characterizing open subgroups of profinite groups I am studying Brian Osserman's notes on infinite Galois theory and i am a little bit confused in his proof of Lemma 2.2. In particular, we have a profinite group $G$ being the inverse limit of $\left\{G_i \right\}$ and $H$ an open subgroup. Then $H$ is compact and so it is a finite union of sets of the form $H \cap \prod_{i \in I} U_i$, where $I$ is finite and $U_i=G_i$. Then he says that each of these sets can be written as some $U_i \neq G_i$ and i am not sure at all what he means by that. Any suggestions?
 A: This is not a complete answer, just a thought. I apologize in case my notation is not the one you are used to.
I am almost sure that to prove the claim one has to use the following property of directed sets: said informally,

for any couple of indices, there always is a third one which is $\geq$
  than both.

We have
$$H\cap \prod_{i\in I}U_i=H\cap \Bigl(\Bigl(\prod_{i\in I-J} G_i\Bigr)\times \prod_{j\in J}U_{j}\Bigr),$$ where $J$ is a finite set of indexes. Here, we have collected together the $j\in I$ such that $U_j\neq G_j$.
We want to rewrite it in the same form as above, but with just a single $j$ such that $U_j\neq G_j$.
Now, by definition of directed system, we can find an index $k\in I$ such that $j\leq k$ for every $j\in J$ (remember that $J$ is finite). So for every $j\in J$ we can look at the preimages $$\pi_{kj}^{-1}(U_j)\subset G_k,$$ where $\pi_{lm}:G_l\to G_m$ makes part of the definition of directed system whenever $m\leq l$.
By (a possible) definition,
$$(\star)\,\,\,\,\,\,\,\,\,\,\,\,\,\,G=\varprojlim G_i=\Bigl\{(x_h)\in\prod G_i\,:\,\pi_{lm}(x_l)=x_m \textrm{ for all }m\leq l\Bigr\}.$$
Now, setting $$V_k=\bigcap_{j\in J}\pi_{kj}^{-1}(U_j)\subset G_k$$ I think (but did not check the details) that by definition $(\star)$ of $G$ we can actually replace the finite product over $J$ with the single $V_k$. Hope this helps, while we wait for an expert on profinite groups.
A: As noted in the comments, you’ve misunderstood the statement that

$H$ is covered by a finite number of open sets of the form $H\cap\prod_{i\in I}U_i$, where $U_i=G_i$ for all but finitely many $i$.

The point here is that $G$, being the inverse limit of the $G_i$, is a subspace of the Tikhonov product $\prod_{i\in I}G_i$, and 
$$\mathscr{B}=\left\{\prod_{i\in I}U_i:\text{each }U_i\subseteq G_i\text{ and }\{i\in I:U_i\ne G_i\}\text{ is finite}\right\}$$
is a base for the topology on this product. And $H\subseteq G$, so $\mathscr{B}_H=\{H\cap B:B\in\mathscr{B}\}$ is a base for $H$. $H$ is compact, and $\mathscr{B}_H$ certainly covers $H$, so there is some finite $\{B_1,\ldots,B_n\}\subseteq\mathscr{B}$ such that $\{H\cap B_1,\ldots,H\cap B_n\}$ covers $H$.
Let $k\in\{1,\ldots,n\}$, and let $B_k=\prod_{i\in I}U_i^k$. Let $I_k=\{i\in I:U_i\ne G_i\}$. There is an $i_k\in I$ such that $i\le i_k$ for each $i\in I_k$. Let $$V_{i_k}=\bigcap_{i\in I_k}\pi_{i,i_k}^{-1}[U_i]\;,$$ where $\pi_{i,j}:G_j\to G_i$ is the bonding map. Then $V_{i_k}$ is an open subset of $G_{i_k}$. For $i\in I\setminus\{i_k\}$ let $V_i=G_i$, and let $B=\prod_{i\in I}V_i$; then $x\in H\cap B_k$ iff $x_i\in U_i$ for each $i\in I_k$ iff $x_{i_k}\in\pi_{i,i_k}^{-1}[U_i]$ for each $i\in I_k$ iff $x_{i_k}\in V_{i_k}$ iff $x\in H\cap B$, so $H\cap B_k=H\cap B$. Thus, we can replace $B_k$ by $B=\prod_{i\in I}V_i$, where $V_i=G_i$ for all $i\in I$ except possibly $i_k$. This justifies the following sentence:

By the definition of inverse limits, each such set can actually be written with only a single $U_i\ne G_i$.

Thus, we may assume that each $B_k$ for $k\in\{1,\ldots,n\}$ has this form, i.e., that each $I_k$ is actually a singleton.
A similar argument then shows that there is a $B\in\mathscr{B}$ of this restricted form such that $H=H\cap(B_1\cup\ldots\cup B_n)=H\cap B$. If $i$ is the unique index on which $\pi_i[B]\ne G_i$, let $U_i=\pi_i[B]$; then $H=\pi_i^{-1}[U_i]$, and $\pi_i:G\to G_i$ is a homomorphism, so $U_i$ is a subgroup of $G_i$, and $H$ is the preimage of $U_i$ under $\pi_i$.
