Section2.4Cauchy Sequences and Complete Metric Spaces
“Within you, you will find everything you need to be complete.”
―Bryant H. McGill
Definition2.41.Cauchy Sequence.
Let (M, d) be a metric space. A sequence {xn} in M is a Cauchy sequence if for every ε > 0, there exists a positive integer N such that if m,n ≥ N, then d(xm, xn) < ε.
A sequence \((x_n)_{n=1}^{\infty}\subseteq \mathbb{C}\) is Cauchy if for all \(\varepsilon>0\,\exists N\in\mathbb{Z}, N\geq 1, \forall n,m\geq N\text{,}\)\(\left|{x_n-x_m}\right|<\varepsilon\text{.}\)
Example2.42.
\((x_n)_{n=1}^{\infty}\subseteq \mathbb{Q}\) is an increasing sequence with \(\sup_{n\geq 1} x_n=\sqrt{2}\text{.}\) (eg \(x_1=1,x_2=1.4,x_3=1.41,\dots\)) Then \((x_n)_{n=1}^{\infty}\subseteq \mathbb{Q}\) is Cauchy (by the next result) but it will not converge in \(\mathbb{Q}\) (In \(R\text{,}\)\(\lim_{n\rightarrow\infty}x_n=\sqrt{2}\text{.}\)
Lemma2.43.
Every Cauchy sequence in \(\mathbb{C}\) is bounded.
Fix \(N\geq 1\) such that for all \(n,m\geq N\text{,}\)\(\left|{x_n-x_m}\right|<1\text{.}\) Set \(C=\max\{\left|{x_1}\right|,\left|{x_2}\right|,\dots,\left|{x_{N-1}}\right|,\left|{x_N}\right|+1\}\) then of \(1\leq n<N, \left|{x_n}\right|\leq C\) and if \(n\geq N, \left|{x_n}\right|\leq \left|{x_n-x_N}\right|+\left|{x_N}\right|<1+\left|{x_N}\right|\leq C\text{.}\)
Theorem2.44.
Every Cauchy sequence in \(\mathbb{C}\) converges.
Every cauchy sequence is bounded by the lemma, so has a convergent subsequence by the Bolzano-Weierstrass theorem, and then converges by assignment 3 number 1.
Warning2.45.
A subsequence \((x_n)_{n=1}^{\infty}\subseteq \mathbb{C}\) which satisfies \(\lim_{n\rightarrow\infty}|x_n-x_{n+1}|=0\) does not imply \((x_n)_{n=1}^{\infty}\) is Cauchy.
Example2.46.
(short version: \(\sum_{k=1}^{\infty}\frac{1}{k}\) diverges) Consider \((S_n)=\sum_{k=1}^{n}\frac{1}{k}\text{.}\) Then \(|S_{n+1}-S_n|=\frac{1}{n+1}\rightarrow 0\text{.}\) We will show \((S_n)_{n=1}^{\infty}\) is unbounded, and hence it is not Cauchy. Note that \(S_{2^n}=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+\dots+\frac{1}{2^n}\geq 1+\frac{1}{2}+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8})+\dots +\frac{1}{2^n}=1+\frac{n}{2}\) So \(\sup_{n\geq 1}S_n=\infty\text{.}\)
Theorem2.47.
If {xn} is a convergent sequence in a metric space, then {xn} is a Cauchy sequence.
Definition2.48.Complete Metric Space.
Let M be a metric space. If every Cauchy sequence in M is convergent, we say that M is a complete metric space.
Theorem2.49.\(\R^n\) is Complete.
The metric space Rn is complete.
Example2.50.
The metric space \Q (viewed as a subspace of \R) is not complete.
