Namesake Property of Completions (3.24(c))

Walter Rudin Principles of Mathematical Analysis, chapter 3, exercise 24(c):

Let $X$ be a metric space, and let $X^*$ be the metric space of equivalence classes of Cauchy sequences under the relation $\{p_n\} \text{~} \{q_n\}$ if $\lim_{n→∞} d(p_n,q_n)=0$ and under the distance metric $Δ(P,Q)=\lim_{n→∞} d(p_n,q_n)$ for any Cauchy sequences $\{p_n\}$, $\{q_n\}$ in $P,Q$ respectively. Prove $X^*$ is complete.

Proof: Lemma: Let $s$ be a Cauchy sequence in some metric space. Say $s$ is a quick Cauchy sequence if $n,m≤N$ implies $d(s(n),s(m)) < \dfrac{1}{N}$. Then a subsequence of $s$ is a quick Cauchy sequence, so particularly the class $S$ of $s$ under the above relation contains a representative that is a quick Cauchy sequence. Proof: Since $s$ is Cauchy, for all $n∈ℕ^+$ let $N_n$ be such that $a,b≥N_n$ implies $d(s(a),s(b)) < \dfrac{1}{n}$. Let $s'$ be the subsequence such that $s'(n)=s(N_n)$, and it's seen that $n,m≥N$ implies $d(s'(n),s'(m))=d(s(N_n),s(N_m)) < \text{max}(\dfrac{1}{n},\dfrac{1}{m}) ≤ \dfrac{1}{N}$.

Now, let $\{S_n\}$ be a Cauchy sequence in $X^*$, with $s_n$ a quick Cauchy sequence of $S_n$ for all $n$. Let $s$ be the sequence with $s(n)=s_n(n)$ for all $n$.

We prove $s$ is Cauchy: Let $ε > 0$. Let $α∈ℕ$ be such that $\dfrac{2}{α} < ε$, and let $N'$ be such that $n,m ≥ N'$ implies $Δ(S_n,S_m) = \lim_{z→∞} d(s_n(z),s_m(z)) < ε - \dfrac{2}{α}$. Then if $n,m ≥ \text{max}(α,N')$ we see $$d(s(n),s(m)) = d(s_n(n),s_m(m))$$$$≤ d(s_n(n),s_n(z))+d(s_n(z),s_m(z))+d(s_m(z),s_m(m)) < \dfrac{1}{α}+ε-\dfrac{2}{α}+\dfrac{1}{α} = ε$$ for $z$ sufficiently large.

We prove $S$, the class of $s$, is the point of convergence of $\{S_n\}$: Let $ε > 0$. Let $α$ be such that $\dfrac{1}{α} < ε$, and let $N'$ be such that $n,m≥N'$ implies $d(s(n),s(m)) < ε-\dfrac{1}{α}$ since $s$ is Cauchy, then choose $N ≥ \text{max}(α,N')$. Given $n≥N$, observe $Δ(S,S_n) = \lim_{z→∞} d(s(z),s_n(z))$. For $z≥N$ we bound $$d(s(z),s_n(z))=d(s_z(z),s_n(z))$$$$≤d(s_z(z),s_n(n))+d(s_n(n),s_n(z)) < ε-\dfrac{1}{α}+\dfrac{1}{α} = ε$$ so $Δ(S,S_n)≤ε$ when $n≥N$, and we are done.$~\square$