Does completion preserve openness?

Question

Let $\hat{A}$ be the completion of $A$ along the ideal $I$. Let $f$ be some non-zero divisor in $A$ that is not in $I$. $\text{Spec}(A_f)$ is an open subscheme of $\text{Spec}(A)$. Does that imply that $\text{Spec}(\widehat{A_f})$ is an open subscheme of $\text{Spec}(\hat{A})$? Or is the image of morphism from $\text{Spec}(\widehat{A_f})$ to $\text{Spec}(\hat{A})$ an open sub-scheme?