Topology on space of eventually-zero sequences Let $R^\infty = \{ (x_i)_{i=1}^\infty | \exists n : x_j = 0, \forall j \geq n \}$. Then the topology on $R^\infty$ is such that a subset $U$ is open if and only if $U \cap \mathbb{R}^n$ is open for each $n$. 
I think that if a sequence $x_n \rightarrow x$ in this topology then there is  some $m$ such that $x, x_i \in \mathbb{R}^m$ for large enough enough $n$. Here I'm thinking of $\mathbb{R}^k \subset \mathbb{R}^j$ if $k \leq j$. Then we would have $x_n \rightarrow x$ in the topology of $\mathbb{R}^m$. But I cannot see why this is true. Can someone give me a hint about how to show it? 
 A: Hint: try a contrario.
For $x \in \mathbb{R}^{\infty}$ let $h(x) = \min \{ m \in \mathbb{N} : (\forall n > m) \, x(n) = 0 \}$ so it is the smallest number $m \in \mathbb{N}$ such that $x \in \mathbb{R}^m$. 
Now suppose that $x_n \to x$ but $h(x_n)$ is unbounded. Without loss of generality $x = 0$ and (by passing to a subsequence) $h(x_n)$ strictly increases. Construct a neighborhood of $0$ such that all terms of $x_n$ lie outside.
A: I apologise for earlier, misread this as the product topology and not the box topology!
Suppose $x_n \rightarrow x$ in the box topology. I will assume $x=(0,0, \ldots)$ as we may always shift the set that we use later. Suppose that for all $k \in \mathbb{N}$ and $N \in \mathbb{N}$ there exists $n \geq N$ such that $x_n(k) \neq 0$. Define the strictly increasing sequence $(n_k)_{k = 1}^\infty \in \mathbb{N}^\mathbb{N}$ where $x_{n_k}(k) \neq 0$. We now define the open $U = \prod_{i=1}^\infty I_i$ where $I_i = (-1,1)$ if $i \neq n_k$ for all $k \in \mathbb{N}$ and $I_i = (-|x_{n_k}(k)|, |x_{n_k}(k)|)$ if $i=n_k$ and note that for all $N \in \mathbb{N}$ there exists $n_k \geq N$ such that $x_{n_k} \notin U$.
