Prove distance generates the product topology of $\mathbb{N}$ x $\mathbb{N}$ ... with discrete topology on each $\mathbb{N}$ I have that $X_i = \mathbb{N}$ for $i = 1,2...$ and let $\tau$ be the discrete topology in $\mathbb{N}$. In the set $X_1 \times X_2 \times ...$ we define the metric d:
$
d((i_1,i_2...), (j_1,j_2...)) = \begin{cases}
\frac{1}{n} \text{ if }i_n \neq j_n \text{ and } i_k = j_k \text{ for } k<n \\
0 \text{ if } i_k = j_k \text{ } \forall k
\end{cases}
$
I have to prove this metric generates the Cartesian product topology in $X_1 \times X_2 \times ...$ I really don't know how to approach this problem, as I tried imagining the open balls but haven't found the way to use it to my advantage.
 A: This is indeed a metric on $\mathbb N^ \mathbb N$ (for the triangle inequality, show that $d(x,y) \leq \max \{ d(x,z),d(z,y)\}$ ). It's not hard to see that
$$ d(x,y) = \begin{cases}  0  \text{ if } x=y\\ \frac{1}{\min{ \{ n\in \mathbb N  \colon \ x(n) \neq y(n)  \} } }
\end{cases}$$
for any $x=(x (n) )$ and $y=(y(n))$. Furthermore, a sequence $(x_n)$ converges to a point $x $, with respect to $d$ if and only if for any $k \in \mathbb N$, there exists $n_0 \in \mathbb N$ such that for $n \geq n_0$, the initial $k-$segment of $(x_n)$ is identical to that of $x$. In other words, for any $k$, it's true that   $ x_n(1) =x(1), \cdots,  x_n(k)=x(k)$ eventually.
It is known that a countable product of first countable spaces is first countable (with the product topology). In fact, we can say more: the countable product of metric spaces is a metrizable space, thus $\mathbb N^\mathbb N$ is actually metrizable with the product topology.
So, in order to show that the product topology is the same as the one generated by the metric $d$, it suffices to show that they have the same convergent sequences. To this end, let $x_n \xrightarrow{d}x $ and let $U= V_1 \times V_2 \times \cdots \times V_k \times \mathbb N^{\mathbb N \setminus \{ k\} } $ be an open set in the product topology that contains $x$. We want to show that  $x_n \in V$ eventually, or equivalently, $x_n(1) \in V_1,  \cdots, x_n(k) \in V_k$ eventually. This is true however, since by the metric convergence, $x_n(1) =x(1), \cdots ,x_n(k) =x(k)$ eventually.
On the other hand, if $x_n \to x$ in the product topology, then for any given  $ε>0$, we can find a $k \in \mathbb N$ such that $1/k<ε$ and since the set $V= \{ x(1) \} \times \cdots \times \{ x(k) \} \times \mathbb N^{\mathbb N \setminus \{k \}} $ is open in the product topology and it contains $x$, it must be true that $x_n \in V$ eventually. Hence, $x_n(1) =x(1), \cdots ,x_n(k) =x(k) $ eventually and so $ \{ m \in \mathbb N \colon \ x_n(m) \neq x(m) \} \geq k$, or equivalently, $ d(x_n,x) < ε$.

A: HINT: Let the product be $X$. Verify that if
$$x=\langle x_n:n\in\Bbb Z^+\rangle,y=\langle y_n:n\in\Bbb Z^+\rangle\in X\,,$$
then $d(x,y)<\frac1n$ if and only if $x_k=y_k$ for $k=1,\ldots,n$. Thus, if for each $x\in X$ and $n\in\Bbb Z^+$ we let
$$B_n(x)=\{y\in X:y_k=x_k\text{ for }k=1,\ldots,n\}\,,$$
then
$$B_d\left(x,\frac1n\right)=B_n(x)$$
for each $x\in X$ and $n\in\Bbb Z^+$. Let
$$\mathscr{B}=\left\{B_d\left(x,\frac1n\right):x\in X\text{ and }n\in\Bbb Z^+\right\}\,;$$
you know that $\mathscr{B}$ is a base for the metric topology on $X$, and we’ve just shown that
$$\mathscr{B}=\{B_n(x):x\in X\text{ and }n\in\Bbb Z^+\}\,.$$
so all you need to do to finish the proof is show that $\mathscr{B}$ is a base for the product topology on $X$. This requires a little work but is pretty straightforward.
