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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 $B_D (x, \epsilon) \subseteq U$.

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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? –  Martin Jan 13 '13 at 17:01
    
@Martin Yes, $\delta \le 1$ and $B_d (x, \epsilon) = B_\delta (x, \frac{\epsilon}{1 + \epsilon})$. –  user54938 Jan 13 '13 at 17:26
    
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 $\mathbb{R}$). –  Martin Jan 13 '13 at 17:40
    
@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. –  user54938 Jan 13 '13 at 18:51
    
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 $2^{-K}$. –  Martin Jan 13 '13 at 18:59
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1 Answer

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 $$ \lim_{n\rightarrow+\infty}D(x^{(n)},x)=\sum_{k\geq 0}\frac{1}{2^k}\lim_{n\rightarrow+\infty}d'(x^{(n)}_k,x_k)=0. $$

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