Showing a sequence converges in product topology. I am looking at this problem. Show that the sequence $y_1=(0,0,...),y_2=(\frac{1}{2},\frac{1}{2},0,0...),y_3=(\frac{1}{3},\frac{1}{3},\frac{1}{3},...)$ converges to $0$ in the product topology and does not converge in the box topology.
I was struggling to figure this out and noticed: if I let $\prod U_i$ denote a basis element that contains $0$. Then $U_i \neq \mathbb{R}$ for finitely many $i$. So one of these $U_i's$ should contain a minimum value $w=\text{sup}\{d(0,x)|x>0\}$ out of all the other $U_i's$ such that $U_i \neq \mathbb{R}$ . So if I choose $N \in \mathbb{N}$ such that $\frac{1}{N}<w$, then $y_n \in \prod U_i$ for all $n>N$. Would this be the correct way to solve this problem?
In the box topology I know it would have to converge to $0$, since the product topology is coarser than the box. So if I take $\prod(-y_{ii},y_{ii})$ this is an open set containing $0$ and no points of $y_n$. Would this be correct?
 A: Yes, those approaches are correct: In fact $(y_n)_n$ converges in the product topology on $\Bbb R^{\Bbb N}$ iff for each coordinate $k$ the real sequence $(y_{n,k})_n$ converges in $\Bbb R$.
Or here directly: if $\prod_{i} U_i$ is a basic product open neighbourhood of $(0,0,0,\ldots)$, then $\exists N$ so that $0 \in U_i, i < N$ and $U_i = \Bbb R$ for $i \ge N$. Then for each $i < N$ we find $N_i$ so that $$\forall n > N_i: y_{n,i}=\frac{1}{n} \in U_i$$ using convergence of $\frac{1}{n} \to 0$ finitely many times. But then for $n > \max(N_i\mid i < N)$ all these conditions apply and so for those $n$: $$y_n \in \prod_i U_i$$ as required.
For the box topology we only have to observe that $$(0,0,0,0,\ldots) \in U=\prod_{n \in \Bbb N} (-\frac1n, \frac1n)$$ contains no point of this sequence at all, as the $n$-th coordinate of $y_n$ fails to be in $U_n$, for each $n$.
Indeed $(0,0,0,\ldots)$ is the only candidate for a limit in the box topology as it is already the unique limit in a coarser topology, so there is no convergence at all.
