실해석학 hw2

A. [과제 수령 및 확인 리포트]

확정 문항 리스트:

• Chapter 3: 1, 3, 5, 14, 20, 21, 23
• Chapter 4: 3, 5, 11, 13, 18, 23, 24

총 문항 수: \[ 7+7=\boxed{14} \]

---

B. [문항별 상세 솔루션]

Chapter 3 — Exercise 1

[문제 원문 및 번역]

Original: Prove that convergence of \(\{s_n\}\) implies convergence of \(\{\vert s_n\vert \}\). Is the converse true?

번역: 수열 \(\{s_n\}\)이 수렴하면 \(\{\vert s_n\vert \}\)도 수렴함을 증명하라. 그 역도 참인가?

[요청 사항]

\[ s_n\to s \implies \vert s_n\vert \to \vert s\vert ,\qquad \text{converse 여부 판정} \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\ \{s_n\}\subset \mathbb C,\quad s_n\to s \]

\[ Step\ 1:\quad \big\vert \vert s_n\vert -\vert s\vert \big\vert \le \vert s_n-s\vert \] \[ \because \text{triangle inequality in }\mathbb C\simeq\mathbb R^2,\ \text{교재 Ch.3, Thm 3.4/sequence norm logic} \]

\[ Step\ 2:\quad s_n\to s \iff \forall\varepsilon>0,\exists N,\ n\ge N\Rightarrow \vert s_n-s\vert <\varepsilon \] \[ \because \text{교재 Ch.3, Def 3.1} \]

\[ Step\ 3:\quad n\ge N \Rightarrow \big\vert \vert s_n\vert -\vert s\vert \big\vert \le \vert s_n-s\vert <\varepsilon \]

\[ \therefore\ \vert s_n\vert \to \vert s\vert \]

Converse test:

\[ Def:\quad s_n=(-1)^n \]

\[ \vert s_n\vert =1\quad\forall n \Rightarrow \vert s_n\vert \to 1 \]

\[ s_{2n}=1,\quad s_{2n-1}=-1 \Rightarrow \text{subsequential limits }1,-1 \Rightarrow s_n\ \text{diverges} \] \[ \because \text{limit uniqueness, 교재 Ch.3, Thm 3.2(b)} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad s_n\to s\Rightarrow \vert s_n\vert \to \vert s\vert . \ \text{역은 거짓: }s_n=(-1)^n. } \]

---

Chapter 3 — Exercise 3

[문제 원문 및 번역]

Original: If \(s_1=\sqrt2\), and \[ s_{n+1}=\sqrt{2+s_n}\quad(n=1,2,3,\ldots), \] prove that \(\{s_n\}\) converges, and that \(s_n<2\) for \(n=1,2,3,\ldots\).

번역: \(s_1=\sqrt2\), \[ s_{n+1}=\sqrt{2+s_n}\quad(n=1,2,3,\ldots) \] 일 때, \(\{s_n\}\)이 수렴함과 모든 \(n\)에 대해 \(s_n<2\)임을 증명하라.

[요청 사항]

\[ \forall n,\ s_n<2,\qquad \{s_n\}\text{ converges} \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad s_1=\sqrt2,\quad s_{n+1}=\sqrt{2+s_n} \]

Bounded above by \(2\):

\[ Step\ 1:\quad s_1=\sqrt2<2 \]

\[ Step\ 2:\quad s_n<2 \Rightarrow 2+s_n<4 \Rightarrow \sqrt{2+s_n}<2 \Rightarrow s_{n+1}<2 \]

\[ \therefore \forall n,\ s_n<2 \] \[ \because \text{induction + order preservation of }\sqrt{x} \]

Monotonic increasing:

\[ Step\ 3:\quad s_{n+1}\ge s_n \iff \sqrt{2+s_n}\ge s_n \]

\[ \iff 2+s_n\ge s_n^2 \iff s_n^2-s_n-2\le0 \]

\[ \iff (s_n-2)(s_n+1)\le0 \]

\[ Step\ 4:\quad 0<s_n<2 \Rightarrow s_n+1>0,\ s_n-2<0 \Rightarrow (s_n-2)(s_n+1)<0 \]

\[ \therefore s_{n+1}>s_n \]

\[ \therefore \{s_n\}\ \text{monotonically increasing and bounded above} \]

\[ Step\ 5:\quad \{s_n\}\text{ monotone}+\text{bounded} \Rightarrow \{s_n\}\text{ converges} \] \[ \because \text{교재 Ch.3, Thm 3.14: monotonic sequence converges iff bounded} \]

Limit value:

\[ Def:\quad \lim s_n=L \]

\[ s_{n+1}=\sqrt{2+s_n} \Rightarrow L=\sqrt{2+L} \] \[ \because \text{limit algebra, 교재 Ch.3, Thm 3.3} \]

\[ L^2=L+2 \Rightarrow L^2-L-2=0 \Rightarrow (L-2)(L+1)=0 \]

\[ s_n>0\Rightarrow L\ge0 \Rightarrow L\ne -1 \Rightarrow L=2 \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad \forall n,\ s_n<2,\quad s_n\uparrow 2,\quad \lim_{n\to\infty}s_n=2. } \]

---

Chapter 3 — Exercise 5

[문제 원문 및 번역]

Original: For any two real sequences \(\{a_n\}\), \(\{b_n\}\), prove that \[ \limsup_{n\to\infty}(a_n+b_n) \le \limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n, \] provided the sum on the right is not of the form \(\infty-\infty\).

번역: 임의의 두 실수열 \(\{a_n\},\{b_n\}\)에 대하여, 오른쪽 합이 \(\infty-\infty\) 꼴이 아닐 때 \[ \limsup_{n\to\infty}(a_n+b_n) \le \limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n \] 임을 증명하라.

[요청 사항]

\[ \limsup(a_n+b_n)\le \limsup a_n+\limsup b_n \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad A=\limsup a_n,\quad B=\limsup b_n \]

\[ Def:\quad c_n=a_n+b_n \]

\[ Step\ 1:\quad x>A \Rightarrow \exists N_1,\ n\ge N_1\Rightarrow a_n<x \] \[ \because \text{교재 Ch.3, Thm 3.17(b): upper limit characterization} \]

\[ Step\ 2:\quad y>B \Rightarrow \exists N_2,\ n\ge N_2\Rightarrow b_n<y \] \[ \because \text{교재 Ch.3, Thm 3.17(b)} \]

\[ Step\ 3:\quad N=\max(N_1,N_2) \Rightarrow n\ge N \Rightarrow c_n=a_n+b_n<x+y \]

\[ Step\ 4:\quad \limsup c_n\le x+y \] \[ \because \text{교재 Ch.3, Thm 3.17(b) 또는 Thm 3.19 comparison of limsup} \]

\[ Step\ 5:\quad x>A,\ y>B\ \text{arbitrary} \Rightarrow x\downarrow A,\ y\downarrow B \]

\[ \therefore \limsup(a_n+b_n)\le A+B \]

\[ \therefore \limsup(a_n+b_n) \le \limsup a_n+\limsup b_n \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad \limsup_{n\to\infty}(a_n+b_n) \le \limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n } \]

---

Chapter 3 — Exercise 14

[문제 원문 및 번역]

Original: If \(\{s_n\}\) is a complex sequence, define its arithmetic means \(\sigma_n\) by \[ \sigma_n=\frac{s_0+s_1+\cdots+s_n}{n+1}\quad(n=0,1,2,\ldots). \] \(a\) If \(\lim s_n=s\), prove that \(\lim \sigma_n=s\). \(b\) Construct a sequence \(\{s_n\}\) which does not converge, although \(\lim \sigma_n=0\). \(c\) Can it happen that \(s_n>0\) for all \(n\) and that \(\limsup s_n=\infty\), although \(\lim \sigma_n=0\)? \(d\) Put \(a_n=s_n-s_{n-1}\), for \(n\ge1\). Show that \[ s_n-\sigma_n=\frac1{n+1}\sum_{k=1}^n ka_k. \] Assume that \(\lim(na_n)=0\) and that \(\{\sigma_n\}\) converges. Prove that \(\{s_n\}\) converges. \(e\) Derive the last conclusion from a weaker hypothesis: Assume \(M<\infty\), \(\vert na_n\vert \le M\) for all \(n\), and \(\lim\sigma_n=\sigma\). Prove that \(\lim s_n=\sigma\).

