Probs. 24 (c) and (e), Chap. 3, in Baby Rudin: Completion of a metric space

Question

Here is Prob. 24, Chap. 3, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $X$ be a metric space.

(a) Call two Cauchy sequences $\left\{ p_n \right\}$, $\left\{ q_n \right\}$ in $X$ equivalent if $$ \lim_{n \to \infty} d \left( p_n, q_n \right) = 0.$$ Prove that this is an equivalence relation. [ This is easy.]

(b) Let $X^*$ be the set of all equivalence classes so obtained. If $P \in X^*$, $Q \in X^*$, $\left\{ p_n \right\} \in P$, $\left\{ q_n \right\} \in Q$, define $$ \Delta (P, Q) = \lim_{n \to \infty} d \left( p_n, q_n \right); $$ by Exercise 23, this limit exists. Show that the number $\Delta (P, Q)$ is unchanged if $\left\{ p_n \right\}$ and $\left\{ q_n \right\}$ are replaced by equivalent sequences, and hence that $\Delta$ is a distance function in $X^*$. [ I don't have any problem with that either.]

(c) Prove that the resulting metric space $X^*$ is complete. [ How to do this? ]

(d) For each $p \in X$, there is a Cauchy sequence all of whose terms are $p$; let $P_p$ be the element of $X^*$ which contains this sequence. Prove that $$ \Delta \left( P_p, P_q \right) = d(p, q) $$ for all $p, q \in X$. In other words, the mapping $\varphi$ defined by $\varphi (p) = P_p$ is an isometry (i.e. a distance-preserving mapping) of $X$ into $X^*$. [ This isn't a problem either. ]

(e) Prove that $\varphi(X)$ is dense in $X^*$, and that $\varphi(X) = X^*$ if $X$ is complete. [ How to? ] By (d), we may identify $X$ and $\varphi(X)$ and thus regard $X$ as embedded in the complete metric space $X^*$. We call $X^*$ the completion of $X$.

My effort:

Prob. 24 (c):

Let $\left\{ P_k \right\}$ be a Cauchy sequence in $X^*$. Then, given any $\varepsilon > 0$, we can find a natural number $K$ such that $$\Delta \left( P_k, P_r \right) < \varepsilon$$ for all $k, r \in \mathbb{N}$ such that $k > K$ and $r > K$.

For each $k \in \mathbb{N}$, let $\left\{ p_{kn} \right\}$ be an element of $P_k$. Then, for each $k \in \mathbb{N}$, the sequence $\left\{ p_{kn} \right\}$ is a Cauchy sequence in $(X, d)$ and also $$ \lim_{n \to \infty } d \left( p_{kn}, p_{rn} \right) < \varepsilon$$ for all $k, r \in \mathbb{N}$ such that $k > K$ and $r > K$.

What next?

Prob. 24 (e):

Let $P \in X^*$ and let $\varepsilon > 0$ be given. We need to find an element $x \in X$ such that $$ \Delta (P, \varphi(x) ) < \varepsilon.$$ Let $\left\{ p_n \right\}$ be a Cauchy sequence in $P$. Then there exists a natural number $N$ such that $$ d \left( p_m , p_n \right) < \frac{\varepsilon}{2} $$ for all $m, n \in \mathbb{N}$ such that $m > N$ and $n > N$.

Let $P_{p_{N+1}}$ be the element of $X^*$ that contains the Cauchy sequence $$ \left\{ p_{N+1}, p_{N+1}, p_{N+1}, \ldots \right\}.$$ Then we see that $$ \Delta \left( P, P_{p_{N+1}} \right) = \lim_{n \to \infty} d \left( p_n, p_{N+1} \right) \leq \frac{\varepsilon}{2} < \varepsilon. $$

And, $P_{p_{N+1}} = \varphi(p_{N+1})$. Is this argument correct? If not, then where lies the flaw? What it the correct proof?

An afterthought:

Here's a proof of Prob. 24 (c) that I propose.

Let $\left\{ P_k \right\}_{k \in \mathbb{N}}$ be a Cauchy sequence in $X^*$. Then, given a real number $\varepsilon > 0$, we can find a natural number $K$ such that $$ \Delta \left( P_k, P_r \right) < \frac{\varepsilon}{2}$$ for all $r, k \in \mathbb{N}$ such that $r > K$ and $k > K$.

For each $k \in \mathbb{N}$, since $P_k \in X^*$, each $P_k$ is an equivalence class of Cauchy sequences in $X$. Let $\left\{ p_{kn} \right\}_{n \in \mathbb{N}}$ be a Cauchy sequence in the equivalence class $P_k$, for each $k \in \mathbb{N}$. Then we can conclude that $$ \Delta \left( P_k, P_r \right) = \lim_{n \to \infty} d \left( p_{kn}, p_{rn} \right) < \frac{\varepsilon}{2}$$ for all $r, k \in \mathbb{N}$ such that $r > K$ and $k > K$.

Therefore we can find a natural number $N$ such that $$ d \left( p_{kn}, p_{rn} \right) < \frac{\varepsilon}{2}$$ for all $r, k, n \in \mathbb{N}$ such that $r > K$, $k > K$, and $n > N$. So $$ d \left( p_{kn}, p_{K+1, \ N+1} \right) < \frac{\varepsilon}{2}$$ for all $k, n \in \mathbb{N}$ such that $k > K$ and $n > N$.

