metrics of Product space

Question

Let for each $i\in\mathbb{N}$ $(X_{i},\rho_{i})$ be discrete metric spaces, the metrics defined on product space $\prod_{i\in\mathbb{N}}X_{i}$ \ $d(x,y)=\sum_{i=1}^{\infty}\frac{\rho_{i}(x_{i},y_{i})}{2^{i}}$ and $d^*(x,y)=\frac{1}{n+1}$ with $n=min\{i:x(i)\neq y(i)\}$ are equivalent I tried so hard for Days but I could not proceed it, i guess special choices is required, any hint or help would be appreciated

Answer 1

In this answer I gave full details as to why the sum metric $d(x,y)=\sum_{i=1}^\infty \frac{\rho_i(x_i,y_i)}{2^i}$ induces the product topology on $\prod_i X_i$ (under the mild assumption that the $\rho_i$ are bounded by $1$ to ensure convergence of the series; this is true for all metric spaces generally).

It's also not too hard to see that $d^\ast(x,y)=\frac{1}{n+1}$ with $n = \min\{i\mid x_i \neq y_i\}$ induces the product topology too if each $X_i$ has the discrete topology, as is the case here:

Let $O$ be product open and $x \in O$. We need to show that $O$ contains a $d^\ast$-ball around $x$, so that $O$ is open in the $d^\ast$-metric too.

First we find a basic open set of the product topology, which is of the form $\prod_i O_i$ where all $O_i$ are open in $X_i$ and $F = \{i\mid O_i \neq X_i\}$ is finite, such that $x \in \prod_i O_i \subseteq O$. This is just by definition of the product topology.

Let $n_0 = \max F \in \Bbb N$. Note that for any $y \in \prod_i X_i$: if $d^\ast(x,y) < \frac{1}{n_0+1}$ then we know that $y_i = x_i$ for all $i \le n_0$ (otherwise $i_0 := \min\{i\mid x_i \neq y_i\} \le n_0$ and $d^\ast(x,y)=\frac{1}{i_0+1} \ge \frac{1}{n_0+1}$, which is not the case). And this in turn means that $x_i = y_i$ on all $i \in F$, so:

$$x \in B_{d^\ast}(x,\frac{1}{n_0+1}) \subseteq \prod_i O_i \subseteq O$$

which thus shows that $O$ is $d^\ast$-open.

If $O$ is $d^\ast$-open in $\prod_i X_i$ we need to show it is product open as well: let $x \in O$ and find $r>0$ such that $B_{d^\ast}(x,r) \subseteq O$. Also find $n \in \Bbb N$ such that $\frac{1}{n} < r$. Now define $O'=\prod_i O_i$ by $O_i = X_i$ when $i > n$ and $O_i = \{x_i\}$ for $i \le n$. This is a basic open subset of the product topology on $\prod_i X_i$ when all $X_i$ are discrete: all singleton sets in $X_i$ are indeed open and all but finitely many coordinates have trivial coordinate set $X_i$.

Now, as before, if $x_i = y_i$ for $i \le n$ we know that $i_0 = \min\{i\mid x_i \neq y_i\} > n$ and so $d^\ast(x,y) = \frac{1}{i_0+1} \le \frac{1}{n} < r$, hence:

$$x \in O' \subseteq B_{d^\ast}(x,r) \subseteq O$$

which shows that $x$ is an interior point of $O$ in the product topology and as $x$ was arbitrary, $O$ is product open.

So we have $\mathcal{T}_{d^\ast} = \mathcal{T}_{\text{prod}} = \mathcal{T}_d$ and both metrics induce the same topology, namely the natural product topology on $\prod_i X_i$.