번역: 복소수열 \(\{s_n\}\)에 대하여 산술평균 \(\sigma_n\)을 \[ \sigma_n=\frac{s_0+s_1+\cdots+s_n}{n+1}\quad(n=0,1,2,\ldots) \] 로 정의한다. \(a\) \(\lim s_n=s\)이면 \(\lim\sigma_n=s\)임을 증명하라. \(b\) \(\lim\sigma_n=0\)이지만 \(\{s_n\}\)은 수렴하지 않는 예를 구성하라. \(c\) 모든 \(n\)에 대해 \(s_n>0\), \(\limsup s_n=\infty\)이면서도 \(\lim\sigma_n=0\)일 수 있는가? \(d\) \(n\ge1\)에 대해 \(a_n=s_n-s_{n-1}\)라 하자. \[ s_n-\sigma_n=\frac1{n+1}\sum_{k=1}^n ka_k \] 임을 보여라. 또한 \(\lim(na_n)=0\)이고 \(\{\sigma_n\}\)이 수렴하면 \(\{s_n\}\)이 수렴함을 증명하라. \(e\) 더 약한 가정 \(\vert na_n\vert \le M<\infty\), \(\lim\sigma_n=\sigma\) 아래에서 \(\lim s_n=\sigma\)임을 증명하라.

[요청 사항]

\[ (a)\ s_n\to s\Rightarrow \sigma_n\to s \] \[ (b)\ \sigma_n\to0,\ s_n\not\to\text{limit 예시} \] \[ (c)\ s_n>0,\ \limsup s_n=\infty,\ \sigma_n\to0\ ? \] \[ (d)\ s_n-\sigma_n=\frac1{n+1}\sum_{k=1}^nka_k,\quad na_n\to0+\sigma_n\to\sigma\Rightarrow s_n\to\sigma \] \[ (e)\ \vert na_n\vert \le M,\ \sigma_n\to\sigma\Rightarrow s_n\to\sigma \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad \sigma_n=\frac1{n+1}\sum_{j=0}^n s_j \]

(a)

\[ s_n\to s \Rightarrow t_n=s_n-s\to0 \] \[ \sigma_n-s=\frac1{n+1}\sum_{j=0}^n(s_j-s)=\frac1{n+1}\sum_{j=0}^nt_j \]

\[ \varepsilon>0\Rightarrow \exists N,\ j\ge N\Rightarrow \vert t_j\vert <\varepsilon \] \[ \because \text{교재 Ch.3, Def 3.1} \]

\[ \left\vert \sigma_n-s\right\vert \le \frac1{n+1}\sum_{j=0}^{N-1}\vert t_j\vert + \frac1{n+1}\sum_{j=N}^n\vert t_j\vert \]

\[ \le \frac{C}{n+1} + \varepsilon \quad \left(C=\sum_{j=0}^{N-1}\vert t_j\vert \right) \]

\[ n\to\infty \Rightarrow \limsup\vert \sigma_n-s\vert \le\varepsilon \]

\[ \varepsilon\downarrow0 \Rightarrow \sigma_n\to s \]

\[ \because \text{finite initial part negligible + 교재 Ch.3, Thm 3.20(a): }1/n\to0 \]

(b)

\[ Def:\quad s_n=(-1)^n \]

\[ \sum_{j=0}^n(-1)^j= \begin{cases} 1,& n\text{ even}
0,& n\text{ odd} \end{cases} \]

\[ \Rightarrow \sigma_n= \frac1{n+1}\sum_{j=0}^n(-1)^j \in \left\{ 0,\frac1{n+1} \right\} \Rightarrow \sigma_n\to0 \]

\[ s_{2k}=1,\quad s_{2k+1}=-1 \Rightarrow s_n\not\to\text{limit} \] \[ \because \text{교재 Ch.3, Thm 3.2(b): limit uniqueness} \]

(c)

\[ s_n>0\ \forall n \Rightarrow \sigma_n=\frac{s_0+\cdots+s_n}{n+1}>0 \]

\[ \sigma_n\to0 \centernot\Rightarrow s_n\text{ bounded} \]

Construction:

\[ Def:\quad s_n= \begin{cases} k,& n=k^3,\ k=1,2,3,\ldots
\dfrac1n,& \text{otherwise} \end{cases} \]

\[ s_n>0\quad\forall n \]

\[ s_{k^3}=k\to\infty \Rightarrow \limsup s_n=\infty \] \[ \because \text{교재 Ch.3, Def 3.16: limsup via subsequential limits} \]

\[ \sum_{j=1}^n s_j \le \sum_{j=1}^n\frac1j+\sum_{k\le n^{1/3}}k \]

\[ \le n+\frac{n^{1/3}(n^{1/3}+1)}2 \quad\text{too weak} \]

더 정밀하게, \(\sum_{j=1}^n\frac1j=o(n)\), 그리고 \[ \sum_{k\le n^{1/3}}k = O(n^{2/3}) =o(n) \]

\[ \Rightarrow \frac1n\sum_{j=1}^ns_j\to0 \]

\[ \therefore \sigma_n\to0 \]

\[ \boxed{\text{가능하다.}} \]

(d)

\[ a_k=s_k-s_{k-1} \Rightarrow s_j=s_0+\sum_{k=1}^ja_k \]

\[ \sum_{j=0}^ns_j = (n+1)s_0+\sum_{j=1}^n\sum_{k=1}^ja_k \]

\[ (n+1)s_0+\sum_{k=1}^n(n-k+1)a_k \]

\[ \sigma_n = s_0+\frac1{n+1}\sum_{k=1}^n(n-k+1)a_k \]

\[ s_n=s_0+\sum_{k=1}^na_k \]

\[ s_n-\sigma_n = \sum_{k=1}^na_k-\frac1{n+1}\sum_{k=1}^n(n-k+1)a_k \]

\[ \frac1{n+1}\sum_{k=1}^nka_k \]

\[ \therefore s_n-\sigma_n = \frac1{n+1}\sum_{k=1}^nka_k \]

이제 \[ na_n\to0 \Rightarrow ka_k\to0 \]

\[ \frac1{n+1}\sum_{k=1}^nka_k\to0 \] \[ \because \text{part (a): convergent sequence }ka_k\to0\Rightarrow \text{its arithmetic means }\to0 \]

\[ s_n-\sigma_n\to0 \]

\[ \sigma_n\to\sigma \Rightarrow s_n=(s_n-\sigma_n)+\sigma_n\to0+\sigma=\sigma \] \[ \because \text{교재 Ch.3, Thm 3.3(a): limit of sums} \]

(e)

\[ Def:\quad \vert na_n\vert \le M,\quad \sigma_n\to\sigma \]

교재의 outline에 따라 \(m<n\)에 대해 \[ s_n-\sigma_n = \frac{m+1}{n-m}(\sigma_n-\sigma_m) + \frac1{n-m}\sum_{i=m+1}^n(s_n-s_i) \]

\[ \because \text{algebraic identity from Exercise 14(e)} \]

\[ s_n-s_i=\sum_{k=i+1}^na_k \]

\[ \vert ka_k\vert \le M \Rightarrow \vert a_k\vert \le \frac Mk \]

\[ \vert s_n-s_i\vert \le \sum_{k=i+1}^n\frac Mk \le M\frac{n-i}{i+1} \]

choose \(m=m(n)\) with \[ n-m\to\infty,\qquad \frac{m+1}{n-m}\to0 \] 예: \(m=\lfloor n-\sqrt n\rfloor\).

\[ \sigma_n-\sigma_m\to0 \Rightarrow \frac{m+1}{n-m}(\sigma_n-\sigma_m)\to0 \]

또한 \(i\in[m+1,n]\Rightarrow i\sim n\), hence \[ \vert s_n-s_i\vert \le M\frac{n-i}{i+1} \le M\frac{n-m}{m+1}\to0 \]

\[ \Rightarrow \left\vert \frac1{n-m}\sum_{i=m+1}^n(s_n-s_i) \right\vert \to0 \]

\[ \therefore s_n-\sigma_n\to0 \]

\[ \sigma_n\to\sigma \Rightarrow s_n\to\sigma \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.} \begin{aligned} &(a)\ s_n\to s\Rightarrow \sigma_n\to s.
&(b)\ s_n=(-1)^n.
&(c)\ \text{Yes; e.g. }s_{k^3}=k,\ s_n=1/n\text{ otherwise}.
&(d)\ s_n-\sigma_n=\frac1{n+1}\sum_{k=1}^nka_k,\quad na_n\to0,\ \sigma_n\to\sigma\Rightarrow s_n\to\sigma.
&(e)\ \vert na_n\vert \le M,\ \sigma_n\to\sigma\Rightarrow s_n\to\sigma. \end{aligned} } \]

---

Chapter 3 — Exercise 20

[문제 원문 및 번역]

Original: Suppose \(\{p_n\}\) is a Cauchy sequence in a metric space \(X\), and some subsequence \(\{p_{n_i}\}\) converges to a point \(p\in X\). Prove that the full sequence \(\{p_n\}\) converges to \(p\).

