Convergence in the Box Topology. Given the sequence in $\mathbb{R}^\omega$:
$$y_1=(1,0,0,0\ldots), y_2=(\frac{1}{2},\frac{1}{2},0,0\ldots), y_3=(\frac{1}{3},\frac{1}{3},\frac{1}{3},0,\ldots),\ldots$$
I know that it converges in the product topology because it converges pointwise to $(0,0,\ldots).$ I know that it converges in the uniform topology since it converges to $(0,0,\ldots)$ uniformly. How do I determine whether or not this sequence converges in the box topology?
 A: Let $$U=\prod_{n\ge 2}\left(-\frac1n,\frac1n\right)\;.$$ Clearly $U$ is an open nbhd of $\langle 0,0,0,\dots\rangle$; does it contain a tail of the sequence $\langle y_n:n\in\Bbb Z^+\rangle$?
A: Examples of convergent sequences: For an example of a sequence in $\mathbb{R}^{\omega}$ that does converge to $0^{\omega} := (0, 0, 0, \ldots)$ in the box topology, suppose that $\left(s^i\right)_{i=1}^{\infty}$ and $\left(t^i\right)_{i=1}^{\infty}$ are two sequences of real numbers, each of which converges to $0$. For every positive integer $i$, let
$$y^i := \left(s^i, t^i, 0, 0, 0, \ldots\right).$$
Then $\left(y^i\right)_{i=1}^{\infty}$ converges to $0^{\omega}$ in the box topology.
This example can be generalized from $2$ convergent real sequences to any finite collection of $n$ convergent real sequence (but not to an infinite collection). $\blacksquare$
In fact, the "generalized version" of the above example (involving $n$ convergent real sequences) is the archetypal example of a sequence in $\mathbb{R}^{\omega}$ that converges to $0^{\omega}$ in the box topology, in the sense that any such convergent sequence $\left(y^i\right)_{i=1}^{\infty}$ is eventually of this form  ("eventually" means after possibly ignoring at most the first $I$ points $y^1, y^2, \ldots, y^I$ of the sequence, where $I$ is some integer).
Characterizing sequence converge: Let $\left(y^i\right)_{i=1}^{\infty}$ be a sequence in $\mathbb{R}^{\omega}$ and for every $i,$ write $y^i = \left(y^i_1, \,y^i_2, \,y^i_3, \,\ldots\right)$. Then $\left(y^i\right)_{i=1}^{\infty}$ converges to $0^{\omega} := (0, 0, 0, \ldots)$ in the box topology if and only if it converges to $0^{\omega}$ in the product topology AND there exist positive integers $I$ and $N$ such that if $i > I$ and $n > N$ then $y^i_n = 0$.
To see why the integers $I$ and $N$ need to exist, suppose that they did not exist, which happens if and only if there exist two strictly increasing sequences of integers $i_1, i_2, \ldots$ and $n_1, n_2, \ldots$ such that $y^{i_k}_{n_k} \neq 0$ for all $k = 1, 2, \ldots$.
We will now define a sequence $r_1, r_2, \ldots$ of positive real numbers that will be used to define the following set:
$$U := \left(-r_1, r_1\right) \times \left(-r_2, r_2\right) \times \cdots$$
which will be an open neighborhood of $0^{\omega}$ in the box topology.
For every positive integer $n$ such that $n \not\in \left\{ n_1, n_2, \ldots \right\},$ let $r_n$ be any positive real number (say $r_n = 1$ for concreteness).
For every $k = 1, 2, \ldots,$ let $r_{n_k}$ be any positive real number strictly less than $\left|y^{i_k}_{n_k}\right|$ (say $r_{n_k} := \left|\, y^{i_k}_{n_k} \,\right| \,/\, 2$ for concreteness). So given any positive integer $k,$ $\;y^{i_k}_{n_k} \not\in \left(-r_{n_k}, r_{n_k}\right)$ because $r_{n_k} < \left|y^{i_k}_{n_k}\right|$ and consequently
$$y^{i_k} = \left(y^{i_k}_1, \,y^{i_k}_2, \,\ldots, \,y^{i_k}_{n_k}, \,\ldots\right) ~\;\not\in\;~ \left(-r_1, r_1\right) \times \left(-r_2, r_2\right) \times \cdots \times \left(-r_{n_k}, r_{n_k}\right) \times \cdots$$
which states exactly that $y^{i_k} \not\in U$ (because the Cartesian product of intervals on the right hand side is just $U$).
This shows that $y^{i_1}, y^{i_2}, y^{i_3}, \ldots$ is an infinite subsequence of $\left(y^i\right)_{i=1}^{\infty}$ that never belongs to the neighborhood $U$ of $0^{\omega}.$ Consequently, there can not exist any integer $i_0$ such that $y^{i_0}, y^{i_0+1}, y^{i_0+2}, \ldots$ all belong to $U,$ which is required (by definition) for the sequence $\left(y^i\right)_{i=1}^{\infty}$ to converge to $0^{\omega}$ in the box topology.
A: The box topology is finer than the product or the uniform topology.   That makes it harder for sequences to converge.
As a result, it's not a "categorical product", as for instance compactness isn't preserved.
We don't have  $y_n\to0\in\mathbb R^\omega $, in the box topology,  as one can construct a nbhd $U $ of $0$ which doesn't contain $y_n $ for all $n\ge N $, for any given  $N $.  (See the accepted answer.)
