Compare the final (weak) topology and the box topology on $\mathbb{R}^\infty$ $\mathbb{R}^\infty$ is the space of sequences of real numbers $(r_1,r_2,\ldots)$ that are eventually $0$. That is, there's an $N$ so that $r_n=0$ for all $n>N$. We can consider this as a "limit" $\displaystyle\lim_{n \rightarrow \infty} \mathbb{R}'^n$, where $\mathbb{R}'^n$ is the space of sequences of real numbers $(r_1,r_2,\ldots)$ such that $r_m=0 \forall m>n$ (it's basically $\mathbb{R}^n$ but every point is given infinitely many $0$ coordinates). We endow $\mathbb{R}'^n$ with the topology it inherits from $\mathbb{R}^n$.
Notice $\mathbb{R}^\infty = \displaystyle\cup_{n=1}^{\infty} \mathbb{R}'^n$, and endow  it with the following topology: $U$ is open iff $U \cap \mathbb{R}'^n$ is open in in $\mathbb{R}'^n$ for all $n$. This is called the weak, final, or coherent topology on $\mathbb{R}^\infty$.
On the other hand,we can consider $\mathbb{R}^\infty$ as a subspace of $\mathbb{R}^\omega$, inheriting the box topology. $\mathbb{R}^\omega$ is simply $\displaystyle\prod_{n=1}^\infty \mathbb{R}$.
Could you help me compare these two topologies? Is one finer than the other?
Thank you for your help.
 A: Let $U=\prod_{m=1}^\infty (a_m,b_m)\cap \mathbb R^\infty$ be an arbitrary set  of a canonical base of the box topology. Since for each $n$ the intersection $U\cap\mathbb R’^n$ is $\prod_{m=1}^n (a_m,b_m) \times\prod_{m=n+1}^\infty \{0\}$ iff $a_m<0<b_m$ for each $m>n$ and is empty, otherwise, we see that the set $U$ is open in $\mathbb  R^\infty$. Thus the final topology is stronger than the box topology. 
A: 
Let $U=\prod_{m=1}^\infty (a_m,b_m)\cap \mathbb R^\infty$ be an arbitrary set  of a canonical base of the box topology. Since for each $n$ the intersection $U\cap\mathbb R’^n$ is $\prod_{m=1}^n (a_m,b_m) \times\prod_{m=n+1}^\infty \{0\}$ iff $a_m<0<b_m$ for each $m>n$ and is empty, otherwise, we see that the set $U$ is open in $\mathbb  R^\infty$. Thus the final topology is stronger than the box topology.

This shows that the box topology is contained in the final topology. Nevertheless the other containment can also be proven, thus the final and box topology in this case are equal. Indeed: 
Recall that $\mathbb R’^n$ has the topology induced by $\mathbb{R^n}$ thus the open sets of  $\mathbb R’^n$ have the form $\prod_{i=1}^n O_i \times\prod_{i=n+1}^\infty \{0\}$ where the $O_i, $ are open sets of $\mathbb{R}$ for $i=1,2,...,n$. 
Let $U$ be open in the the final topology of $\mathbb R^\infty$. Thus $U\cap\mathbb R’^n$ is open in $\mathbb R’^n$ for all $n$. 
For $n=k$ we get that $U\cap\mathbb R’^k$ is open. This together with the form of the open sets in $\mathbb R’^k$ discussed above implies that the projection of $U$ into the $k$ coordinate is an open set of $\mathbb{R}$. which implies that $U=\prod_{i=1}^\infty O_i$ where the $O_i, $ are open sets of $\mathbb{R}$ for $i=1,2,...$. 
Thus $U$ is in the box topology of $\mathbb{R^\infty}$ 
