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Let $(X_n,d_n)_{n\in\mathbb{N}}$ is numerable family of pseudo-metric spaces. The function

$d:\prod_\limits{k\subset\mathbb{N}}^{}X_k\times\prod_\limits{k\subset\mathbb{N}}^{}X_k\to\mathbb{R}\\(x(k)_{k\subset\mathbb{N}},y(k)_{k\subset\mathbb{N}}\to\sum_{k=1}^{\infty}\frac{1}{2^k}\frac{d_k(x(k),y(k))}{1+d_k(x(k),y(k))}$

is well defined as pseudo-metric in $ \ $ $X=\prod_\limits{k\subset\mathbb{N}}^{}X_k$.

Let $(x_n)_{n\in\mathbb{N}}\in X$ be a Cauchy sequence iff for each $k\in\mathbb{N}$ the sequence $(x_n(k))_{n\in\mathbb{N}}$ is Cauchy in $(X_k,d_k)$.

Proof: If a sequence is Cauchy in $X,d$, for a given $\epsilon>0$ there exists $p(\epsilon)\in\mathbb{N}$ such as $d(x_n,x_m)\leqslant\epsilon$ if $n,m\geqslant p$, and therefore for each $k\in\mathbb{N}$

$n,m\in\mathbb{N}\:\:\:n\geqslant m\geqslant p(\epsilon)\implies\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}\leqslant 2^k\epsilon$

For $\epsilon<\frac{1}{2^{k+1}}$ we have

$n,m\in\mathbb{N}\:\:\:n\geqslant m\geqslant p(\epsilon)\implies{d_k(x_n(k),x_m(k))}\leqslant 2^{k+1}\epsilon$

Therefore the sequence $(x_n(k))_{n\in\mathbb{N}}$ is Cauchy in $(X_k,d_k)$. Suppose that for each $k\in\mathbb{N}$ th sequence $(x_n(k))_{n\in\mathbb{N}}$ is Cauchy in $(X_k,d_k)$. Let´s fix $\epsilon>0$. We choose $q=q(\epsilon)$ in a way so that $\sum_{k=q+1}^{\infty}\frac{1}{2^k}<\frac{\epsilon}{2}$. So for each $n,m\in\mathbb{N}$,

$d(x_n,x_m)=\sum_{k=1}^{q}\frac{1}{2^k}\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}+\sum_{k=q+1}^{\infty}\frac{1}{2^k}\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}<\sum_{k=1}^{q}\frac{1}{2^k}\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}+\frac{\epsilon}{2}$

Now we can choose $p(\epsilon\in\mathbb{N})$ in a way that :

$n\geqslant m\geqslant p(\epsilon)\implies d_k(x_n(k),x_m(k))<\frac{\epsilon}{2}\:\:\:\:\forall k\in\{1,...q(\epsilon)\}$.

Then

$n\geqslant m\geqslant p(\epsilon)\implies d(x_n,x_m)<\epsilon$.

I have been stuck at this proof for two days. That is I posted this lengthy proof. I would be grateful if someone clarified me the following points.

Questions:

1) I understood that in the first part of the proof $\epsilon<\frac{1}{2^{k+1}}$ was chosen with the purpose to ensure that $2^k\epsilon$ was smaller than one which would converge to 0 if we applied the infinite sum. However the author writes immediately afterwards "$n,m\in\mathbb{N}\:\:\:n\geqslant m\geqslant p(\epsilon)\implies{d_k(x_n(k),x_m(k))}\leqslant 2^{k+1}\epsilon$".

How can the author write"$n,m\in\mathbb{N}\:\:\:n\geqslant m\geqslant p(\epsilon)\implies{d_k(x_n(k),x_m(k))}\leqslant 2^{k+1}\epsilon$"? What for?

2) The author writes $\sum_{k=q+1}^{\infty}\frac{1}{2^k}<\frac{\epsilon}{2}$. How can the author be sure that by assuming $\sum_{k=q+1}^{\infty}\frac{1}{2^k}<\frac{\epsilon}{2}$, that $\sum_{k=q+1}^{\infty}\frac{1}{2^k}\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}<\frac{\epsilon}{2}$ is true?

Thanks in advance!

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  • $\begingroup$ It should exists a global bound for all the metrics. $\endgroup$
    – Math.mx
    Oct 13 '17 at 17:00
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For your question 1: The function $\frac{t}{1+t}$ is strictly increasing. If we take $t=2^{k+1} \epsilon$, then $$ \frac{2^{k+1}\epsilon}{1+2^{k+1}\epsilon}>\frac{2^{k+1}\epsilon}{1+1}=2^{k}\epsilon, $$ where we use the condition that $ 2^{k+1}\epsilon\leq 1$. So $d_k(x_{n}(k),x_{m}(k))\leq 2^{k+1}\epsilon$. Otherwise, $d_k(x_{n}(k),x_{m}(k))> 2^{k+1}\epsilon$ would imply that $\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}>2^{k}\epsilon$. A contradiction.

For your question 2: It is easy to see that $\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}\leq \frac{1+d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}= 1$. Hence, we have $$ \sum_{k=q+1}^\infty \frac{1}{2^k}\frac{d_k(x_n(k),x_m(k))}{1+d_k(x_n(k),x_m(k))}\leq \sum_{k=q+1}^\infty \frac{1}{2^k} $$

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  • $\begingroup$ Excuse me but there is one problem that I am going through when I am trying to understand point 1) which I hope you could clarify me. If you take t=2^{k+1} \epsilon, then $\frac{2^{k+1}\epsilon}{1+2^{k+1}\epsilon}>2^{k+1}\epsilon$. Imagine pick $t=5$, then $\frac{5}{1+5}>5$, which is not true once $\frac{5}{6}<1<5$. How can you have stated it only considering the fact the function is strictly increasing? $\endgroup$ Oct 14 '17 at 15:42
  • $\begingroup$ @PedroGomes Please see the update. It is a misprint, and I have make it clearer now. Hope you understand now $\endgroup$
    – Ice sea
    Oct 14 '17 at 15:46
  • $\begingroup$ Thanks! Now I am getting it. $\endgroup$ Oct 14 '17 at 16:16

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