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Prove that $(X,d)$ is complete iff every $(X_n,d_n) \ n \in \mathbb{N}$ is complete.

Let $(X_n, d_n) \ n \in \mathbb{N}$ be a family of metric spaces. Define

$$ X = \prod_{n=1}^{\infty} X_n = \{(x_n)_{n \in \mathbb{N}} : x_j \in X_j\} $$

Furthermore let $(\gamma_n)_{n \in \mathbb{N}} \subset (0, \infty)$ be a sequence s.t $\sum_{n=1}^{\infty} \gamma_n < \infty$. Let $x,y \in X$ where $x = (x_n)_{n \in \mathbb{N}}$, $y = (y_n)_{n \in \mathbb{N}}$ and define

$$ d(x,y) = \sum_{n=1}^{\infty} \gamma_n \frac{d_n(x_n,y_n)}{1 + d_n(x_n,y_n)} $$ where $d(x,y)$ is a metric on $X$.

I need a hint on how to get started on this problem.

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Olba12
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2 Answers2

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$(x_n)$ is Cauchy in $(X,d)\implies d(x_n,x_m)<\epsilon \forall n,m\ge p$

$\implies \sum_{n=1}^\infty \gamma_n\dfrac{d_n(x_m,x_n)}{1+d_n(x_n,x_m)} <\epsilon$

$\implies \lim_{n\to \infty }\gamma _n\dfrac{d_n(x_m,x_n)}{1+d_n(x_n,x_m)} =0$ (If the series $\sum a_n$converges then $\lim a_n=0$)

Hence $\gamma _n\dfrac{d_n(x_m,x_n)}{1+d_n(x_n,x_m)} <\epsilon$

Since $\gamma_n$ is convergent so $\dfrac{d_n(x_m,x_n)}{1+d_n(x_n,x_m)} <\epsilon$

Now the function $\frac{x}{1+x}$ is increasing so $d_n(x_m,x_n)<\epsilon$

Can you take it now?

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  • Can you explain in more detail the part "$\gamma_n $ is convergent so..."? – Olba12 Nov 17 '16 at 12:24
  • The choice of $m$ makes it all confusing for me since, $d$ is defined to be a sum over $n$ – Olba12 Nov 17 '16 at 12:50
  • Since the series $\sum \gamma_n$ is convergent so $\lim \gamma_n=0$ and hence $\gamma_n$ is bounded so whenever the product of $\gamma_n $ and that $\dfrac{d_n}{1+d_n}<\epsilon \implies \dfrac{d_n}{1+d_n}<\epsilon$ – Learnmore Nov 17 '16 at 13:53
  • you will have to fill up the rest – Learnmore Nov 17 '16 at 13:56
  • I dont see it, looking at $0.1 \cdot 10 < 2 \not \Rightarrow 10 < 2$. So if $\gamma_n = 0.1$ how can we conclude that $\frac{d_n}{1+d_n} < \epsilon$. Must we no atleast say $\frac{d_n}{1+d_n} < \frac{\epsilon}{\gamma_n}$? – Olba12 Nov 17 '16 at 17:22
  • Neither did I mean nor did I do it.I said if for two sequences $a_n,b_n$ you have $|a_nb_n|<\epsilon \forall \epsilon >0$ and $|a_n|\le M$ ,thencan't you conclude that $|b_n|<\epsilon$? – Learnmore Nov 18 '16 at 01:59
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Hint: You can prove that $\{ x^k \}_{k \in \mathbb{N}} \subset X $ is Cauchy sequence iff $\{ x^k_n\}_{k \in \mathbb{N}} \subset X_n$ is Cauchy sequence for every $n \in \mathbb{N}$.

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