((A, \rho)\).

Suppose that no two of the sets \(A_{i}(i \in I)\) are disjoint. Prove that if all \(A_{i}\) are connected, so is \(A=\bigcup_{i \in I} A_{i}\) [Hint: If not, let \(A=P \cup Q(P, Q\) as in Definition 3). Let \(P_{i}=A_{i} \cap P\) and \(Q_{i}=A_{i} \cap Q,\) so \(A_{i}=P_{i} \cup Q_{i}, i \in I\) That is, onto a two-point set \\{0\\}\(\cup\\{1\\}\). \S10. Arcs and Curves. Connected Sets At least one of the \(P_{i}, Q_{i}\) must be \(\emptyset\) (why?); say, \(Q_{j}=\emptyset\) for some \(j \in I\). Then \((\forall i) Q_{i}=\emptyset,\) for \(Q_{i} \neq \emptyset\) implies \(P_{i}=\emptyset,\) whence $$ A_{i}=Q_{i} \subseteq Q \Longrightarrow A_{i} \cap A_{j}=\emptyset\left(\text { since } A_{j} \subseteq P\right) $$ contrary to our assumption. Deduce that \(Q=\bigcup_{i} Q_{i}=\emptyset\). (Contradiction!)]

Prove that if \(f: S \rightarrow T\) is uniformly continuous on \(B \subseteq S,\) and \(g: T \rightarrow U\) is uniformly continuous on \(f[B]\), then the composite function \(g \circ f\) is uniformly continuous on \(B\).

Let \(f_{n} \rightarrow f\) (uniformly) on \(B\). Prove the equivalence of the following statements: (i) Each \(f_{n}\), from a certain \(n\) onward, is bounded on \(B\). (ii) \(f\) is bounded on \(B\). (iii) The \(f_{n}\) are ultimately uniformly bounded on \(B ;\) that is, all function values \(f_{n}(x), x \in B,\) from a certain \(n=n_{0}\) onward, are in one and the same globe \(G_{q}(K)\) in the range space. For real, complex, and vector-valued functions, this means that $$ \left(\exists K \in E^{1}\right)\left(\forall n \geq n_{0}\right)(\forall x \in B) \quad\left|f_{n}(x)\right|

Prove Theorem 2 , with (i) replaced by the weaker assumption ("subuniform limit") $$ (\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in G_{\neg p}(\delta)\right)\left(\forall y \in G_{\neg q}(\delta)\right) \quad \rho(g(x), f(x, y))<\varepsilon $$ and with iterated limits defined by $$ s=\lim _{x \rightarrow p} \lim _{y \rightarrow q} f(x, y) $$ iff \((\forall \varepsilon>0)\) $$ \left(\exists \delta^{\prime}>0\right)\left(\forall x \in G_{\neg p}\left(\delta^{\prime}\right)\right)\left(\exists \delta_{x}^{\prime \prime}>0\right)\left(\forall y \in G_{\neg q}\left(\delta_{x}^{\prime \prime}\right)\right) \quad \rho(f(x, y), s)<\varepsilon $$

Let \(A^{\prime}\) be the set of all cluster points of \(A \subseteq(S, \rho) .\) Let \(f: A \rightarrow\left(T, \rho^{\prime}\right)\) be uniformly continuous on \(A,\) and let \(\left(T, \rho^{\prime}\right)\) be complete. (i) Prove that \(\lim _{x \rightarrow p} f(x)\) exists at each \(p \in A^{\prime}\). (ii) Thus define \(f(p)=\lim _{x \rightarrow p} f(x)\) for each \(p \in A^{\prime}-A,\) and show that \(f\) so extended is uniformly continuous on the set \(\bar{A}=A \cup A^{\prime} .6\) (iii) Consider, in particular, the case \(A=(a, b) \subseteq E^{1},\) so that $$ \bar{A}=A^{\prime}=[a, b] $$

Prove that if the functions \(f_{n}\) and \(g_{n}\) are real or complex (or if the \(g_{n}\) are vector valued and the \(f_{n}\) are scalar valued), and if $$ f_{n} \rightarrow f \text { and } g_{n} \rightarrow g \text { (uniformly) on } B \text { , } $$ then $$ f_{n} g_{n} \rightarrow f g \text { (uniformly) on } B $$ provided that either \(f\) and \(g\) or the \(f_{n}\) and \(g_{n}\) are bounded on \(B\) (at least from some \(n\) onward \() ;\) cf. Problem 11 . Disprove it for the case where only one of \(f\) and \(g\) is bounded.