I was trying to solve the following problem:
Let $\mathbb{Z}/n\mathbb{Z}$ denote the set of congruence classes of integers mod n, endowed with discrete topology. Give the cartesian product $X=\prod_{n\geq 2}\mathbb{Z}/n\mathbb{Z}$ the product topology. Denote by $[x]_n$ the congruent class of $x\in \mathbb{Z}$ mod n.
Fix $k\in \mathbb{Z}$, and let $F_k\subset X$ denote the set of elements x$=([x_n]_n)_{n\geq 2}$ such that $[x_n]_n$ is a multiple of $[k]_n$ for all $n>k$. Show that $F_k$ is closed in $X$.
I have tried to compute $F_2$. The result is $F_2=\mathbb{Z/2Z}\times \mathbb{Z/3Z}\times\{0,2\}\times\mathbb{Z/5Z}\times \{0,2,4\}\times...$. How can I show that this set is closed?
Basically what is going on here is you have a product of topological spaces $X=\prod_{n=2}^\infty X_n$, and a subset $A$ of $X$ of the form $A=\prod_{n=2}^\infty A_n$, where each $A_n$ is a closed subset of $X_n$.
Such a set is always a closed subset of $A$. Indeed $A=\bigcap_{n=2}^\infty B_n$ where $B_n=\pi_n^{-1}(A_n)$ and $\pi_n:X\to X_n$ is the projection map. Then $\pi_n$ is continuous, so $B_n$ is closed, and an intersection of closed subsets is closed.