Convergence in the box topology A sequence in $R^{\omega}$ converges in the box topology iff it converges for each coordinate and is eventually constant in all but finitely many of the coordinates.
I know for the direction $\Rightarrow$ we can use the fact that since $\pi_{\alpha}$ is continous so $\pi_{\alpha}(x_n) \rightarrow \pi_{\alpha}(x)$, but how can I show it is eventually constant in all but finitely many of the coordinates? And what about the other direction $\Leftarrow$?
 A: I have edited my answer to reflect the fact that I believe I misinterpreted the condition "$(x_n)$ is eventually constant in all but finitely many coordinates." This condition means, properly, that there exists some $N$ large enough such that for all $n > N$, there are only finitely many coordinates $\alpha$ for which $\pi_\alpha(x_n) \neq \pi_\alpha(x)$. This is somewhat different than the statement that there are only finitely many coordinates $\alpha$ such that $\pi_\alpha(x_n)$ is not eventually constant, because we are also assuming that we can find a single $N$ which works for almost all coordinates.
Now, suppose that it is not the case that all but finitely many coordinates of $(x_n)$ are eventually constant. This means that for every $N$, there exist infinitely many coordinates $\alpha$ such that $\pi_\alpha(x_n) \neq \pi_\alpha(x)$ for some $n > N$. In particular, there is a coordinate $\alpha_1$ so that $\pi_{\alpha_1}(x_n) \neq \pi_{\alpha_1}(x)$ for some $n > 1$. Inductively, for every $i = 1, 2, 3, \ldots$, we can always find a coordinate $\alpha_i$ such that $\alpha_i \neq \alpha_j$ for all $j < i$, and $\pi_{\alpha_i}(x_n) \neq \pi_{\alpha_i}(x)$ for some $n > i$. Here we are making use of the hypothesis that there are infinitely many coordinates which eventually differ from $x$ to make sure we can find a new coordinate $\alpha_i$. Let $\epsilon_i = |\pi_{\alpha_i}(x_n) - \pi_{\alpha_i}(x)|/2$.
Now let $R$ be the open rectangle such that $\pi_{\alpha_i}(R) = (\pi_{\alpha_i}(x) -\epsilon_i, \pi_{\alpha_i}(x) + \epsilon_i)$ for $i=1, 2, \ldots$, and $\pi_\alpha(R) = \Bbb R$ for all other coordinates $\alpha$. Then $R$ is an open set in the box topology which contains $x$ but which does not contain any tail of the sequence $(x_n)$. In particular, for any $N>0$, by construction there is an $n > N$ such that $\pi_{\alpha_N}(x_n) \notin (\pi_{\alpha_N}(x) -\epsilon_N, \pi_{\alpha_N}(x) + \epsilon_N)$, so $x_n \notin R$.
(My thought process here was to find something akin to the diagonalization argument you use to prove that $|\Bbb R| > |\Bbb N|$.)
For the other direction, remember that the box topology has a basis consisting of open rectangles, so given a sequence which is eventually constant in all but finitely many coordinates and which converges in the other coordinates, show that any open rectangle centered at the limit point contains a tail of the sequence (this is straightforward).
