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I have this exercise:

Let $X_i, i\in I$ be non-empty topological spaces with I some index set. Let $(x_n)_{n\in\mathbb{N}}$ be a sequence of points in the product space $\prod_{i \in I}X_i$. Let x be another point in $\prod_{i \in I}X_i$. For every $i\in I$ let $\pi_{X_i}:\prod_{j\in J}X_j\to X_i$ denote the projection onto the the $X_i$ component.

I have to show that $(x_n)_{n\in\mathbb{N}}$ converges to x (wrt. the product topology) if and only if for every $i\in I$ the sequence $(\pi_{X_i}(x_n))_{n\in\mathbb{N}}$ converges to $\pi_{X_i}(x)$ in $X_i$. I have shown $\Rightarrow$, but I have a hard time showing $\Leftarrow$. I am thinking that I have to show that for every neighborhood U of $\pi_{X_i}(x)$ there exists an $N\in\mathbb{N}$, such that for all $n\geq N$ we have that $\pi_{X_i}(x_n)\in U$, but I am not sure how to do so.

Furthermore I have to show which one of the directions holds when we equip $\prod_{i \in I}X_i$ with the box topology, and then find an example to show that the other direction does not always hold when we use the box topology. I am thinking that it's the direction $\Rightarrow$ which holds for the box topology, but I have a hard time showing this, and for which example wouldn't $\Leftarrow$ hold, when we have a box topology?

Sara
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  • Start with a open $U$ of $\prod_{i\in I} X_i$ that contains $x$. Then there is an open set $V$ containing $x$, and contained in $U$, that is of the form $V=\prod_{i\in I}U_i$, such that there is a finite $J\subset I$ such that $U_i=X_i$ for $i\in I\setminus J$. By the assumption, for all $i\in J$, there is $N_i$ such that for all $n>N_i$ we have that $\pi_i x_n\in U_i$. Then, for all $n>\max_{i\in J}(N_i)$ we have $x_n\in V$. Hence $x_n\to x$. – plop May 21 '21 at 14:53
  • As you can see, it was crucial that we can reduce the argument to look at those neighborhoods in which almost all factors are the entire $X_i$. This might help you suspect that this is the direction that can fail in the box topology. – plop May 21 '21 at 14:57
  • I am not sure about the arrows. Just to be clear the argument above is proving that if for all $i\in I$ we have that $\pi_i x_n$ converges to $\pi_i x$, then $x_n$ converges to $x$. Is that the direction you were missing, or the converse? – plop May 21 '21 at 14:59

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