Let $A$ be a ring, $I\subset A$ an ideal and $\alpha:A\to\hat{A}$ the natural homomorphism defined by the $I$-adic completion. Defining $\hat{I}:=\{(a_i)_{i\geq 1}\in\hat{A}\mid a_1\in I\}$, show that $\hat{I}$ equals the topological closure of $\alpha(I)$ in $\hat{A}$ and that $\hat{I}$ is an ideal of $\hat{A}$.
I guess proving that $\hat{I}$ is an ideal is straight-forward, because if $(a_i)_{i\geq 1}\in\hat{I}$ and $(b_i)_{i\geq 1}\in\hat{A}$, then $a_1b_1\in I$, so $(a)_{i\geq 1}(b)_{i\geq 1}=(a_ib_i)_{i\geq 1}\in\hat{I}$.
For the topological completion of $\alpha(I)$, I'm not sure I understand what topology I'm supposed to consider. I know what the $I$-adic topology on $A$ means, but not on $\hat{A}$.
And how do I find the closure of something in this topology?