번역: \(\{p_n\}\)이 거리공간 \(X\)에서 코시 수열이고, 어떤 부분수열 \(\{p_{n_i}\}\)이 \(p\in X\)로 수렴한다고 하자. 전체 수열 \(\{p_n\}\)도 \(p\)로 수렴함을 증명하라.

[요청 사항]

\[ \{p_n\}\text{ Cauchy},\quad p_{n_i}\to p \Rightarrow p_n\to p \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad (X,d)\text{ metric space} \]

\[ \{p_n\}\text{ Cauchy} \iff \forall\varepsilon>0,\exists N_1,\ m,n\ge N_1 \Rightarrow d(p_n,p_m)<\varepsilon/2 \] \[ \because \text{교재 Ch.3, Def 3.8} \]

\[ p_{n_i}\to p \iff \forall\varepsilon>0,\exists I,\ i\ge I \Rightarrow d(p_{n_i},p)<\varepsilon/2 \] \[ \because \text{교재 Ch.3, Def 3.1} \]

\[ n_i\to\infty \Rightarrow \exists i\ge I,\ n_i\ge N_1 \]

\[ n\ge N_1 \Rightarrow d(p_n,p) \le d(p_n,p_{n_i})+d(p_{n_i},p) \] \[ \because \text{triangle inequality} \]

\[ <\varepsilon/2+\varepsilon/2=\varepsilon \]

\[ \therefore \forall\varepsilon>0,\exists N_1,\ n\ge N_1 \Rightarrow d(p_n,p)<\varepsilon \]

\[ \therefore p_n\to p \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad {p_n}\text{ Cauchy and }p_{n_i}\to p \Rightarrow p_n\to p. } \]

---

Chapter 3 — Exercise 21

[문제 원문 및 번역]

Original: Prove the following analogue of Theorem 3.10\(b\): If \(\{E_n\}\) is a sequence of closed nonempty and bounded sets in a complete metric space \(X\), if \[ E_n\supset E_{n+1}, \] and if \[ \lim_{n\to\infty}\operatorname{diam}E_n=0, \] then \[ \bigcap_1^\infty E_n \] consists of exactly one point.

번역: 정리 3.10\(b\)의 다음 유사 명제를 증명하라. 완비 거리공간 \(X\)에서 \(\{E_n\}\)이 닫힌, 공집합이 아닌, 유계 집합들의 열이고 \[ E_n\supset E_{n+1}, \] 또한 \[ \lim_{n\to\infty}\operatorname{diam}E_n=0 \] 이면 \[ \bigcap_{n=1}^{\infty}E_n \] 은 정확히 하나의 점으로 이루어진다.

[요청 사항]

\[ X\text{ complete},\quad E_n\neq\varnothing,\ E_n\text{ closed bounded},\quad E_n\supset E_{n+1},\quad \operatorname{diam}E_n\to0 \] \[ \Rightarrow \bigcap_{n=1}^\infty E_n=\{p\} \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad \operatorname{diam}E=\sup{d(x,y):x,y\in E} \] \[ \because \text{교재 Ch.3, Def 3.9} \]

Existence:

\[ E_n\neq\varnothing \Rightarrow \forall n,\ \exists x_n\in E_n \]

\[ m\ge n \Rightarrow x_m\in E_m\subset E_n,\quad x_n\in E_n \]

\[ \Rightarrow d(x_m,x_n)\le \operatorname{diam}E_n \]

\[ \operatorname{diam}E_n\to0 \Rightarrow \forall\varepsilon>0,\exists N,\ n\ge N \Rightarrow \operatorname{diam}E_n<\varepsilon \]

\[ m,n\ge N,\ m\ge n \Rightarrow d(x_m,x_n)\le\operatorname{diam}E_n<\varepsilon \]

\[ \therefore \{x_n\}\text{ Cauchy} \] \[ \because \text{교재 Ch.3, Def 3.8} \]

\[ X\text{ complete} \Rightarrow \exists p\in X,\ x_n\to p \] \[ \because \text{교재 Ch.3, Def 3.12: complete metric space} \]

\[ Fix\ k.\quad n\ge k\Rightarrow x_n\in E_n\subset E_k \]

\[ E_k\text{ closed},\quad x_n\to p,\quad x_n\in E_k\ \text{eventually} \Rightarrow p\in E_k \]

\[ \because \text{closed set contains limits of convergent sequences; 교재 Ch.3, Thm 3.2(d) logic / Ch.2 closed characterization} \]

\[ \forall k,\ p\in E_k \Rightarrow p\in\bigcap_{k=1}^\infty E_k \]

Uniqueness:

\[ p,q\in\bigcap_{n=1}^\infty E_n \Rightarrow \forall n,\ p,q\in E_n \]

\[ \Rightarrow d(p,q)\le \operatorname{diam}E_n\quad\forall n \]

\[ \operatorname{diam}E_n\to0 \Rightarrow d(p,q)=0 \Rightarrow p=q \]

\[ \because \text{metric axiom: }d(p,q)=0\iff p=q \]

\[ \therefore \bigcap_{n=1}^{\infty}E_n \text{ consists of exactly one point} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad \bigcap_{n=1}^{\infty}E_n=\{p} \text{ for a unique }p\in X. \} \]

---

Chapter 3 — Exercise 23

[문제 원문 및 번역]

Original: Suppose \(\{p_n\}\) and \(\{q_n\}\) are Cauchy sequences in a metric space \(X\). Show that the sequence \(\{d(p_n,q_n)\}\) converges. Hint: For any \(m,n\), \[ d(p_n,q_n)\le d(p_n,p_m)+d(p_m,q_m)+d(q_m,q_n); \] it follows that \[ \vert d(p_n,q_n)-d(p_m,q_m)\vert \] is small if \(m\) and \(n\) are large.

번역: 거리공간 \(X\)에서 \(\{p_n\}\), \(\{q_n\}\)이 코시 수열이라고 하자. 수열 \(\{d(p_n,q_n)\}\)이 수렴함을 보여라. 힌트: 임의의 \(m,n\)에 대해 \[ d(p_n,q_n)\le d(p_n,p_m)+d(p_m,q_m)+d(q_m,q_n) \] 이고, 따라서 \(m,n\)이 충분히 크면 \[ \vert d(p_n,q_n)-d(p_m,q_m)\vert \] 이 작아진다.

