Show limit point of product topology and sequence converging to $0$ I have been really struggling with the product topology. May I have a solution verification and get some pointers on how I should approach these product topology questions?
Let $\mathbb{R}^{\omega}$ be the countable product of copies of $\mathbb{R}$.So every point in $\mathbb{R}^{\omega}$ is a sequence $(x_1,x_2,...)$.Let $A \subset \mathbb{R}^{\omega}$ be the set consisting of all points with positive coordinates. Show that in the product topology $0=(0,0,...)$ is a limit point of the set $A$ and there is a sequence of points in $A$ converging to $0$. Then show that in the box topology, $0=(0,0,...)$ is a limit point of the set $A$, but there is no sequence of points in $A$ converging to $0$.
Attempt:
Let $U$ be a basic open set containing $(0,0,...)$. Then $U=\prod U_i$.If $U_i \neq \mathbb{R}$ since $U_i$ is open in $\mathbb{R}$ for each $U_i \neq \mathbb{R}$ and $\epsilon_i>0$ there is an interval $(-\epsilon_i,\epsilon_i) \subset U_i$. So $U_i \cap R^{+}\neq \varnothing$. Choose $\epsilon=\text{min}\{\epsilon_i\}$. Then $(-\epsilon,\epsilon)\subset U_i$ for each $i$ and so $U \cap A \neq \varnothing$, and $(0,0,...)$ is a limit point of $A$.
Consider the sequence of points in $A$ given by $x_n=(\frac{1}{n},\frac{1}{n},...)$.Let $U=\prod U_i$ be a basic open set containing $(0,0,...)$ if $U_i \neq \mathbb{R}$ there is an $N_i \in \mathbb{N}$ such that $\frac{1}{n} \in U_i$ for each $n \geq N_i$.Choose $N=\text{max}\{N_i\}$ Then $n \geq N \implies x_n \in U$, so $x_n \rightarrow (0,0,...)$.
In the box topology $(0,0,...)$ is a limit point of $A$ since if $U=\prod U_i$ is an open set containing $(0,0,...)$ for all $i$ there is an $\epsilon>0$ such that $(-\epsilon,\epsilon) \subset U$. So $U_i \cap R^{+} \neq \varnothing$ for each $i$ and $(0,0,...)$ is a limit point of $U$.
Comment:I do not know if any of this is correct. I used some help I had learned previously to try and give this solution a shot. However I really think this is incorrect, but would like to know how it is supposed to be done properly, or if I am in the right direction. Also I do not understand why there is no sequence converging to $(0,0,...)$ in the box topology, so I omit it. Does it have anything to do with not being able to take the maximum of the $N_i's$ since there might be an infinite number of $U_i's$ that are not equal to $\mathbb{R}$
 A: Your proof that the zero sequence is a limit point of $A$ in the product topology is correct, and you correctly found a sequence in $A$ converging to the zero sequence. In proving that it does so, however, you could be clearer. Let $I=\{i:U_i\ne\Bbb R\}$, and for $i\in I$ define $N_i$ as you did; then when you take $N=\max_{i\in I}N_i$, it’s explicit that you’re taking the maximum of a finite set of positive integers.
In your argument that the zero sequence is also a limit point of $A$ in the box topology it would be better to say that for each $i$ there is an $\epsilon_i>0$ such that $(-\epsilon_i,\epsilon_i)\subseteq U_i$, to emphasize that the choice of $\epsilon_i$ really does depend on the coordinate factor involved. In fact, this is basically why in the box topology the zero sequence is not the limit of a sequence in $A$.
To show this, let $\sigma=\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ be any sequence of points in $A$, where $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle$ for each $n\in\Bbb N$. For each $k\in\Bbb N$ let $r_k=\frac12x_k^{(k)}$, and let $U_k=(-r_k,r_k)$. Let $U=\prod_{k\in\Bbb N}U_k$, and show that in the box topology $U$ is an open nbhd of the zero sequence that contains no points of the sequence $\sigma$. (The intuitive basis for this construction is that we can construct $U$ to exclude the $k$-th term of $\sigma$ on the $k$-th coordinate.)
A: (a) For the part related to product topology.
Claim 1: $0$ is an accumulation point of $A$: Clearly $\vec{0}\notin A$.
