The question belongs to elementary category theory, so please forgive me if this is trivial. I think I even read a proof for this some weeks ago, but I can't find it.

In topology, you have the equation $\overline{\overline{A}}=\overline{A}$ for subsets $A$ of a topological space $X$. An analogous theorem in category theory would be: let $X$ be a complete category, and $A$ be a full subcategory of $X$. define $\overline{A}$ as the full subcategory of $X$ consisting of those objects that are limits of small diagrams in $C$, whose objects are in $A$. Do we have $\overline{\overline{A}}=\overline{A}$?

Let's try it: assume $x_i$ is a diagram in $\overline{A}$ and $(y_{ij})_j$ is a diagram in $A$ such that $x_i = \lim_j y_{ij}$. then $\lim_i x_i = \lim_i \lim_j y_{ij}$ and we want to interchange limits. But to do this, we have to make $y_{ij}$ to a diagram in two parameters $i,j$. Perhaps the claim can't be proved in that naive way?

It is easy to see that $\overline{A}$ is closed under products: with the notation above, a morphism $(i,j) \to (i',j')$ corresponds to $i=i'$ and a morphism $j \to j'$. Then $y_{ij}$ is a diagram in two parameters with limit $lim_i x_i$. so it remains to consider equalizers, but how? The problem is here that morphisms between two limits cannot be described in terms of the factors.

Perhaps one should look at examples. Let $X$ be the category of groups, and $A$ the category of finite groups. Then $\overline{A}$ consists of the groups which come from profinite groups, which are exactly the compact, hausdorff, totally disconnected topological groups. Now the category of profinite groups is complete (due to this description) and the forgetful functor to groups preserves limits. Thus in this case, $\overline{\overline{A}} = \overline{A}$. if we put $A=\{\mathbb{Z}/n\}$, $X$ as before, then a similar argument works with the help of $\mathbb{Z}/n$-modules.

EDIT: ok david has given a counterexample. does anyone have an idea how to "fix" this? So what is the "right" definition of $\overline{A}$, so that $\overline{\overline{A}} = \overline{A}$? also, Harry asked for a condition in order this becomes true with my definition.