# Questions about proof for metric space completion

Theorem: Let $$(M, d)$$ be a metric space. Then there exists a metric space $$(M^{∗},\Delta)$$ and a map $$\varphi: M \to M^{∗}$$ such that:

(1) $$\varphi$$ is one–to–one. That is, $$\varphi(x)=\varphi(y)$$ if and only if $$x=y$$.

(2) $$\Delta(\varphi(x),\varphi(y) = d(x, y)$$ for all $$x,y \in M$$.

(3) $$M^{∗}$$ is complete.

(4) $$\varphi(M)$$ is dense in $$M^{∗}$$. This means that each element of $$M^{∗}$$ is a limit of elements of $$\varphi(M) = \{\varphi(x): x \in > M$$}. Equivalently, for each $$P \in M^{∗}$$ and each $$\epsilon>0$$, there is a $$p \in M$$ with $$\Delta(\varphi(p),P)<\epsilon$$.

(5) If $$M$$ is complete, then $$\varphi(M)=M^{∗}$$.

Definitions:

First define $$M^{′} = \{(p_{n})_{n \in \mathbb{N}} | (p_{n})_{n \in \mathbb{N}}$$ is a Cauchy sequence in $$M\}$$ and define $$(p_{n})_{n \in \mathbb{N}}$$ and $$(q_{n})_{n \in \mathbb{N}}$$ in $$M$$ to be “equivalent”, written $$(p_{n})_{n \in \mathbb{N}} \sim (q_{n})_{n \in \mathbb{N}}$$, if and only if $$\lim_{n \to \infty} d(p_{n},q_{n}) = 0$$.

Now let $$[(p_{n})_{n \in \mathbb{N}}]=\{(q_{n})_{n \in \mathbb{N}} \in M^{′} | (q_{n})_{n \in \mathbb{N}} \sim (p_{n})_{n \in \mathbb{N}}$$ be the equivalence class of $$(p_{n})_{n \in \mathbb{N}}$$ under this equivalence relation.

If $$P,Q \in M^{*}$$ and $$(p_{n})_{n \in \mathbb{N}}$$ is a representative of $$P$$ and $$(q_{n})_{n \in \mathbb{N}}$$ is a representative of $$Q \in M^{*}$$, we define $$\Delta(P,Q)=\lim_{n \to \infty} d(p_{n},q_{n})$$.

Finally, we define $$\varphi: M \to M^{∗}$$ by $$\varphi(p) = [(p,p,p,...)]$$.

(4) is proven in a separate lemma and this is the one I'm having trouble with.

Lemma 8: $$\varphi(M)$$ is dense in $$M^{∗}$$.

Proof: Let $$P = [(p_{n})_{n \in \mathbb{N}}] \in M^{∗}$$. Then I claim that the sequence $$(\varphi(p_{m})_{m \in \mathbb{N}}$$ converges in $$M^{∗}$$ to $$P$$. To check this, it suffices to observe that $$\Delta(P,\varphi(p_{m})=\lim_{n \to \infty} > d(p_{n},\varphi(p_{m})_{n}=\lim_{n \to \infty} d(p_{n},p_{m})$$.

Since $$(p_{n})_{n \in \mathbb{N}}$$ is Cauchy, this converges to zero as $$m \to \infty$$.

First of all the notation is a bit confusing since $$\varphi(p_{m})$$ is an equivalence class, but $$(\varphi(p_{m})_{n}$$ is also used as a the nth element of a representative sequence of $$\varphi(p_{m})$$. Second of all I struggle with the last sentence.

Given $$\epsilon>0$$ we need to show that $$\exists N$$ s.t. $$m \geq N \implies \Delta(P,\varphi(p_{m}))<\epsilon$$. I understand that if we can show that $$\lim_{m \to \infty} \Delta(P,\varphi(p_{m})=0$$ as a real number, then $$\Delta(P,\varphi(p_{m})=\lvert \Delta(P,\varphi(p_{m})-0\rvert<\epsilon$$ follows. But I'm really struggling with how the author tries to show this.

My thoughts are as follows:

Since $$p_{n}$$ is Cauchy, we know that for all $$\epsilon$$ there exists an $$N$$ s.t. $$m,n \geq N \implies d(p_{n},p_{m})<\epsilon$$. Now let m=N, then $$d(p_{n},p_{N})<\epsilon$$ for all $$n \geq N$$, so $$\Delta(P,\varphi(p_{N})=\lim_{n \to \infty} d(p_{n},p_{N})<\epsilon$$. This is also true if we pick any other $$m>N$$ and we'll get a similar statement as anove, but with $$N$$ replaced by $$m$$. But then we have found the desired $$N$$ s.t. $$m \geq N \implies \Delta(P,\varphi(p_{m})<\epsilon$$.

Are there any issues with this idea? To me it feels easier to prove it directly than showing that the limit is $$0$$. Any help and suggestions are welcome. Thanks!

• The notation is indeed confusing: $\varphi(p_m)$ is the class of the constant sequence $p_m$, so $\varphi(p_m)_n$ is well-defined up to the choice of a representative. Then again here the canonical representative is explicitly given so it should be clear. Commented Jan 22, 2020 at 13:30
• As for the last question, you could try and prove that in general a sequence $(x_n)_{n \geq 1} \subset (M, d)$ is Cauchy if and only if $\lim\limits_{n \to \infty} \lim\limits_{m \to \infty} d(x_n, x_m) = 0$. Commented Jan 22, 2020 at 13:31
• Yeah, I think this is the way the author does it but with $m$ and $n$ reversed. If we fix one of the variables, say $n$, then $\lim_{m \to \infty} d(x_{n},x_{m})<\epsilon$ where we view $d(x_{n},x_{m})$ as a sequence of real numbers. Now we can view this limit as a sequence of $n$ again and do the same trick again and note that the distance must be positve, so it can only be $0$. Do you agree? Commented Jan 22, 2020 at 14:25