[요청 사항]

\[ \{p_n\},\{q_n\}\text{ Cauchy in }X \Rightarrow {d(p_n,q_n)}\text{ converges in }\mathbb R \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad r_n=d(p_n,q_n)\in\mathbb R \]

\[ \{p_n\}\text{ Cauchy} \Rightarrow \forall\varepsilon>0,\exists N_1,\ m,n\ge N_1 \Rightarrow d(p_n,p_m)<\varepsilon/2 \] \[ \because \text{교재 Ch.3, Def 3.8} \]

\[ \{q_n\}\text{ Cauchy} \Rightarrow \forall\varepsilon>0,\exists N_2,\ m,n\ge N_2 \Rightarrow d(q_n,q_m)<\varepsilon/2 \] \[ \because \text{교재 Ch.3, Def 3.8} \]

\[ N=\max(N_1,N_2) \]

\[ m,n\ge N \Rightarrow d(p_n,q_n) \le d(p_n,p_m)+d(p_m,q_m)+d(q_m,q_n) \]

\[ \Rightarrow r_n-r_m \le d(p_n,p_m)+d(q_m,q_n) <\varepsilon \]

교환하면 \[ r_m-r_n \le d(p_m,p_n)+d(q_n,q_m) <\varepsilon \]

\[ \therefore \vert r_n-r_m\vert <\varepsilon \]

\[ \therefore \{r_n\}\text{ is Cauchy in }\mathbb R \]

\[ \mathbb R\text{ complete} \Rightarrow \{r_n\}\text{ converges} \] \[ \because \text{교재 Ch.3, Thm 3.11(c): every Cauchy sequence in }\mathbb R^k\text{ converges} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad {d(p_n,q_n)}\text{ is Cauchy in }\mathbb R, \text{ hence converges.} } \]

---

Chapter 4 — Exercise 3

[문제 원문 및 번역]

Original: Let \(f\) be a continuous real function on a metric space \(X\). Let \(Z(f)\) \(the zero set of (f)\) be the set of all \(p\in X\) at which \(f(p)=0\). Prove that \(Z(f)\) is closed.

번역: 거리공간 \(X\) 위의 연속 실함수 \(f\)가 주어졌다고 하자. \(Z(f)\)를 \(f(p)=0\)인 모든 \(p\in X\)의 집합, 즉 \(f\)의 영점집합이라 하자. \(Z(f)\)가 닫힌 집합임을 증명하라.

[요청 사항]

\[ f:X\to\mathbb R\text{ continuous} \Rightarrow Z(f)=\{p\in X:f(p)=0\}\text{ closed} \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad Z(f)=\{p\in X:f(p)=0\} \]

\[ Step\ 1:\quad \{0\}\subset\mathbb R\text{ is closed} \]

\[ \because \mathbb R\text{ metric space, singleton closed} \]

\[ Step\ 2:\quad Z(f)=f^{-1}(\{0\}) \]

\[ Step\ 3:\quad f\text{ continuous} \Rightarrow f^{-1}(C)\text{ closed in }X\text{ for every closed }C\subset\mathbb R \] \[ \because \text{교재 Ch.4, Thm 4.8 Corollary} \]

\[ Step\ 4:\quad C=\{0\} \Rightarrow f^{-1}(\{0\})\text{ closed} \]

\[ \therefore Z(f)\text{ closed} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad Z(f)=f^{-1}(\{0\})\text{이고 }{0}\text{은 closed이므로 }Z(f)\text{ is closed.} } \]

---

Chapter 4 — Exercise 5

[문제 원문 및 번역]

Original: If \(f\) is a real continuous function defined on a closed set \(E\subset R^1\), prove that there exist continuous real functions \(g\) on \(R^1\) such that \(g(x)=f(x)\) for all \(x\in E\). Such functions \(g\) are called continuous extensions of \(f\) from \(E\) to \(R^1\). Show that the result becomes false if the word “closed” is omitted. Extend the result to vector-valued functions. Hint: Let the graph of \(g\) be a straight line on each of the segments which constitute the complement of \(E\).

번역: \(f\)가 닫힌 집합 \(E\subset R^1\) 위에 정의된 연속 실함수라고 하자. 그러면 \(R^1\) 전체에서 연속이고 모든 \(x\in E\)에 대해 \(g(x)=f(x)\)인 실함수 \(g\)가 존재함을 증명하라. 이러한 \(g\)를 \(f\)의 \(E\)에서 \(R^1\)로의 연속 확장이라 한다. “닫힌”이라는 조건을 생략하면 명제가 거짓이 됨을 보여라. 또한 벡터값 함수로 결과를 확장하라. 힌트: \(E\)의 여집합을 이루는 각 구간에서 \(g\)의 그래프를 직선으로 잡아라.

[요청 사항]

\[ E\subset\mathbb R\text{ closed},\quad f:E\to\mathbb R\text{ continuous} \Rightarrow \exists g:\mathbb R\to\mathbb R\text{ continuous},\ g\vert _E=f \]

\[ E\text{ not closed}\Rightarrow\text{false example} \]

\[ f:E\to\mathbb R^k\text{ continuous}\Rightarrow\exists g:\mathbb R\to\mathbb R^k\text{ continuous extension} \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad E\subset\mathbb R\text{ closed} \Rightarrow E^c=\mathbb R\setminus E\text{ open} \]

\[ \because \text{closed }\iff\text{ complement open} \]

\[ Step\ 1:\quad E^c=\bigcup_\alpha I_\alpha \]

\[ I_\alpha\text{ are disjoint open intervals} \] \[ \because \text{open subsets of }\mathbb R\text{ are countable unions of disjoint open intervals} \]

각 성분 \(I=(a,b)\), \((-\infty,b)\), \((a,\infty)\), \(\mathbb R\)에 대해 \(g\) 정의.

---

Case 1: (I=(a,b)), (a,b\in E)

\[ Def:\quad g(x)= \frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b), \quad a<x<b \]

\[ Step\ 2:\quad g(a)=f(a),\quad g(b)=f(b) \]

\[ Step\ 3:\quad g\vert _{(a,b)}\text{ linear} \Rightarrow g\text{ continuous on }(a,b) \] \[ \because \text{polynomial/rational-linear functions continuous, 교재 Ch.4, Examples 4.11} \]

\[ x\to a+ \Rightarrow g(x)\to f(a) \]

\[ x\to b- \Rightarrow g(x)\to f(b) \]

\[ \because \text{limit algebra, 교재 Ch.4, Thm 4.4} \]

---

Case 2: (I=(a,\infty)), (a\in E)

\[ Def:\quad g(x)=f(a)\quad(x>a) \]

\[ \Rightarrow g\text{ continuous on }(a,\infty),\quad x\to a+\Rightarrow g(x)\to f(a) \]

---

Case 3: (I=(-\infty,b)), (b\in E)

\[ Def:\quad g(x)=f(b)\quad(x<b) \]

\[ \Rightarrow g\text{ continuous on }(-\infty,b),\quad x\to b-\Rightarrow g(x)\to f(b) \]

---

Case 4: (E=\varnothing)

\[ Def:\quad g(x)=0\quad\forall x\in\mathbb R \]

\[ \Rightarrow g\text{ continuous} \]

---

Global definition

\[ g(x)= \begin{cases} f(x),&x\in E,
\text{linear/constant as above},&x\in E^c. \end{cases} \]

\[ Step\ 4:\quad x_0\in E^\circ \Rightarrow g=f\text{ near }x_0 \Rightarrow g\text{ continuous at }x_0 \] \[ \because f\text{ continuous on }E \]

\[ Step\ 5:\quad x_0\in E,\ x_0\text{ boundary point} \]

좌우 여집합 성분이 존재하면 위 정의상 \[ x\to x_0,\ x\in E^c \Rightarrow g(x)\to f(x_0) \]

또한 \[ x\to x_0,\ x\in E \Rightarrow f(x)\to f(x_0) \] \[ \because f\text{ continuous on }E,\ \text{교재 Ch.4, Def 4.5} \]

\[ \therefore g(x)\to g(x_0) \Rightarrow g\text{ continuous at }x_0 \] \[ \because \text{교재 Ch.4, Def 4.5 / Thm 4.6} \]

\[ \therefore g\in C(\mathbb R),\quad g\vert _E=f \]

---

“closed” 생략 시 반례

\[ Def:\quad E=(0,1),\quad f(x)=\frac1x \]

\[ f\text{ continuous on }E \] \[ \because \text{rational functions continuous where denominator }\ne0,\ \text{교재 Ch.4, Examples 4.11} \]

가정: \[ \exists g:\mathbb R\to\mathbb R\text{ continuous},\quad g(x)=1/x\ \forall x\in(0,1) \]

\[ x_n=\frac1n\in E,\quad x_n\to0 \]

\[ g(x_n)=n\to\infty \]

