Verifying that a map is a metric inducing the product topology on $X$ Given two distinct configurations $x, y ∈ X$, let #$(x, y)$ denote the smallest $k ∈ \mathbb N$ such that $x_k \not= y_k$.
Let
$β : \mathbb N → \mathbb R$ be a decreasing function with $lim_{n→∞} β(n) = 0$.
Verify that the map $d : X × X → R$ defined by
$d(x, y) :=$$
β($#$(x, y))$ if $x \not= y$, and $0$ if $x = y$, is a metric inducing the product topology on X.
My attempt:
$d$ is a metric iff
$1) d(x,y) \geq 0$ and $d(x,y) = 0$ iff $x=y$.
$2) d(x,y) = d(y,x)$
$3) d(x,y) + d(y,z) \geq d(x,z)$
$1)$ is trivial by the definition of the map $d$.
If $x=y$, then $d(x,y) = 0$, and if $d(x,y) = 0$,then $x=y$.
And if $x \not= y$, then $d(x,y) = β($#$(x, y)) > 0$ since $lim_{n→∞} β(n) = 0$.
$2) d(x,y) = β($#$(x, y))$ and $d(y,x) = β($#$(y,x)) = β($#$(x, y)) = d(x, y)$
$3) d(x, y) + d(y, z) = β($#$(x, y)) + β($#$(y, z)) = k + k_1$
And $d(x,z) = β($#$(x,z)) = k_2$. Hence $k + k_1 \geq k_2$ since $k, k_1, k_2$ are the smallest natural numbers.
 A: There are some problems.
In your proof of $(1)$, the fact that $\beta\big(\#(x,y)\big)>0$ does not follow from the fact that $\lim_{n\to\infty}\beta(n)=0$; it follows from the fact that the sequence $\langle\beta(n):n\in\Bbb N\rangle$ is strictly decreasing with limit $0$, so that every $\beta(n)$ is positive. The argument is that $\{n\in\Bbb N:x_n\ne y_n\}\ne\varnothing$ if $x\ne y$, so there is a smallest $k\in\Bbb N$ such that $x_k\ne y_k$, and then $d(x,y)=\beta(k)>0$.
The proof of $(2)$ is correct, though it would be better if you said explicitly that $\#(x,y)=\#(y,x)$: that’s the fact that actually makes it work.
What you’ve written for $(3)$ simply doesn’t make sense. On the one hand you apparently intend that $k=d(x,y)$, $k_1=d(y,z)$, and $k_2=d(x,z)$, but on the other hand you say that $k,k_1$, and $k_2$ are natural numbers. These statements are inconsistent: the distances are numbers $\beta(n)$ for some $n\in\Bbb N$, and those number certainly don’t have to be natural numbers. Indeed, for large enough $n$ they cannot be natural numbers, since they are positive and approach $0$ as a limit. It appears that you are confusing $d(x,y)$ and $\#(x,y)$ here. Proving that $d$ satisfies the triangle inequality is going to take some actual work, like this.

If $d(x,z)=0$, then certainly $d(x,z)\le d(x,y)+d(y,z)$ no matter what $y$ is, so suppose that $d(x,z)\ne 0$, and let $k=\#(x,z)$. If $y=x$ or $y=x$, then $d(x,y)+d(y,z)=d(x,z)$, and the triangle inequality is satisfied, so suppose that $x\ne y\ne z$, let $k_x=\#(x,y)$, and let $k_z=\#(y,z)$; we want to show that $\beta(k)\le\beta(k_x)+\beta(k_z)$. Suppose that $k<k_x$; then $x_n=y_n$ for all $n\le k$. However, $\#(x,z)=k$, so $x_n=z_n$ for all $n<k$, but $x_k\ne z_k$. This shows that $y_n=x_n=z_n$ for all $n<k$, but $y_k=x_k\ne z_k$, so $k_z=\#(y,z)=k$. Thus, $$\begin{align*}d(x,z)&=\beta(k)\\&\le\beta(k_x)+\beta(k)\\&=\beta(x_k)+\beta(x_z)\\&=d(x,y)+d(y,z)\,,\end{align*}$$ as desired.

Once all of this is done, you know that $d$ is a metric on $X$, but you have not yet shown that it induces the product topology on $X$. In order to do that, you should show two things:

*

*if $x\in X$ and $\epsilon>0$, there is an open set $U$ in the product topology such that $x\in U\subseteq B(x,\epsilon)$, where $B(x,\epsilon)=\{y\in X:d(x,y)<\epsilon\}$; and

*if $U$ is an open set in the product topology, and $x\in U$, then there is an $\epsilon>0$ such that $B(x,\epsilon)\subseteq U$.

