Vtyjkyuk

Let $bar d(a,b)=min,1$ the standard bounded metric on $mathbb R$. Define $$D=supleftfracbar d(x_i,y_i)iright.$$

I proved it's indeed a metric on $mathbb R^mathbb N$ but in the Munkres they say that it induce the product topology, but I don't understand why. I mean, why for all $$V=prod_i=1^infty U_i$$
open where $U_i=mathbb R$ but finitely many $i$, then $V=B_D(x,varepsilon)$ for an $xin mathbb R^mathbb N$ and an $varepsilon>0$ ?

Let us fix some notation first. I will denote by $tau_d$ the topology on $mathbbR^mathbbN$ generated by the metric $D$, and by $tau_p$ the product topology on $mathbbR^mathbbN$. The first topology has as a basis the set of open balls, and the second one has as a basis the sets given by the product of open subsets of $mathbbR$, with only finitely many of the factors being proper subsets.

Your question at the end is if these basis are the same. The answer is no, and this is not a problem. We want to show that $tau_d = tau_p$, so we need to show that an element of $tau_d$ also belongs to $tau_p$, and vice-versa. This is not related to equality between bases. For example, the euclidean metric and the taxicab metric both induce the same topology on $mathbbR^k$, but an open ball in the euclidean metric cannot be given by an open ball in the taxicab metric.

Let us then show that $tau_d=tau_p$. We will show first that $tau_p subseteq tau_d$. The product topology on $mathbbR^mathbbN$ has as subbasis the sets of the form $pi_j^-1(U)$, where $U subseteq mathbbR$ is open and $pi_j:mathbbR^mathbbN to mathbbR$ is the projection onto the $j$-th factor. To show that $tau_p subseteq tau_d$, it is sufficient to show that every subbasic element of $tau_p$ belongs to $tau_d$. So, if $x=(x_i)_i in mathbbN in pi_j^-1(U)$, then $x_j in U$ and so there exists an $r>0$ such that $B(x_j,r) subseteq U$. Notice that, by replacing $r$ by a smaller one if necessary, we may assume that $r<1$. Now we claim that $B_D(x,r/j) subseteq pi_j^-1(U)$. If $y in B_D(x,r/j)$, then $D(x,y)<r/j$ and by the definition of $D$ we conclude that $overlined(x_j,y_j)/j < r/j$, so $overlined(x_j,y_j) < r$. By the choice of $r$ we made before, $overlined(x_j,y_j) = lvert x_j-y_j rvert$, so we conclude that $y_j in B(x_j,r) subset U$, i.e., $y in pi_j^-1(U)$. It follows that $B_D(x,r/j) subset pi_j^-1(U)$, so $pi_j^-1(U)$ is also open in the topology given by $D$.

For the converse, consider a ball $B_D(x,r) subset mathbbR^mathbbN$ and fix $r'<r$. Notice that if $i_0$ is the smallest index satisfying $r'i_0 geq 1$, then we have $overlined(x_i,y_i) leq r'i$ trivially for $i geq i_0$. So, intuitively, being in the ball $B_D(x,r)$ imposes conditions on the coordinates $y_i$ only for $i<i_0$. Motivated by this, consider a ball $B(y_i,varepsilon_i) subset mathbbR$ for each $i<i_0$, with $varepsilon_i=r'i-overlined(x_i,y_i)$. We claim that
$$
U := B(x_1,varepsilon_1)times dotsm times B(x_i_0-1,varepsilon_i_0-1) times mathbbR times mathbbR times dotsm subset B_D(x,r).
$$

Indeed, if $z in U$, by the observations we made above, we have $overlined(z_i,x_i)<r'i$ for $i geq i_0$. For $i<i_0$, by the choice of $varepsilon_i$, we have
$$
overlined(z_i,x_i) leq overlined(z_i,y_i)+overlined(y_i,x_i) leq r'i-overlined(x_i,y_i)+overlined(x_i,y_i) = r'i.
$$

From the calculation above we conclude that
$$
fracoverlined(z_i,x_i)i leq r' forall i in mathbbN implies D(z,x) = supleftfracoverlined(z_i,x_i)iright leq r' < r implies z in B_D(x,r).
$$

This shows that $U subset B_D(x,r)$, and since $U$ is open in $tau_p$, the ball $B_D(x,r)$ is also open in this topology. Since the balls form a basis for $tau_d$, we conclude that $tau_d subseteq tau_p$.