하지만 \[ g\text{ continuous at }0 \Rightarrow g(x_n)\to g(0)\in\mathbb R \] \[ \because \text{교재 Ch.4, Thm 4.2: sequential characterization of limits} \]

\[ \Rightarrow \text{contradiction} \]

\[ \therefore E\text{ closed condition essential} \]

---

Vector-valued extension

\[ Def:\quad f:E\to\mathbb R^k,\quad f=(f_1,\ldots,f_k) \]

\[ f\text{ continuous} \iff f_j\text{ continuous for all }j=1,\ldots,k \] \[ \because \text{교재 Ch.4, Thm 4.10(a)} \]

각 \(f_j:E\to\mathbb R\)에 대해 위 실수값 확장 적용:

\[ \exists g_j:\mathbb R\to\mathbb R\text{ continuous},\quad g_j\vert _E=f_j \]

\[ Def:\quad g=(g_1,\ldots,g_k):\mathbb R\to\mathbb R^k \]

\[ g\text{ continuous} \iff g_j\text{ continuous }\forall j \] \[ \because \text{교재 Ch.4, Thm 4.10(a)} \]

\[ g\vert _E=f \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad E\text{ closed}\Rightarrow f\text{ has a continuous extension }g\text{ to }\mathbb R. } \]

\[ \boxed{ \boldsymbol{Counterexample:}\quad E=(0,1),\ f(x)=1/x \Rightarrow \text{no continuous extension to }\mathbb R. } \]

\[ \boxed{ \boldsymbol{Vector\ case:}\quad f=(f_1,\ldots,f_k),\ \text{extend each }f_j,\ \text{then }g=(g_1,\ldots,g_k). } \]

---

Chapter 4 — Exercise 11

[문제 원문 및 번역]

Original: Suppose \(f\) is a uniformly continuous mapping of a metric space \(X\) into a metric space \(Y\), and prove that \(\{f(x_n)\}\) is a Cauchy sequence in \(Y\) for every Cauchy sequence \(\{x_n\}\) in \(X\). Use this result to give an alternative proof of the theorem stated in Exercise 13.

번역: \(f\)가 거리공간 \(X\)에서 거리공간 \(Y\)로 가는 균등연속 사상이라고 하자. \(X\)의 임의의 코시 수열 \(\{x_n\}\)에 대해 \(\{f(x_n)\}\)이 \(Y\)에서 코시 수열임을 증명하라. 이 결과를 사용하여 Exercise 13에 진술된 정리의 대안적 증명을 제시하라.

[요청 사항]

\[ f:X\to Y\text{ uniformly continuous},\quad \{x_n\}\text{ Cauchy in }X \Rightarrow {f(x_n)}\text{ Cauchy in }Y \]

그리고 Exercise 13:

\[ E\text{ dense in }X,\quad f:E\to\mathbb R\text{ uniformly continuous} \Rightarrow \exists!\ g:X\to\mathbb R\text{ continuous extension} \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad f:X\to Y\text{ uniformly continuous} \]

\[ \iff \forall\varepsilon>0,\exists\delta>0,
d_X(p,q)<\delta \Rightarrow d_Y(f(p),f(q))<\varepsilon \] \[ \because \text{교재 Ch.4, Def 4.18} \]

\[ Def:\quad \{x_n\}\text{ Cauchy} \]

\[ \iff \forall\delta>0,\exists N,\ m,n\ge N \Rightarrow d_X(x_m,x_n)<\delta \] \[ \because \text{교재 Ch.3, Def 3.8} \]

\[ Step\ 1:\quad \varepsilon>0 \Rightarrow \exists\delta>0,\ d_X(u,v)<\delta \Rightarrow d_Y(f(u),f(v))<\varepsilon \]

\[ Step\ 2:\quad \{x_n\}\text{ Cauchy} \Rightarrow \exists N,\ m,n\ge N \Rightarrow d_X(x_m,x_n)<\delta \]

\[ Step\ 3:\quad m,n\ge N \Rightarrow d_Y(f(x_m),f(x_n))<\varepsilon \]

\[ \therefore \{f(x_n)\}\text{ Cauchy in }Y \]

---

Exercise 13 alternative proof

\[ Def:\quad E\subset X\text{ dense} \Rightarrow \forall p\in X,\ \exists \{x_n\}\subset E,\ x_n\to p \] \[ \because \text{density + 교재 Ch.3, Thm 3.2(d) style sequential characterization} \]

\[ Def:\quad f:E\to\mathbb R\text{ uniformly continuous} \]

For each \(p\in X\), choose \[ x_n\in E,\quad x_n\to p \]

\[ x_n\to p \Rightarrow \{x_n\}\text{ Cauchy} \] \[ \because \text{교재 Ch.3, Thm 3.11(a)} \]

\[ f\text{ uniformly continuous} \Rightarrow \{f(x_n)\}\text{ Cauchy in }\mathbb R \]

\[ \mathbb R\text{ complete} \Rightarrow f(x_n)\to L_p\in\mathbb R \] \[ \because \text{교재 Ch.3, Thm 3.11(c)} \]

\[ Def:\quad g(p)=L_p \]

---

Well-definedness

Let \[ x_n,y_n\in E,\quad x_n\to p,\quad y_n\to p \]

\[ d_X(x_n,y_n)\le d_X(x_n,p)+d_X(p,y_n)\to0 \]

\[ f\text{ uniformly continuous} \Rightarrow \vert f(x_n)-f(y_n)\vert \to0 \]

\[ \Rightarrow \lim f(x_n)=\lim f(y_n) \]

\[ \therefore g(p)\text{ is well-defined} \]

---

Extension property

\[ p\in E \]

Choose \[ x_n=p\quad\forall n \]

\[ x_n\to p,\quad f(x_n)=f(p) \Rightarrow g(p)=f(p) \]

\[ \therefore g\vert _E=f \]

---

Continuity of (g)

\[ p_n\to p\text{ in }X \]

목표: \[ g(p_n)\to g(p) \]

\[ \varepsilon>0 \Rightarrow \exists\delta>0,\ d_X(u,v)<\delta \Rightarrow \vert f(u)-f(v)\vert <\varepsilon/3 \] \[ \because f\text{ uniformly continuous} \]

For each \(n\), by density choose \[ x_n\in E,\quad d_X(x_n,p_n)<\frac1n \quad\text{and}\quad \vert f(x_n)-g(p_n)\vert <\frac1n \]

또한 choose \(x\in E\) with \[ d_X(x,p)<\delta/3,\quad \vert f(x)-g(p)\vert <\varepsilon/3 \]

\[ p_n\to p \Rightarrow \exists N,\ n\ge N \Rightarrow d_X(p_n,p)<\delta/3,\quad 1/n<\min(\delta/3,\varepsilon/3) \]

\[ n\ge N \Rightarrow d_X(x_n,x) \le d_X(x_n,p_n)+d_X(p_n,p)+d_X(p,x) <\delta \]

\[ \Rightarrow \vert f(x_n)-f(x)\vert <\varepsilon/3 \]

\[ \vert g(p_n)-g(p)\vert \le \vert g(p_n)-f(x_n)\vert +\vert f(x_n)-f(x)\vert +\vert f(x)-g(p)\vert \]

\[ <\varepsilon/3+\varepsilon/3+\varepsilon/3 = \varepsilon \]

\[ \therefore g(p_n)\to g(p) \]

\[ \therefore g\text{ continuous} \] \[ \because \text{교재 Ch.4, Thm 4.2: sequential criterion} \]

---

Uniqueness

\[ g,h:X\to\mathbb R\text{ continuous},\quad g\vert _E=h\vert _E=f \]

\[ p\in X,\quad \exists x_n\in E,\ x_n\to p \]

\[ g(x_n)=h(x_n) \]

\[ g(x_n)\to g(p),\quad h(x_n)\to h(p) \] \[ \because \text{continuity + 교재 Ch.4, Thm 4.2} \]

\[ \therefore g(p)=h(p) \]

\[ \forall p\in X,\ g(p)=h(p) \Rightarrow g=h \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad f\text{ uniformly continuous}\Rightarrow {x_n}\text{ Cauchy}\Rightarrow {f(x_n)}\text{ Cauchy}. } \]

\[ \boxed{ \boldsymbol{Exercise\ 13:}\quad E\text{ dense},\ f:E\to\mathbb R\text{ uniformly continuous} \Rightarrow \exists!\text{ continuous extension }g:X\to\mathbb R. } \]

---

Chapter 4 — Exercise 13

[문제 원문 및 번역]

Original: Let \(E\) be a dense subset of a metric space \(X\), and let \(f\) be a uniformly continuous real function defined on \(E\). Prove that \(f\) has a continuous extension from \(E\) to \(X\). \(Uniqueness follows from Exercise 4.\) Hint: For each \(p\in X\) and each positive integer \(n\), let \(V_n(p)\) be the set of all \(q\in E\) with \(d(p,q)<1/n\). Use Exercise 9 to show that the intersection of the closures of the sets \(f(V_1(p)), f(V_2(p)),\ldots\), consists of a single point, say \(g(p)\), of \(R^1\). Prove that the function \(g\) so defined on \(X\) is the desired extension of \(f\). Could the range space \(R^1\) be replaced by \(R^k\)? By any compact metric space? By any complete metric space? By any metric space?

