In which of the three topologies does the following sequence converge? Can someone please verify my proof or offer suggestions for improvement?
Notations:
$d(x, y) = |x-y|$: Standard metric on $\mathbb{R}$
$\bar d(x, y) = \operatorname{min}\{1, d(x, y)\}$: Standard bounded metric on $\mathbb{R}$
$\bar \rho (x, y) = \operatorname{sup}\{\bar d(x_i, y_i): 1 \leq i \leq n\}$: Uniform metric on $\mathbb{R}^\omega$
$B_d(x, \epsilon) = \{y: d(x, y) < \epsilon\}$: $\epsilon$-Ball centered at $x$.

Consider the product, uniform, and box topologies on $\mathbb{R}^\omega$, the countably infinite cartesian product of $\mathbb{R}$ with itself. In which of the three topologies does the following sequence converge?
  \begin{eqnarray}
w_1 &=& (1,1,1,1, \ldots) \\
w_2 &=& (0,2,2,2, \ldots) \\
w_3 &=& (0,0,3,3, \ldots) \\
&\ldots&
\end{eqnarray}

Let $U = \displaystyle{\prod_{i=1}^\infty U_i}$ be a basis element of the product topology on $\mathbb{R}^\omega$ containing $(0,0,0, \ldots)$.
Let $a_1, a_2, \ldots, a_n$ be the indices for which $U_i \neq \mathbb{R}$. Then, for all $k > \operatorname{max}\{a_1, \ldots a_n\}$, $w_k \in U$. Therefore, $(w_n)$ converges to $(0,0,0, \ldots)$ in the product topology.
Now, suppose for the sake of contradiction that $(w_n)$ converges to some point $x$ in the box topology. Let $U = \displaystyle{\prod_{i=1}^\infty U_i}$ be a basis element for the box topology on $\mathbb{R}^\omega$ such that $x \in U$. Suppose $(0,0,0, \ldots) \notin U$. Let $\alpha$ be an index for which $0 \notin U_\alpha$. Then, clearly, for all $n > \alpha, w_n \notin U$. So, it must be the case that $(0,0,0, \ldots) \in U$. Since $\mathbb{R}^\omega$ is Hausdorff under the box topology, the only point to which $(w_n)$ can possibly converge is $(0,0,0, \ldots)$. However, this is not the case. To see this, set $S = \displaystyle{\prod_{i=1}^\infty (-1, 1)}$. Clearly, $w_k \notin S$ for all $k \in \mathbb{N}$. So, it cannot be the case that $(w_n)$ converges in the box topology.
Now, suppose for the sake of contradiction that $(w_n)$ converges to some element $x = (x_n)_{n \in \mathbb{N}}$ of $\mathbb{R}^\omega$ in the uniform topology. If there exists an index $\alpha$ such that $x_\alpha \neq 0$, then for all $n > \alpha$, $w_n \notin B_{\bar \rho}(x, \operatorname{min}(1, |x_\alpha|))$. Therefore, the only point to which $(w_n)$ may converge is $(0, 0, 0, \ldots)$. However, it is clearly not the case that $(w_n)$ converges to $y = (0, 0, 0, \ldots)$. To see this, note that $w_n \notin B(y, \frac{1}{3})$ for all $n \in \mathbb{N}$, since there always exists an index $\beta$ such that $\bar d(w_\beta, 0) = 1$. Therefore, $(w_n)$ does not converge in the uniform topology.
 A: Your arguments are fine. They can, however, be simplified. Let $\tau_0,\tau_1$, and $\tau_2$ be the product, uniform, and box topologies, respectively, and note that $\tau_0\subseteq\tau_1\subseteq\tau_2$. Suppose that $\langle w_n:n\in\Bbb Z^+\rangle$ converges to some $p$ in $\tau_i$, where $i\in\{1,2\}$; then clearly $\langle w_n:n\in\Bbb Z^+\rangle$ converges to $p$ in $\tau_0$ as well. Thus, once you’ve shown that $\langle w_n:n\in\Bbb Z^+\rangle$ converges to the zero sequence $z$ in $\tau_0$, you can use the fact that all three topologies are Hausdorff to conclude immediately that $z$ is the only possible limit of $\langle w_n:n\in\Bbb Z^+\rangle$ in $\tau_1$ and $\tau_2$. Then prove, as you did, that $\langle w_n:n\in\Bbb Z^+\rangle$ does not in fact converge to $z$ in $\tau_1$, and immediately conclude (since $\tau_2\supseteq\tau_1$) that it cannot converge to $z$ in $\tau_2$ either.
(Of course you may well feel that you get more insight into the topologies by working within each one separately.)
