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 $\varphi(M)=M^{∗}$.


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!

  • $\begingroup$ 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. $\endgroup$
    – frafour
    Commented Jan 22, 2020 at 13:30
  • $\begingroup$ 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 $\lim\limits_{n \to \infty} \lim\limits_{m \to \infty} d(x_n, x_m) = 0$. $\endgroup$
    – frafour
    Commented Jan 22, 2020 at 13:31
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Jan 22, 2020 at 14:25


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