Relationship between topology of a compact group and the topology of its profinite completion Suppose $G$ is a compact topological group.  We can construct the profinite completion of $G$; let's call this $\Gamma$.
My questions are:
1) Assuming that we know nothing about the (original) topology of $G$ other than that it is compact, is there anything we can say linking the topology of $G$ to the topology of $\Gamma$?
I assume the answer to this question is "no" since it seems to me that we usually regard $G$ as an abstract group when thinking about constructing its profinite completion.
2) If not (and, like I said, I assume the answer to (1) is "no") is there a way to construct a profinite completion (or something like this) of $G$ that takes into account the topology we already have?
 A: In principle, the answer to question (1) should be "no".
As for question (2), it seems that a basic construction would be to form a "profinite completion" of the topological group $G$ by taking the inverse limit of all quotients $G/N$, where $N$ ranges over all finite-index open normal subgroups of $G$.  The resulting completion $\hat{G}$ will have the property that the canonical function $G\to\hat{G}$ is continuous, and it should be somehow universal with respect to this property.
For example, let $\mathbb{Z}_2^\infty$ be the direct sum of infinitely many copies of $\mathbb{Z}_2$, which is a subspace of the infinite product $\mathbb{Z}_2^\omega$.  Then $\mathbb{Z}_2^\infty$ has many finite-index subgroups that are not open (e.g. the subgroup of elements with an even number of $1$'s), but it seems to me that every finite-index open subgroup should be the intersection of an open subgroup of $\mathbb{Z}_2^\omega$ with $\mathbb{Z}_2^\infty$.  Then the resulting topological profinite completion of $\mathbb{Z}_2^\infty$ ought to be $\mathbb{Z}_2^\omega$.
