The topology of $\mathbb{Z}_p$

Question

I don't know much about topology, but anyway...

Assuming $\displaystyle\prod{A_n} =\prod_{n\geq 1}{A_n}$, why is $\mathbb{Z}_p$ closed in a product of compact spaces?

Googling I found Tychonoff's theorem on ProofWiki, so I see why the product is compact.

Answer 1

It is closed, because its complement is open in the product topology.

Let $x=(\ldots,x_3,x_2,x_1)\in\prod_n A_n$ be an arbitrary sequence that is not in $\Bbb{Z}_p$. This implies that there exists an index $i\ge2$ such that $\phi_i(x_i)\neq x_{i-1}$. Consider the set $U$ $$\begin{aligned}U&=\{y=(\ldots,y_3,y_2,y_1)\in\prod_n A_n\mid y_i=x_i, y_{i-1}=x_{i-1}\}\\ &=\prod_{n>i}A_n\times \{x_i\}\times \{x_{i-1}\}\times\prod_{n<i-1}A_n. \end{aligned}$$ No element $y\in U$ satisfies the condition $\phi_i(y_i)=y_{i-1}$, so $U$ is contained in the complement of $\Bbb{Z}_p$. The latter description shows that $U$ is open w.r.t. the product toplogy. Because $x$ was an arbitrary element of the complement the claim follows.