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Let $(X_n, d_n)$ be a family of complete metric spaces. Prove that $(X,d)$ is a complete metric space.

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

Furthermore let $(\gamma_n)_{n \in \mathbb{N}} \subset (0, \infty)$ be a seq. s.t $\sum_{n=1}^{ \infty} \gamma_n < \infty$. For $x,y \in X$ where $x = (x_n)_{n \in \mathbb{N}}$, $y = (y_n)_{n \in \mathbb{N}}$ 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 $d_n$ is the metric on $X_n$

This is what I have come up with.

Let $(x^m)_{m \in \mathbb{N}}$ be Cauchy seq in $X$. Thus we know that $(x_n^m)_{m \in \mathbb{N}}$ is Cachy in $X_n$ (I proved that earlies problem). Hence $(x_n^m)_{m \in \mathbb{N}}$ converges to some $x_n \in X_n$. Then

$$ d(x_n^m, x_n) = \sum_{n=1}^{\infty} \gamma_n \frac{d_n(x_n^m, x_n)}{1 + d_n(x_n^m,x_n)} < \sum_{n=1}^{\infty} \gamma_n \frac{\epsilon}{1 + \epsilon} = \frac{\epsilon}{1 + \epsilon} \sum_{n=1}^{\infty} \gamma_n < \frac{\epsilon}{1 + \epsilon} \cdot M < \epsilon \cdot M $$

Where $M = \sum_{n=1}^{\infty} \gamma_n$ since $\sum_{n=1}^{\infty} \gamma_n < \infty$

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The step $$\tag{*} \sum_{n=1}^{\infty} \gamma_n \frac{d_n\left(x_n^m, x_n\right)}{1 + d_n\left(x_n^m, x_n\right)} < \sum_{n=1}^{\infty} \gamma_n \frac{\epsilon}{1 + \epsilon}$$ need to be more detailed; in particular, we have to precise for which $m$'s it holds. It seems that you use the fact that $d_n\left(x_n^m, x_n\right) \lt\varepsilon$ for $m$ large enough, but the "large enough" may depend on the considered $m$. Nevertheless, an inequality similar to (*) does hold for $m$ large enough. Indeed, fix a positive $\epsilon$. There exists $M$ such that if $m,m'\geqslant M$, then $d\left(x^m,x^{m'} \right) \lt\epsilon$. We thus have for any fixed $N$ and $m,m'\geqslant M$ that $$\sum_{n=1}^{N} \gamma_n \frac{d_n\left(x_n^m, x_n^{m'} \right)}{1 + d_n\left(x_n^m, x_n\right)} \lt \epsilon.$$ In particular, letting $m'$ going to infinity, we get for any $N\geqslant 1$ $$\sum_{n=1}^{N} \gamma_n \frac{d_n\left(x_n^m, x_n\right)}{1 + d_n\left(x_n^m, x_n\right)} \leqslant \epsilon .$$ This implies by letting $N$ going to $+\infty$ that $d\left(x^{(m)},x\right)\leqslant \varepsilon$ for each $m\geqslant M$.

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  • $\begingroup$ Since $X_n$ is complete, I know that every Cauchy seq. converge. Thus if $x_n$ is the limit of the Cauchy seq $(x_n^m)$ then I know that there exists $n_0 \in \mathbb{N}$ s.t $\forall m > n_0 \Rightarrow d_n(x_n^m,x_n) < \epsilon$. That is how I resonead. Not just "large enough". $\endgroup$ – Olba12 Nov 17 '16 at 21:23
  • $\begingroup$ But this $n_0$ may be dependent on $n$. $\endgroup$ – Davide Giraudo Nov 17 '16 at 21:27
  • $\begingroup$ Let $p_1, p_2, \dots \in \mathbb{N}$ be those numbers such that if $m_i>p_i$ then $d_i1(x_1^{m_1},x_1), d_2(x_2^{m_{2}}, x_2) \dots < \epsilon$, then let $m = \text{max}\{m_i\}$? $\endgroup$ – Olba12 Nov 17 '16 at 21:31
  • $\begingroup$ Sorry I missed in "if $m, m' \geq M$, then $d(x^m,x^{m'}) < \epsilon$" that it was $d$ and not $d_n$! $\endgroup$ – Olba12 Nov 17 '16 at 21:34
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    $\begingroup$ It is in general not trivial to take a limit into an infinite sum, but we can do this into a finite sum. $\endgroup$ – Davide Giraudo Nov 17 '16 at 21:40

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