Closure of product equals product of closures

Theorem
Let: $\{X_i\}_{i \in I}$ be an arbitrary family of topological spaces and:
 * $A_i \subset X_i$ for each $i$.

If:
 * $X = \prod_{i \in I}X_i$

has the product topology, then:
 * $\prod_{i\in I} \overline{A_i} = \overline{\prod_{i\in I}A_i}$

Proof
We proceed by double inclusion.

Suppose:
 * $(x_i)_{i\in I}$ $\in \prod_{i\in I} \overline{A_i}$.

This means:
 * $x_i \in \overline{A_i}$

for each $i$.

Let $U$ be an open set containing $(x_i)_{i \in I}$. Since:
 * $x_i \in \overline{A_i}$

we can choose:
 * $y_i \in U_i \cap A_i$

and then:
 * $(y_i)_{i\in I} \in U \cap \prod_{i\in I} A_i$

But $U$ was arbitrary, hence:
 * $(x_i)_{i\in I} \in \overline{\prod_{i\in I}A_i}$

Conversely, suppose:
 * $(x_i)_{i\in I} \in \overline{\prod_{i\in I}A_i}$

We want to show that:


 * $x_j \in \overline{A_j}$ for all $j$

Fixed $j$, we take $V_j$ an open set in $X_j$ such that:
 * $x_j \in V_j$

Since:
 * $\pi_{j}^{-1}(V_j)$ is open in $\prod_{i\in I} X_i$

we can pick:
 * $(y_i)_{i\in I} \in \prod_{i\in I} A_i$

Then:
 * $y_j \in V_j \cap A_j$

so it follows that:
 * $x_j \in \overline{A_j}$

namely:
 * $(x_i)_{i\in I} \in \prod_{i\in I} \overline{A_i}$

$\blacksquare$