번역: \(E\)가 거리공간 \(X\)의 조밀한 부분집합이고, \(f\)가 \(E\)에서 정의된 균등연속 실함수라고 하자. \(f\)가 \(E\)에서 \(X\)로의 연속 확장을 가짐을 증명하라. 유일성은 Exercise 4에서 따른다. 힌트: 각 \(p\in X\), 양의 정수 \(n\)에 대해 \(V_n(p)\)를 \(d(p,q)<1/n\)인 모든 \(q\in E\)의 집합으로 두라. Exercise 9를 사용하여 \(f(V_1(p)), f(V_2(p)),\ldots\)의 폐포들의 교집합이 \(R^1\)의 단 하나의 점, 이를 \(g(p)\)라 하자, 으로 이루어짐을 보여라. 이렇게 정의된 \(g:X\to R^1\)가 원하는 \(f\)의 확장임을 증명하라. 치역 \(R^1\)을 \(R^k\)로 바꿀 수 있는가? 임의의 compact metric space로? 임의의 complete metric space로? 임의의 metric space로?

[요청 사항]

\[ E\subset X\text{ dense},\quad f:E\to\mathbb R\text{ uniformly continuous} \Rightarrow \exists!g:X\to\mathbb R,\quad g\vert _E=f,\quad g\text{ continuous} \]

또한

\[ \mathbb R^1\to \mathbb R^k,\quad compact,\quad complete,\quad arbitrary\ metric\ space \]

가능 여부 판정.

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad V_n(p)=\{q\in E:d_X(p,q)<1/n\} \]

\[ E\text{ dense} \Rightarrow V_n(p)\neq\varnothing\quad(\forall p\in X,\forall n) \]

\[ \because \text{dense subset: every neighborhood of }p\text{ meets }E \]

\[ V_{n+1}(p)\subset V_n(p) \Rightarrow \overline{f(V_{n+1}(p))}\subset \overline{f(V_n(p))} \]

\[ Def:\quad F_n(p)=\overline{f(V_n(p))}\subset\mathbb R \]

\[ \Rightarrow F_n(p)\supset F_{n+1}(p),\quad F_n(p)\neq\varnothing,\quad F_n(p)\text{ closed} \]

---

Diameter control

\[ f\text{ uniformly continuous} \Rightarrow \forall\varepsilon>0,\exists\delta>0,\ d_X(u,v)<\delta \Rightarrow \vert f(u)-f(v)\vert <\varepsilon \] \[ \because \text{교재 Ch.4, Def 4.18} \]

\[ n>\frac{2}{\delta},\quad u,v\in V_n(p) \Rightarrow d_X(u,v)\le d_X(u,p)+d_X(p,v)<\frac1n+\frac1n<\delta \]

\[ \Rightarrow \vert f(u)-f(v)\vert <\varepsilon \]

\[ \Rightarrow \operatorname{diam}f(V_n(p))\le\varepsilon \]

\[ \Rightarrow \operatorname{diam}F_n(p)=\operatorname{diam}\overline{f(V_n(p))}\le\varepsilon \] \[ \because \text{diameter unchanged by closure, 교재 Ch.3, Thm 3.10(a)} \]

\[ \therefore \operatorname{diam}F_n(p)\to0 \]

---

Existence of one intersection point

\[ F_n(p)\text{ closed in }\mathbb R,\quad F_n(p)\text{ bounded for large }n,\quad F_n(p)\supset F_{n+1}(p) \]

\[ \operatorname{diam}F_n(p)\to0 \]

\[ \mathbb R\text{ complete} \Rightarrow \bigcap_{n=1}^\infty F_n(p)=\{g(p)\} \] \[ \because \text{교재 Ch.3, Exercise 21 result / complete nested closed sets with diameters }\to0 \]

\[ Def:\quad g(p)=\text{the unique point in }\bigcap_{n=1}^\infty F_n(p) \]

---

Extension property

\[ p\in E \]

\[ f(p)\in f(V_n(p))\quad\forall n \]

\[ \Rightarrow f(p)\in F_n(p)\quad\forall n \]

\[ \Rightarrow f(p)\in\bigcap_{n=1}^\infty F_n(p)=\{g(p)\} \]

\[ \therefore g(p)=f(p) \]

\[ \therefore g\vert _E=f \]

---

Continuity of (g)

\[ \varepsilon>0 \Rightarrow \exists\delta>0,\ d_X(u,v)<\delta \Rightarrow \vert f(u)-f(v)\vert <\varepsilon/3 \]

Choose \(n\) such that

\[ \frac1n<\frac{\delta}{3} \]

\[ q\in X,\quad d_X(p,q)<\frac{\delta}{3} \]

\[ u\in V_n(p),\quad v\in V_n(q) \Rightarrow d_X(u,v) \le d_X(u,p)+d_X(p,q)+d_X(q,v) <\frac\delta3+\frac\delta3+\frac\delta3=\delta \]

\[ \Rightarrow \vert f(u)-f(v)\vert <\varepsilon/3 \]

\[ g(p)\in\overline{f(V_n(p))},\quad g(q)\in\overline{f(V_n(q))} \]

\[ \Rightarrow \vert g(p)-g(q)\vert \le\varepsilon \]

더 정확히, \(u,v\)를 \(g(p),g(q)\)에 충분히 가까운 점으로 잡으면

\[ \vert g(p)-g(q)\vert \le \vert g(p)-f(u)\vert +\vert f(u)-f(v)\vert +\vert f(v)-g(q)\vert <\varepsilon \]

\[ \therefore d_X(p,q)<\delta/3 \Rightarrow \vert g(p)-g(q)\vert <\varepsilon \]

\[ \therefore g\text{ uniformly continuous, hence continuous} \] \[ \because \text{uniform continuity }\Rightarrow\text{ continuity, 교재 Ch.4, Def 4.18 논의} \]

---

Uniqueness

\[ g,h:X\to\mathbb R\text{ continuous},\quad g\vert _E=h\vert _E=f \]

\[ E\text{ dense in }X \Rightarrow g=h\text{ on }X \] \[ \because \text{교재 Ch.4, Exercise 4: continuous mapping determined by dense subset} \]

---

Range space replacements

\[ \mathbb R^k:\quad \text{Yes} \]

\[ \because \mathbb R^k\text{ complete, and same nested-closed-diameter argument works} \] \[ \because \text{교재 Ch.3, Thm 3.11(c): }\mathbb R^k\text{ complete} \]

\[ \text{compact metric space }Y:\quad \text{Yes} \]

\[ \because \text{compact metric space is complete} \] \[ \because \text{교재 Ch.3, Thm 3.11(b): Cauchy sequence in compact metric space converges} \]

\[ \text{complete metric space }Y:\quad \text{Yes} \]

\[ \because \text{same proof using completeness + nested closed sets} \]

\[ \text{arbitrary metric space }Y:\quad \text{No} \]

Counterexample:

\[ X=\mathbb R,\quad E=\mathbb Q,\quad Y=\mathbb Q,\quad f:E\to Y,\ f(x)=x \]

\[ f\text{ uniformly continuous} \]

\[ E\text{ dense in }X \]