Let $U$ be an open neighborhood of $0$, then there exists open subset
$U_{i}\subseteq\mathbb{R}$ such that $U_{i}=\mathbb{R}$ excepts
for at most finitely many $i$ and $0\in\prod_{i=1}^{\infty}U_{i}\subseteq U.$
Choose $N\in\mathbb{N}$ such that $U_{i}=\mathbb{R}$ whenever $i>N$.
Since $U_{i}$ is an open neighborhood of $0$, we may choose $x_{i}\in U_{i}$
with $x_{i}>0$. Let $\vec{x}=(x_{1},x_{2},\ldots)$, then $\vec{x}\in A\cap\prod_{i=1}^{\infty}U_{i}\subseteq A\cap U$.
This shows that $A\cap U\neq\emptyset$ and hence $\vec{0}$ is an
accumulation of $A$.
Claim 2: There exists a sequence $(\vec{x}_{n})$ in $A$ such that
$\vec{x}_{n}\rightarrow0:$ For each $n$, let $\vec{x}_{n}=(\frac{1}{n},\frac{1}{n},\ldots)\in A$.
We assert that $\vec{x}_{n}\rightarrow\vec{0}$. Let $U$ be an arbitrary
open neighborhood of $0$. Then there exist open sets $U_{i}$ such
that $\vec{0}\in\prod_{i=1}^{\infty}U_{i}\subseteq U$ and $U_{i}=\mathbb{R}$
excepts for at most finitely many $i$. Choose $N$ such that $U_{i}=\mathbb{R}$
whenever $i>N$. For $i=1,\ldots,N$, $U_{i}$ is an open neighborhood
of $0\Rightarrow$ there exists $\delta_{i}>0$ such that $(-\delta_{i},\delta_{i})\subseteq U_{i}$.
Let $\delta=\min(\delta_{1},\delta_{2},\ldots,\delta_{N})>0$. Choose
$n_{0}\in\mathbb{N}$ such that $\frac{1}{n_{0}}<\delta$. Clearly,
for any $n\geq n_{0}$, $\vec{x}_{n}\in\prod_{i=1}^{\infty}U_{i}$.
It follows that $\vec{x}_{n}\rightarrow\vec{0}$.

(b) For the part related to box topology.
Claim 3: $0$ is an accumulation point of $A$ (with respect to the
box topology): Clearly $\vec{0}\notin A$. Let $U$ be an open neighborhood
of $0$, then there exist open sets $U_{i}$ such that $\vec{0}\in\prod_{i=1}^{\infty}U_{i}\subseteq U.$
For each $i$, $U_{i}$ is an open neighborhood of $0$, so we can
choose $x_{i}>0$ such that $x_{i}\in U_{i}$. Define $\vec{x}=(x_{1},x_{2}\ldots)$,
then $\vec{x}\in A\cap\prod_{i=1}^{\infty}U_{i}\subseteq A\cap U$.
This shows that $A\cap U\neq\emptyset$ and hence $\vec{0}$ is an
accumulation point of $A$.
Claim 4: There is no sequence in $A$ that converges to $\vec{0}$
(with respect to the box topology): Prove by contradiction. Suppose the contrary that there exists a sequence $(\vec{x}_{n})$ in $A$ such that $\vec{x}_{n}\rightarrow\vec{0}$
with respect to the box topology. For each $n$, denote $\vec{x}_{n}=(x_{n1},x_{n2},\ldots)$.
For each $i=1,2,\ldots$, define $U_{i}=(-\frac{1}{2}x_{ii},\frac{1}{2}x_{ii})$.
Clearly $U_{i}$ is an open neighborhood of $0$, so $U:=\prod_{i=1}^{n}U_{i}$
is an open neighborhood of $\vec{0}$ (of course with respect to the
box topoogy). For that open neighborhood, there exists $N$ such that
$\vec{x}_{n}\in U$ whenever $n\geq N$. In particular $\vec{x}_{N}\in U$.
Hence, $\vec{x}_{N}(N)\in U_{N}$, i.e., $x_{NN}\in(-\frac{1}{2}x_{NN},\frac{1}{2}x_{NN})$,
which is a contradiction.