Now $$\left\{ p_{K+1, \ N+1}, \ p_{K+1, \ N+1}, \ p_{K+1, \ N+1}, \ \ldots \right\}$$ is a Cauchy sequence in $X$. Let $P \in X^*$ be the equivalence class of this Cauchy sequence.

Now, for each $k \in \mathbb{N}$, since $\left\{ p_{kn} \right\}_{n \in \mathbb{N}}$ is a Cauchy sequence in the equivalence class $P_k$, we can conclude that, for all $k \in \mathbb{N}$ such that $k > K$, the following is true: $$ \Delta \left( P_k, P \right) = \lim_{n \to \infty} d \left( p_{kn}, p_{K+1, \ N+1} \right) \leq \frac{\varepsilon}{2} < \varepsilon.$$ Since $\varepsilon > 0$ was arbitrary, we can conclude that our original sequence $\left\{ P_k \right\}_{k \in \mathbb{N}}$ converges in $\left( X^*, \Delta \right)$ to the point $P \in X^*$. Hence $\left( X^*, \Delta \right)$ is a complete metric space.

Is this proof correct? If not, then where lies the problem? I wonder if the $P\in X^*$ in my argument is dependent on our choice of $\varepsilon$ and is not uniform.

Answer 1

Hint If $x^n=(x^n_m)$ is a Cauchy Sequence in this space, show that there exists a sequence $k_n$ such that $x^n_{k_n}$ is a Cauchy sequence in $X$. Show that the equivalence class of this sequence is the limit of $x^n$.

For $(e)$ the idea is correct.

Answer 2

Late reply, but you can use (e) to solve (c) as follows:

Proof of (e):

Let $P\in X^*$, we will show either $P$ is a limit point of $\phi(X)$ or $P$ is a point in $\phi(X)$.

Choose any sequence $\{p_n\}\in P$. For given $r\gt 0$ assume there exists $Q\neq P$, such that $Q\in N_{r/2}(P)$. There exists $\{q_n\}\in Q$ and some $N_1\in\mathbb{N}$ satisfying $$d(p_n,q_n)\lt \frac{r}{2},\quad\forall n\geq N_1.$$ Since $\{q_n\}$ is Cauchy, there exists $N_2\in\mathbb{N}$ such that $$d(q_n,q_m)\lt \frac{r}{2},\quad\forall n,m\geq N_2.$$ Letting $N=\max\{N_1,N_2\}$, $$d(p_n,q_N)\leq d(p_n,q_n)+d(q_n,q_N)\lt r,\quad\forall n\geq N.$$ Hence $P_{q_N}\in N_r(P)$.

Now assume for some $r\gt 0$ that $N_r(P)=P$. This means $$Q\neq P\implies \Delta(P,Q)\geq r,$$ but $\{p_n\}\in P$ is Cauchy and hence for some $N\in\mathbb{N}$ $$\Delta(P,P_{p_N})\lt r\implies P=P_{p_N}$$ and we conclude $\phi(X)$ is dense in $X^*$.

If $X$ is complete, then for any $\{p_n\}\in P$, there exists $p\in X$ such that $$\lim_{n\rightarrow\infty}d(p_n,p)=0\implies P=P_p.$$

Proof of (c):

Let $\{P_j\}$ be a Cauchy sequence in $X^*$and for each $P_{j}$ fix a sequence $\{p_{j,n}\}$. Since $\phi(X)$ is dense in $X^*$ either $P_j\in\phi(X)$ or $P_j$ is a limit point of $\phi(X)$. In either case, for each $\{p_{j,n}\}$ and $\epsilon\gt 0$, there exists a point $p$ such that $$\lim_{n\rightarrow\infty}d(p_{j,n},p)\lt\epsilon.$$ So, there exists $N_1$ satisfying $$d(p_{1,n},p_1) \lt 1,\quad \forall n\geq N_1$$ and $N_2$ satisfying $$d(p_{2,n},p_2) \lt \frac{1}{2},\quad \forall n\geq N_2.$$

First, we will prove the sequence $\{p_n\}$ thus constructed is Cauchy.

Let $\epsilon\gt 0$ be given, then there exists an integer $M\gt 3/\epsilon$ such that $$\Delta(P_k,P_l)\lt\frac{\epsilon}{3},\quad\forall k,l\geq M$$ and so taking $n$ large enough $$d(p_k,p_l)\leq d(p_k,p_{k,n}) +d(p_{k,n},p_{l,n})+d(p_{l,n},p_l)\lt\epsilon.$$ Hence $\{p_j\}$ is Cauchy and we can write $\{p_j\}\in P$.

Now, we can show the sequence $\{P_j\}$ converges to $P$. This is almost immediate, since

$$\Delta(P_j,P)\leq \Delta(P_j,P_{p_j})+\Delta(P_{p_j},P)=\lim_{n\rightarrow\infty}d(p_{j,n}, p_j)+\lim_{n\rightarrow\infty}d(p_j,p_n)$$ and as $j$ increases this goes to $0$.