If extension \(g:\mathbb R\to\mathbb Q\) existed and continuous with \(g\vert _{\mathbb Q}=id\),

\[ x_n\in\mathbb Q,\quad x_n\to \sqrt2 \]

\[ g(x_n)=x_n\to \sqrt2 \]

\[ g\text{ continuous} \Rightarrow g(x_n)\to g(\sqrt2)\in\mathbb Q \]

\[ \Rightarrow g(\sqrt2)=\sqrt2\notin\mathbb Q \]

\[ \Rightarrow \bot \]

\[ \therefore \text{arbitrary metric space: false} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad \exists!g:X\to\mathbb R,\quad g\vert _E=f,\quad g\text{ continuous}. } \]

\[ \boxed{ \mathbb R^k:\ Yes,\quad compact\ metric\ space:\ Yes,\quad complete\ metric\ space:\ Yes,\quad arbitrary\ metric\ space:\ No. } \]

---

Chapter 4 — Exercise 18

[문제 원문 및 번역]

Original: Every rational \(x\) can be written in the form \(x=m/n\), where \(n>0\), and \(m\) and \(n\) are integers without any common divisors. When \(x=0\), we take \(n=1\). Consider the function \(f\) defined on \(R^1\) by \[ f(x)= \begin{cases} 0,&x\text{ irrational},[2mm] \dfrac1n,&x=\dfrac mn. \end{cases} \] Prove that \(f\) is continuous at every irrational point, and that \(f\) has a simple discontinuity at every rational point.

번역: 모든 유리수 \(x\)는 \(x=m/n\) 꼴로 쓸 수 있으며, 여기서 \(n>0\), \(m,n\)은 서로소인 정수이다. \(x=0\)일 때는 \(n=1\)로 둔다. \(R^1\)에서 정의된 함수 \[ f(x)= \begin{cases} 0,&x\text{가 무리수일 때},[2mm] \dfrac1n,&x=\dfrac mn\text{일 때} \end{cases} \] 를 생각하라. \(f\)가 모든 무리수점에서 연속이고, 모든 유리수점에서 단순 불연속을 가짐을 증명하라.

[요청 사항]

\[ x_0\notin\mathbb Q\Rightarrow f\text{ continuous at }x_0 \]

\[ x_0\in\mathbb Q\Rightarrow f\text{ simple discontinuity at }x_0 \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad f(x)=0\ (x\notin\mathbb Q),\quad f(m/n)=1/n,\quad (m,n)=1,\ n>0 \]

---

Lemma: bounded interval contains finitely many rationals with bounded denominator

\[ Fix\ M>0,\ N\in\mathbb N \]

\[ \left\{\frac mn\in[-M,M]:(m,n)=1,\ 1\le n\le N\right\} \text{ finite} \]

\[ \because 1\le n\le N,\quad \vert m/n\vert \le M \Rightarrow \vert m\vert \le Mn \Rightarrow \text{finite choices} \]

---

Irrational continuity

\[ x_0\notin\mathbb Q,\quad \varepsilon>0 \]

Choose \(N\) such that

\[ \frac1N<\varepsilon \]

유한집합

\[ A=\left\{\frac mn\in[x_0-1,x_0+1]:(m,n)=1,\ 1\le n\le N\right\} \]

\[ A\text{ finite},\quad x_0\notin A \]

\[ \Rightarrow \eta=\min{\vert x_0-r\vert :r\in A}>0 \]

\[ Def:\quad \delta=\min(1,\eta/2) \]

\[ \vert x-x_0\vert <\delta \]

If \(x\notin\mathbb Q\),

\[ f(x)=0=f(x_0) \Rightarrow \vert f(x)-f(x_0)\vert =0<\varepsilon \]

If \(x=m/n\in\mathbb Q\),

\[ \vert x-x_0\vert <\delta\le1 \Rightarrow x\in[x_0-1,x_0+1] \]

\[ x\notin A \Rightarrow n>N \]

\[ \Rightarrow f(x)=1/n<1/N<\varepsilon \]

\[ \therefore \vert f(x)-f(x_0)\vert <\varepsilon \]

\[ \therefore f\text{ continuous at }x_0 \] \[ \because \text{교재 Ch.4, Def 4.5} \]

---

Rational discontinuity

\[ x_0=\frac mn\in\mathbb Q,\quad (m,n)=1,\ n>0 \]

\[ f(x_0)=1/n>0 \]

Take irrational sequence:

\[ y_k=x_0+\frac{\sqrt2}{k} \]

For sufficiently large \(k\), \(y_k\to x_0\), and \(y_k\notin\mathbb Q\).

\[ f(y_k)=0\to0\neq f(x_0) \]

\[ \Rightarrow f\text{ not continuous at }x_0 \] \[ \because \text{교재 Ch.4, Thm 4.2 sequential criterion} \]

---

Simple discontinuity

목표:

\[ f(x_0+)=f(x_0-)=0 \]

Let \(t_k\to x_0\) with \(t_k>x_0\), \(t_k\neq x_0\).

Same finite denominator argument:

\[ \forall\varepsilon>0,\exists\delta>0,\quad 0<\vert t-x_0\vert <\delta \Rightarrow f(t)<\varepsilon \]

\[ \Rightarrow f(t_k)\to0 \]

\[ \therefore f(x_0+)=0 \]

동일하게

\[ f(x_0-)=0 \]

\[ f(x_0+)=f(x_0-)=0\neq f(x_0)=1/n \]

\[ \therefore x_0\text{ is a simple discontinuity} \] \[ \because \text{교재 Ch.4, Def 4.25 and Def 4.26} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad f\text{ is continuous at every irrational }x_0, \text{ and has a simple discontinuity at every rational }x_0. } \]

---

Chapter 4 — Exercise 23

[문제 원문 및 번역]

Original: A real-valued function \(f\) defined in \((a,b)\) is said to be convex if \[ f(\lambda x+(1-\lambda)y)\le \lambda f(x)+(1-\lambda)f(y) \] whenever \(a<x<b\), \(a<y<b\), \(0<\lambda<1\). Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex. For example, if \(f\) is convex, so is \(e^f\). If \(f\) is convex in \((a,b)\) and if \(a<s<t<u<b\), show that \[ \frac{f(t)-f(s)}{t-s} \le \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t}. \]

번역: \((a,b)\)에서 정의된 실함수 \(f\)가 다음을 만족할 때 convex라고 한다: \[ f(\lambda x+(1-\lambda)y)\le \lambda f(x)+(1-\lambda)f(y) \] 단, \(a<x<b\), \(a<y<b\), \(0<\lambda<1\). 모든 convex 함수가 연속임을 증명하라. 증가하는 convex 함수를 convex 함수에 합성하면 convex임을 증명하라. 예를 들어 \(f\)가 convex이면 \(e^f\)도 convex이다. 또한 \(f\)가 \((a,b)\)에서 convex이고 \(a<s<t<u<b\)이면 \[ \frac{f(t)-f(s)}{t-s} \le \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t} \] 임을 보여라.

