\newcommand{\D}{\mathrm{d}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\F}{\mathbb{F}} \newcommand{\U}{\mathbb{U}} \newcommand{\re}{\,\mathrm{Re}\,} \newcommand{\im}{\,\mathrm{Im}\,} \newcommand{\ord}{\mathrm{ord}} \newcommand{\Gal}{\mathrm{Gal}} {\bf Lemma 5.1.15} Every Cauchy sequence $(a_n)_{n=1}^\infty$ is bounded. \begin{proof} Let $(a_n)_{n\in\N}$ be a Cauchy sequence (of rationals). In particular, $(a_n)_{n\in\N}$ is ``eventually $1$-steady''. This means that, taking $\varepsilon = 1$ in the definition, there exists some $N \in \N$ such that $|a_p - a_q| < 1$ for all $p,q\geq N$: $$\exists N \in \N,\ \forall p \in \N,\ \forall q \in \N,\ q \geq p \geq N \Rightarrow |a_p -a_q| \leq 1.$$ In particular, if $p \geq N$, then $|a_p - a_N|<1$, which implies $$|a_p| = |(a_p - a_N) + a_N| \leq |a_p - a_N| + |a_N| \leq 1 + |a_N|.$$We obtain $$\forall p \geq N,\ |a_p| \leq 1 + |a_N|.$$ Since the set $S=\{a_0,a_1,\ldots, a_{N-1}\}$ is finite, $S$ is bounded (Lemma 5.1.14): there is some $M$ such that $$\forall i,\ 0\leq i <N \Rightarrow |a_i| \leq M.$$ Take $M' = M + (1+ |a_N|)$. \begin{itemize} \item If $0 \leq i < N$, then $|a_i| \leq M \leq M + (1+ |a_N|) = M'$. \item If $i \geq N$, then $|a_i| \leq 1 + |a_N| \leq M + (1+|a_N|) = M'$. \end{itemize} Therefore, for all $i \in \N$, $|a_i| \leq M'$. This proves that $(a_i)_{i\in\N}$ is bounded. \end{proof}

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Analysis I

Exercise 5.1.1 (Cauchy sequences are bounded)

Prove Lemma 5.1.15.

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Lemma 5.1.15 Every Cauchy sequence ${(a_{n})}_{n = 1}^{\infty}$ is bounded.

Proof. Let ${(a_{n})}_{n \in ℕ}$ be a Cauchy sequence (of rationals). In particular, ${(a_{n})}_{n \in ℕ}$ is “eventually $1$ -steady”. This means that, taking $𝜀 = 1$ in the definition, there exists some $N \in ℕ$ such that $| a_{p} - a_{q} | < 1$ for all $p, q \geq N$ :

\exists N \in ℕ, \forall p \in ℕ, \forall q \in ℕ, q \geq p \geq N \Rightarrow | a_{p} - a_{q} | \leq 1 .

In particular, if $p \geq N$ , then $| a_{p} - a_{N} | < 1$ , which implies

| a_{p} | = | (a_{p} - a_{N}) + a_{N} | \leq | a_{p} - a_{N} | + | a_{N} | \leq 1 + | a_{N} | .

We obtain

\forall p \geq N, | a_{p} | \leq 1 + | a_{N} | .

Since the set $S = {a_{0}, a_{1}, \dots, a_{N - 1}}$ is finite, $S$ is bounded (Lemma 5.1.14): there is some $M$ such that

\forall i, 0 \leq i < N \Rightarrow | a_{i} | \leq M .

Take $M^{'} = M + (1 + | a_{N} |)$ .

If $0 \leq i < N$ , then $| a_{i} | \leq M \leq M + (1 + | a_{N} |) = M^{'}$ .
If $i \geq N$ , then $| a_{i} | \leq 1 + | a_{N} | \leq M + (1 + | a_{N} |) = M^{'}$ .

Therefore, for all $i \in ℕ$ , $| a_{i} | \leq M^{'}$ . This proves that ${(a_{i})}_{i \in ℕ}$ is bounded. □

richardganaye

2024-06-14 18:10

Exercise 5.1.1 (Cauchy sequences are bounded)

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