I was showing $D(x,y) = \sup_{k \in \mathbb N} \frac{d' (x_k,y_k)}{k}$ induces the product topology on $\mathbb R^{\mathbb N}$. Here $d(x,y)$ is the standard metric on $\mathbb R$ and $d'(x,y) = \min(d(x,y), 1)$.

Next I defined $D(x,y) = \sum_{k=1}^\infty \frac{d(x,y)}{(1 + d(x,y)) 2^k}$ and I was trying to show that $D$ also induces the product topology but I am unable to prove it. How to show that $D$ induces the product topology? Thank you.

P.S.: I showed the first with $$B_D (x, \frac{1}{n}) = B_d (x_1, \frac{1}{n}) \times B_d (x_2, \frac{2}{n}) \times B_d (x_3, \frac{3}{n}) \times \dots \times B_d (x_{n-1}, \frac{n-1}{n}) \times \mathbb R \times \mathbb R \times \dots$$

is open in the product topology. And in the other direction with $U = O_1 \times \dots \times O_n \times \mathbb R \times \dots$ and $\epsilon = \min_{1 \le i \le n} (\frac{\epsilon_i}{i})$ then $B_D (x, \epsilon) \subseteq U$.

  • $\begingroup$ Can you prove that for every metric space $(X,d)$ the expression $\delta(x,y) = \frac{d(x,y)}{1+d(x,y)}$ defines a bounded metric on $X$ equivalent to the original metric? $\endgroup$ – Martin Jan 13 '13 at 17:01
  • $\begingroup$ @Martin Yes, $\delta \le 1$ and $B_d (x, \epsilon) = B_\delta (x, \frac{\epsilon}{1 + \epsilon})$. $\endgroup$ – user54938 Jan 13 '13 at 17:26
  • $\begingroup$ Next, consider the $D$-ball of radius $2^{-K+1}$. Can you find an open set in the product topology containing it? (you should be able to find a finite product of $\delta$-balls times infinitely many copies of $\mathbb{R}$) Then consider a product of finitely many $\delta$-balls (times infinitely many copies of $\mathbb{R}$). $\endgroup$ – Martin Jan 13 '13 at 17:40
  • $\begingroup$ @Martin One direction: An open set in the product topology contained in $B_D (x, \frac{1}{2^{K-1}})$ is $B_\delta (x_1, \frac{1}{2^K}) \times B_\delta (x_2, \frac{1}{2^K}) \times \dots B_\delta (x_K, \frac{1}{2^K}) \times \mathbb R \times \mathbb R \times \dots$. Next I will attempt the other direction. $\endgroup$ – user54938 Jan 13 '13 at 18:51
  • $\begingroup$ Yes. Notice that $\delta(x_k,y_k) \geq 2^{k-K}$ implies $D(x,y) \geq 2^{-K}$ (for $k \leq K$) and that the terms in the definition of $D$ with $k \geq K+1$ sum up to at most $2^{-K}$. $\endgroup$ – Martin Jan 13 '13 at 18:59

This product topology is sequential, so you can simply work with sequences. Indeed, $\mathbb{R}^\mathbb{N}$ is a countable product of metric whence first countable spaces. So the product topology on $\mathbb{R}^\mathbb{N}$ is first countable. Hence we have a sequential space. That is, the topology is characterized by its converging sequences.

Recall that the product topology is also known as the topology of pointwise convergence. A sequence $x^{(n)}=(x_k^{(n)})$ converges to $x=(x_k)$ for the product topology if and only if $x_k^{(n)}\longrightarrow x_k$ in $(\mathbb{R},|\cdot|)$ for every $k$. Therefore:

Showing that the product topology is equal to the topology induced by $D$, is equivalent to proving: $$ D(x^{(n)},x)\rightarrow 0\quad\iff\quad |x^{(n)}_k-x_k|\rightarrow 0 \quad\forall k. $$

This holds whether you take $D(x,y)=\sup_k \frac{\min \{1,d(x_k,y_k)\}}{k}$ or $D(x,y)=\sum_k\frac{d(x_k,y_k)}{(1+d(x_k,y_k))2^k}$. Note that the exact same arguments apply more generally to $X^\mathbb{N}$ for any metric space $(X,d)$. The first $D$ is a bit easier. I'll do the second one. If you already know that $d'$ is topologically equivalent to $d$, you can jump to the last paragraph.

I assume you have checked that $$ D(x,y)=\sum_{k=0}^{+\infty}\frac{d'(x_k,y_k)}{2^k}\qquad\mbox{where }\;d'(s,t)=\frac{|s-t|}{1+|s-t|} $$ defines a metric on $\mathbb{R}^\mathbb{N}$. This essentially requires to check that $d'$ is a metric on $\mathbb{R}$.

Now observe that $d'$ is topologically equivalent to the standard distance $|s-t|$. Indeed, if $|s_n-s|\longrightarrow 0$, then clearly $d'(s_n,s)\longrightarrow 0$. Conversely, assume that $d'(s_n,s)\longrightarrow 0$. First, note that $|s_n-s|$ is bounded, for otherwise there would exist a susbequence such that $|s_{n_k}-s|\longrightarrow +\infty$, whence $d'(s_{n_k},s)\longrightarrow 1$. Second, let $M$ be an upper bound and note $d'(s_n,s)\geq \frac{|s_n-s|}{1+M}$. So $|s_n-s|\longrightarrow 0$.

Let $x^{(n)}=(x_k^{(n)})$ and $x=(x_k)$ in $\mathbb{R}^\mathbb{N}$. We need to show that $D(x^{(n)},x)\longrightarrow 0$ if and only if $x^{(n)}_k\longrightarrow x_k$ in $\mathbb{R}$ for every $k$ (with respect to the usual distance on $\mathbb{R}$). By the previous paragraph, the latter is equivalent to the condition: $d'(x^{(n)}_k,x_k)\longrightarrow 0$ for every $k$. So we need only show: $$ D(x^{(n)},x)\longrightarrow 0\qquad\iff\qquad d'(x^{(n)}_k,x_k)\longrightarrow 0\quad\forall k. $$

Forward direction: First, fix $k$ and observe that $d'(x^{(n)}_k,x_k)\leq 2^k D(x^{(n)},x)$ for every $n$. This direction follows immediately.

Backward direction: The series defining $D(x,y)$ converges normally on $\mathbb{R}^\mathbb{N}\times \mathbb{R}^\mathbb{N}$, as $\sum_{k\geq 0}\frac{1}{2^k}<\infty$. So we can swap sum and limits (dominated convergence for series). In particular $$ \lim_{n\rightarrow+\infty}D(x^{(n)},x)=\sum_{k\geq 0}\frac{1}{2^k}\lim_{n\rightarrow+\infty}d'(x^{(n)}_k,x_k)=0. $$

| cite | improve this answer | |

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.