Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X_n, d_n)$ be a countable family of metric spaces. Show that the distance function on the product topology defined by $$d\big( (x),(y) \big) = \sum_{n\geqslant 1} 2^{-n} \frac{d_n(x_n,y_n)} {1+d_n(x_n,y_n)}$$ is complete, if all spaces $(X_n, d_n)$ are complete.

share|improve this question
2  
That’s pretty straightforward; what have you tried? –  Brian M. Scott Nov 19 '12 at 7:49
    
I first looked at every "coordinate" to get a coordinate-wise cauchy sequences. From this I defined $$x^*$$. Then I split the sum to get a finite one and an arbitrary small one. But how do I now that $$x^*$$ converge to something in the product? –  Johan Nov 19 '12 at 8:34

1 Answer 1

up vote 1 down vote accepted

Here’s a guide; all you have to do is fill in the details.

Start with a Cauchy sequence in $X=\prod_nX_n$, say $\sigma=\langle x^n:n\in\Bbb N\rangle$, where $x^n=\langle x_k^n:k\in\Bbb N\rangle\in X$. Use the fact that $\sigma$ is $d$-Cauchy to show that $\langle x_k^n:n\in\Bbb N\rangle$ is $d_k$-Cauchy in $X_k$ for each $k\in\Bbb N$. Each $X_k$ is complete, so $\langle x_k^n:n\in\Bbb N\rangle\to y_k$ for some $y_k\in X_k$. Let $y=\langle y_k:k\in\Bbb N\rangle\in X$, and show that the sequence $\sigma$ converges to $y$ in $X$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.