Suppose that $X=\operatorname{colim} (A_1 \to A_2 \to \dots)$ and $Y=\operatorname{colim} (B_1 \to B_2 \to \dots)$, where $A_i$ and $B_j$ topological spaces. I want to prove that some set $U$ is open in $X\times Y$ iff $U$ is open in every $A_i \times B_j$. My attempt:
$U$ is open in $X\times Y$ $\Leftrightarrow$ $U=\bigcup _k V_k\times V_k^\prime$ for $V_k$ open in $X$ and $V_k^\prime$ open in $Y$ $\Leftrightarrow$ $V_k$ is open in $A_i$ and $V_k^\prime$ is open in $B_j$ ,$\forall k,i,j$ $\Leftrightarrow$ $V_k \times V_k ^\prime$ is open in $A_i\times B_j$, $\forall k,i,j$
And I stuck there. The next step I wanted to do is to say $V_k \times V_k ^\prime$ is open in $A_i\times B_j$, $\forall k,i,j$ $\Leftrightarrow$ $\bigcup _k V_k\times V_k^\prime$ is an open subset of $A_i\times B_j$, but it's not true.
Any help? Maybe the statement is false?