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.


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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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

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