The closure and the boundary of $\mathbb{R}^{\infty}$ in $\mathbb{R}^{\mathbb{N}}$. I think that in:
Product topology: $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\mathbb{N}}$ and $\partial\mathbb{R}^{\infty}=\mathbb{R}^{\mathbb{N}}$.
Box topology: $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$ and $\partial\mathbb{R}^{\infty}=\mathbb{R}^{\infty}$.
Edit: $\mathbb{R}^{\infty}$ is defined as the set of all the real sequences $(a_n)$ with at most finite elements $a_n\not =0$. 
Edit again:
Product topology: $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\mathbb{N}}$: Take $(x_n)_{n\in \mathbb{N}}\in \mathbb{R}^{\mathbb{N}}$ and $U=\Pi_{n\in\mathbb{N}} U_n$ a basic open of $\mathbb{R}^{\mathbb{N}}$ that contains $(x_n)_{n\in \mathbb{N}}$. Define the element $(y_n)_{n\in\mathbb{N}}$ as $y_n=0$ if $U_n=\mathbb{R}$ and $y_n\in U_n$ if $U_n\not = \mathbb{R}$. There exist only a finite number  $U_n\not =\mathbb{R}$ then $(y_n)_{n\in\mathbb{N}}\in \mathbb{R}^{\infty}$. Then $\bar {\mathbb{R}^{\infty}}=\mathbb{R}^{\mathbb{N}}$. The same argument show that $\partial \mathbb{R}^{\infty}=\mathbb{R}^{\mathbb{N}}$. 
Box topology: $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$: if $(x_n)_{n\in \mathbb{N}}\not\in\mathbb{R}^{\infty} $ then there exist a subsquence $(x_{n_k})_{k\in \mathbb{N}}$ such that $x_{n_k}\not=0$ for every $k$. We can choose for each $k$ a neighbourhood $U_{n_k}$  of $x_{n_k}$ in $\mathbb{R}$ which does not contains $0$.For each of the others $x_n$ we can choose any neighbourhood $U_n$ of $x_n$ in $\mathbb{R}$. The open set $U=\Pi_{n\in\mathbb{N}}U_n$ contains $(x_n)_{n\in \mathbb{N}}$ but does not contains any element of $\mathbb{R}^{\infty}$. This prove that the complement of  $\mathbb{R}^{\infty}$ is open, then $\mathbb{R}^{\infty}$ is closed and $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$. 
$\partial\mathbb{R}^{\infty}=\mathbb{R}^{\infty}$. We have $\partial\mathbb{R}^{\infty}\subset\mathbb{R}^{\infty}$ since $\mathbb{R}^{\infty}$  is closd. Given $(x_n)_{n\in \mathbb{N}}\in\mathbb{R}^{\infty}$ and basic neighborhood $U=\Pi_{n\in\mathbb{N}}U_n$ of $(x_n)_{n\in \mathbb{N}}$. We can defined $(y_n)_{n\in \mathbb{N}}$ such that $y_n\in U_n$ y $y_n\not = 0$ for every $n\in\mathbb{N}$ sincefor each open $U_n$ of $\mathbb{R}$ there exist elements different to $0$. Then we have that  $(y_n)_{n\in \mathbb{N}}\in U$ but $(y_n)_{n\in \mathbb{N}}\not\in \mathbb{R}^{\infty}$ then $(x_n)_{n\in \mathbb{N}}\in \partial \mathbb{R}^{\infty}$. 
Am I right?
Thanks a lot!
 A: Your results are correct, and the arguments are correct or mostly correct.
In the argument for the box topology, I think it would be good to finish the $\partial \mathbb{R}^\infty = \mathbb{R}^\infty$ argument by explicitly stating that you have shown that $(x_n)$ is not an interior point of $\mathbb{R}^\infty$, and therefore it is a boundary point.
In the argument that $\overline{\mathbb{R}^\infty} = \mathbb{R}^\mathbb{N}$ in the product topology, you write "Define ... as ... and $y_n\in U_n$ if $U_n\neq \mathbb{R}$". That doesn't really define $y = (y_n)$. To define a specific $y\in U\cap \mathbb{R}^\infty$, pick a specific point in $U_n$, say $y_n = x_n$ if $U_n\neq\mathbb{R}$ (and of course, as you have, $y_n = 0$ if $U_n = \mathbb{R}$). You could also replace the word "define" there, since an arbitrary choice of $y\in U_n$ for those $n$ with $U_n\neq\mathbb{R}$ (and $y_n = 0$ otherwise) gives you an element of $U\cap \mathbb{R}^\infty$.
Then you write that "The same argument show that $\partial \mathbb{R}^\infty = \mathbb{R}^\mathbb{N}$" (in the product topology). That needs some work. What is "the same argument" here? As I see it, for that you don't use the same argument, but an analogous argument. You need to show that $\mathbb{R}^\infty$ has no interior points. Thus you start with a point $x = (x_n) \in \mathbb{R}^\infty$ and a neighbourhood $U$ of that point, and need to show that $U \not\subset \mathbb{R}^\infty$. For that, you find a point $y = (y_n) \in U\setminus\mathbb{R}^\infty$, with the analogous construction as above, just that instead of picking $y_n = 0$ for those $n$ with $U_n = \mathbb{R}$ you pick $y_n = 1$ (or whatever your favourite nonzero number is).
A: Both $\mathbb{R}^\infty$ and $(\mathbb{R}^\infty)^c$ are dense in $\mathbb{R}^\mathbb{N}$ in the product topology because given $(a_n)\in\mathbb{R}^\mathbb{N}$ we can take $\{(b_n)_m\}\subset\mathbb{R}^\infty$ and $\{(c_n)_m\}\subset(\mathbb{R}^\infty)^c$ both converging to $(a_n)$ defining them as follows:
$$(b_n)_m=(a_0,a_1,\dots,a_m,0,0,\dots)\quad (c_n)_m=(a_0+\frac{1}{m},a_1+\frac{1}{m},a_2+\frac{1}{m},\dots,a_k+\frac{1}{m},\dots)$$
and they both converge because they converge in every component, i.e. for every projection map. And so $\overline{\mathbb{R}^\infty}=\mathbb{R}^\mathbb{N}$ and $\partial\mathbb{R}^\infty=\mathbb{R}^\infty\cap(\mathbb{R}^\infty)^c$
In the box topology any element of $(a_n)\in(\mathbb{R}^\infty)^c$ has an open set containing it disjoint from $\mathbb{R}^\infty$. $$\prod_{i\in\mathbb{N}}\left] a_i-\frac{a_i}{2},a_i+\frac{a_i}{2}\right[$$
We for simplicity of notation consider $a_i>0 \quad\forall i\in\mathbb{N}$.
Now taking any countable collection of balls containing the origin $(U_i)\subset\mathbb{R}\quad 0\in U_i\quad \forall i\in\mathbb{N}$. We note that any finite subcollection of them might be changed to verify $a_i\in U_i$ where $(a_i)\in\mathbb{R}^\infty$. We see that the product of these open balls will have nonempty intersection with both $\mathbb{R}^\infty$ and $(\mathbb{R}^\infty)^c$ and knowing that $\mathbb{R}^\infty$ is closed it must contain its boundary and so must be all the set $\mathbb{R}^\infty$.
