9
$\begingroup$

Let $(X_n, d_n)$ be a sequence of metric spaces. Show that the function $ d: X \times X \to \mathbb R^+$ on the product space $X: = \prod_n X_n$ defined by

$$d ((x_n)_{n = 1}^\infty, (y_n)_{n=1}^\infty ) := \sum_{ n=1}^\infty 2^{-n} \frac{ d_n(x_n,y_n)} { 1+ d_n (x_n,y_n) } $$

is a metric on $X$ which generates the product topology on $X$.


I showed that $d$ is actually a metric, which was easy. To show that this metric generates the product topology I think I need to show:

(i) Each ball $ B((x_n)_{n=1}^\infty , \epsilon )$ is open in the product topology.

(ii) For any $B(x_n , \epsilon) \subset X_n$, $\pi_n ^{-1} (B(x_n , \epsilon)) \subset X$ is the union of finite intersections of balls in $(X, d)$.

But I couldn't even get started to do (i). Any help is appreciated.

$\endgroup$
8
$\begingroup$

Note that if $(X,d)$ is metric space, then $d'=d/(1+d)$ generates same topology of $(X,d)$. So we only prove this proposition:

Let $(X_n,d_n)$ be a sequence of metric spaces, and $d_n(x,y)\le 1$ for all $n$ and $x,y\in X_n$, then $d((x_n),(y_n))=\sum_n 2^{-n} d_n(x_n,y_n)$ generates the product topology of $X=\prod_n X_n$.

At first, we prove that for each $a=(a_n)_{n=1}^\infty\in X$ and $r>0$, there is a open basis $V$ of $X$ satisfy that $a\in V\subset B_d(a,r)$. Let $N$ be a natural number that satisfy $2^{-N}\le r/2$. Consider $$ V= B_1 (a_1,r/2)\times \cdots\times B_N (a_N,r/2)\times X_{N+1}\times X_{N+2}\times\cdots $$ (where $B_i(x,r)$ is a open ball in $X_i$.) If $x\in V$ then $d(a_i,x_i)<r/2$ for $i=1,2,\cdots, N$. Therefore $$ \begin{aligned} d(a,x)=\sum_{n=1}^\infty 2^{-n}d_n(a_n,x_n)&\le \frac{1}{2}\sum_{n=1}^N 2^{-n}r + \sum_{n>N}2^{-n}\\ &=(1-2^{-N})\cdot\frac{r}{2}+2^{-N}\\ &< \frac{r}{2}+\frac{r}{2}=r \end{aligned} $$ so $x\in B_d(a,r)$.

Finally, you prove that for each $a\in X$ and for each basis $V$ of $X$ that contaning $a$ there is $r$ satisfy that $a\in B_d(a,r)\subset V$. It is easy to prove so I leave the proof of this part for yours.

$\endgroup$
  • $\begingroup$ I think I got it! Thank you so much! $\endgroup$ – user112564 Mar 4 '14 at 22:46
  • $\begingroup$ Hello. Does this need any form of choice if you are given that the product is non-empty and is the metric you define complete? $\endgroup$ – user370967 Nov 2 '18 at 8:20
  • 1
    $\begingroup$ @Math_QED Countable choice is necessary to ensure the product $\prod_n X_n$ is non-empty in general. (Some special cases does not. For example, if every $X_n$ is same then we can find a canonical example of an element of the product.) I do not think completeness of the product requires any form of the axiom of choice. $\endgroup$ – Hanul Jeon Nov 4 '18 at 11:08
  • $\begingroup$ In the setting I worked in, we had a product of the form $X^\mathbb{N}$ so this product isn't empty. I managed to prove it was complete when $X$ is completely metrizable. Thanks for your comment. $\endgroup$ – user370967 Nov 4 '18 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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