[요청 사항]

\[ (1)\ f\text{ convex}\Rightarrow f\text{ continuous} \]

\[ (2)\ g\text{ increasing convex},\ f\text{ convex} \Rightarrow g\circ f\text{ convex} \]

\[ (3)\ a<s<t<u<b \Rightarrow \frac{f(t)-f(s)}{t-s} \le \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t} \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad f\text{ convex} \iff f(\lambda x+(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y) \]

---

(3) slope inequalities first

\[ a<s<t<u<b \]

\[ t=\lambda s+(1-\lambda)u \]

Solve:

\[ \lambda=\frac{u-t}{u-s},\quad 1-\lambda=\frac{t-s}{u-s} \]

\[ 0<\lambda<1 \]

Convexity:

\[ f(t)\le \frac{u-t}{u-s}f(s) + \frac{t-s}{u-s}f(u) \]

\[ (u-s)f(t) \le (u-t)f(s)+(t-s)f(u) \]

\[ (u-s)f(t)-(u-s)f(s) \le (t-s)f(u)-(t-s)f(s) \]

\[ (u-s)(f(t)-f(s)) \le (t-s)(f(u)-f(s)) \]

\[ \Rightarrow \frac{f(t)-f(s)}{t-s} \le \frac{f(u)-f(s)}{u-s} \]

Next,

\[ t=\mu u+(1-\mu)s \]

동일하게 위 식을 재배열하면

\[ (u-t)(f(u)-f(s)) \le (u-s)(f(u)-f(t)) \]

\[ \Rightarrow \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t} \]

\[ \therefore \frac{f(t)-f(s)}{t-s} \le \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t} \]

---

(1) Convex implies continuous

Fix

\[ x_0\in(a,b) \]

Choose

\[ a<r<x_0<R<b \]

For \(x\in(x_0,R)\), use slope inequality with

\[ r<x_0<x<R \]

\[ \frac{f(x_0)-f(r)}{x_0-r} \le \frac{f(x)-f(r)}{x-r} \le \frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(R)-f(x_0)}{R-x_0} \]

필요한 부분만:

\[ \frac{f(x)-f(x_0)}{x-x_0} \le \frac{f(R)-f(x_0)}{R-x_0} =:M_+ \]

또한 slope lower bound:

\[ \frac{f(x)-f(x_0)}{x-x_0} \ge \frac{f(x_0)-f(r)}{x_0-r} =:m_+ \]

\[ m_+(x-x_0) \le f(x)-f(x_0) \le M_+(x-x_0) \]

\[ x\to x_0+ \Rightarrow f(x)\to f(x_0) \]

For \(x\in(r,x_0)\), use

\[ r<x<x_0<R \]

\[ \frac{f(x_0)-f(x)}{x_0-x} \le \frac{f(R)-f(x)}{R-x} \quad\text{and bounded slope estimates} \]

More directly, there exist constants \(m_-,M_-\) such that

\[ m_-(x_0-x)\le f(x_0)-f(x)\le M_-(x_0-x) \]

\[ x\to x_0- \Rightarrow f(x)\to f(x_0) \]

\[ \therefore \lim_{x\to x_0}f(x)=f(x_0) \]

\[ \therefore f\text{ continuous at }x_0 \] \[ \because \text{교재 Ch.4, Def 4.5 and Thm 4.6} \]

\[ x_0\in(a,b)\text{ arbitrary} \Rightarrow f\text{ continuous on }(a,b) \]

---

(2) Increasing convex function of convex function

Let

\[ f:(a,b)\to I,\quad f\text{ convex} \]

\[ g:I\to\mathbb R,\quad g\text{ increasing and convex} \]

\[ h=g\circ f \]

For \(x,y\in(a,b)\), \(0<\lambda<1\),

\[ f(\lambda x+(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y) \]

\[ g\text{ increasing} \Rightarrow g(f(\lambda x+(1-\lambda)y)) \le g(\lambda f(x)+(1-\lambda)f(y)) \]

\[ g\text{ convex} \Rightarrow g(\lambda f(x)+(1-\lambda)f(y)) \le \lambda g(f(x))+(1-\lambda)g(f(y)) \]

\[ \therefore h(\lambda x+(1-\lambda)y) \le \lambda h(x)+(1-\lambda)h(y) \]

\[ \therefore h=g\circ f\text{ convex} \]

Example:

\[ g(t)=e^t \]

\[ e^t\text{ increasing and convex} \Rightarrow e^{f}=g\circ f\text{ convex} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad \text{Every convex }f:(a,b)\to\mathbb R\text{ is continuous.} } \]

\[ \boxed{ g\text{ increasing convex},\ f\text{ convex} \Rightarrow g\circ f\text{ convex}; \quad f\text{ convex}\Rightarrow e^f\text{ convex}. } \]

\[ \boxed{ \frac{f(t)-f(s)}{t-s} \le \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t}. } \]

---

Chapter 4 — Exercise 24

[문제 원문 및 번역]

Original: Assume that \(f\) is a continuous real function defined in \((a,b)\) such that \[ f\left(\frac{x+y}{2}\right)\le \frac{f(x)+f(y)}2 \] for all \(x,y\in(a,b)\). Prove that \(f\) is convex.

번역: \(f\)가 \((a,b)\)에서 정의된 연속 실함수이고, 모든 \(x,y\in(a,b)\)에 대해 \[ f\left(\frac{x+y}{2}\right)\le \frac{f(x)+f(y)}2 \] 를 만족한다고 하자. \(f\)가 convex임을 증명하라.

[요청 사항]

\[ f\text{ continuous},\quad f\left(\frac{x+y}{2}\right)\le\frac{f(x)+f(y)}2 \Rightarrow f\text{ convex} \]

즉,

\[ \forall \lambda\in(0,1),\quad f(\lambda x+(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y) \]

[풀이 과정 (Symbolic & Referenced)]

\[ Def:\quad f\left(\frac{x+y}{2}\right)\le\frac{f(x)+f(y)}2 \]

이를 midpoint convexity라 하자.

---

Step 1: dyadic convexity

\[ D=\left\{\frac{k}{2^n}:n\in\mathbb N,\ k=0,1,\ldots,2^n\right\} \]

Claim:

\[ \lambda\in D \Rightarrow f(\lambda x+(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y) \]

Proof by induction on \(n\).

\[ n=1:\quad \lambda=0,\frac12,1 \]

\[ \lambda=\frac12 \Rightarrow f\left(\frac{x+y}{2}\right)\le\frac{f(x)+f(y)}2 \]

\[ \lambda=0,1\Rightarrow \text{equality} \]

Induction step:

\[ \lambda=\frac{k}{2^{n+1}} \]

If \(k\) even,

\[ \lambda=\frac{j}{2^n} \Rightarrow \text{done by induction} \]

If \(k\) odd,

\[ \lambda=\frac12\left(\frac{k-1}{2^n}\right)+\frac12\left(\frac{k+1}{2^n}\right) \]

Define

\[ z_-=\frac{k-1}{2^n}x+\left(1-\frac{k-1}{2^n}\right)y \]

\[ z_+=\frac{k+1}{2^n}x+\left(1-\frac{k+1}{2^n}\right)y \]

\[ \frac{z_-+z_+}{2} = \frac{k}{2^{n+1}}x+\left(1-\frac{k}{2^{n+1}}\right)y \]

Midpoint convexity:

\[ f\left(\frac{z_-+z_+}{2}\right) \le \frac{f(z_-)+f(z_+)}2 \]

Induction:

\[ f(z_-) \le \frac{k-1}{2^n}f(x) + \left(1-\frac{k-1}{2^n}\right)f(y) \]

\[ f(z_+) \le \frac{k+1}{2^n}f(x) + \left(1-\frac{k+1}{2^n}\right)f(y) \]

Average:

\[ f\left(\lambda x+(1-\lambda)y\right) \le \lambda f(x)+(1-\lambda)f(y) \]

\[ \therefore \text{dyadic convexity holds} \]

---

Step 2: pass to arbitrary (\lambda)

\[ 0<\lambda<1 \]

Choose dyadic sequence

\[ \lambda_n\in D,\quad \lambda_n\to\lambda \]

\[ z_n=\lambda_nx+(1-\lambda_n)y \]

\[ z=\lambda x+(1-\lambda)y \]

\[ \lambda_n\to\lambda \Rightarrow z_n\to z \]

\[ f\text{ continuous} \Rightarrow f(z_n)\to f(z) \] \[ \because \text{교재 Ch.4, Thm 4.2 / Def 4.5} \]

For each \(n\),

\[ f(z_n) \le \lambda_nf(x)+(1-\lambda_n)f(y) \]

Take limits:

\[ f(z) \le \lambda f(x)+(1-\lambda)f(y) \] \[ \because \text{limit algebra, 교재 Ch.3, Thm 3.3} \]

\[ \therefore f\text{ convex} \]

[최종 답안]

\[ \boxed{ \boldsymbol{Ans.}\quad \text{continuous midpoint-convex }f \Rightarrow f\text{ convex}. } \]

---

\[ \boxed{ \text{All assigned problems completed: Ch.3 }(1,3,5,14,20,21,23),\quad \text{Ch.4 }(3,5,11,13,18,23,24). } \]