Let $x_1,x_2, \ldots$ be a sequence of points of the product space $\prod X_\alpha$. Show that the sequence converges to the point $x$ if and only if the sequence $\pi_\alpha (x_1), \pi_\alpha (x_2)\ldots$ converges to $\pi_\alpha (x)$ for each $\alpha$. Is this fact true for box topology instead of product topology?
i found the answer here,,,,As i could not able to understand the red lines...
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$1)$Consider the product of $\Bbb N$ copies of $\{0,1\}$ (with the discrete topology). The result is the Cantor space. This is a compact metric space, and therefore it has a countable basis, and only $2^{\aleph_0}$ open sets.
Consider the box topology on the same product, then you get a discrete space of size $2^{\aleph_0}$, which therefore has $2^{2^{\aleph_0}}$ open sets, and is most certainly not compact.
$2)$Now let us consider the sequence the sequence $\mathbf{x}_n \colon= \left( \frac{1}{n } , \frac{ 1 }{ n }, \frac{ 1 }{ n }, \ldots \right)$ in $\mathbb{R}^\omega$. For each $\alpha \in \mathbb{Z}_+$, we note that $\frac{1}{n}$ converges to the point $0$ in $\mathbb{R}$. However, $\mathbf{x}_n$ does not converge to the point $\mathbf{x} \colon= ( 0, 0, 0, \ldots )$ if $\mathbb{R}^\omega$ is given the box topology, for the set $$ U \colon= \left( - \frac{1}{4}, \frac{1}{4} \right) \times \left( - \frac{1}{9}, \frac{1}{9} \right) \times \left( - \frac{1}{ 16 }, \frac{1}{ 16 } \right) \times \cdots $$ is a box topology open set in $\mathbb{R}^\omega$, and so if $\mathbf{x}_n$ belongs to $U$ for some $n \in \mathbb{N}$, then for that $n$, we would also have $\pi_n \left( \mathbf{x}_n \right) \in \pi_n (U)$, that is, $$ \frac{1}{n} \in \left( \frac{1}{ (n+1)^2 }, \frac{1}{ (n+1)^2 } \right), $$ which is not true.
$3)$This is not true if we use the box topology instead of the product topology. Consider the box topology on $\mathbb{R}^\omega$, the countably infinite cartesian product of $\mathbb{R}$ with itself. Let $$x_n = \left( \frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \ldots \right)$$
Clearly, for any $k \in \mathbb{N}$, $\pi_k(x_n) \longrightarrow 0$. However, it is not the case that $x \longrightarrow \textbf{0}$. To see this, consider the basis element $U = \displaystyle{\prod_{k=1}^\infty (-1, 1)}$ containing $\textbf{0}$. Then, for all $x_n$, $x_n \notin U$, since we can always find a $k$ such that $\pi_k(x_n) > 1$.