Existence of deep enough open subgroups in profinite groups

Let $G$ be an infinite profinite group and $\{U_i\}_{i \in I}$ be any family of open subgroups of $G$. Is it possible to choose a family of open subgroups $V_i < U_i$ with the property that, for any finite set of elements $g_1,\ldots,g_n$ and of indices $i_1,\ldots,i_n$, one has that $\bigcap_{k=1}^n g_kV_{i_k}g_k^{-1}$ is a proper subgroup of $\bigcap_{k=1}^n g_kU_{i_k}g_k^{-1}$?

It seems to me that, if we pick each $V_i$ such that the index $(U_i\colon V_i)$ is sufficiently bigger then $(G\colon U_i)$ for each $i \in I$, this should work, but I couldn't find exactly how to do so. I'm also interested in this question when each $U_i$ and each $V_i$ are normal in $G$, and thus simply asking whether we can find such $V_i$ with the property that $\bigcap_{k=1}^n V_{i_k} < \bigcap_{k=1}^n U_{i_k}$.

Solution 1:

Yes, this is always possible. First, as I commented, your question is equivalent to the version with normal subgroups, since any open subgroup contains a normal open subgroup. So, we will prove only the version with normal subgroups.

Now recursively define a descending sequence of closed sets $C_\alpha\subseteq G$ for ordinals $\alpha$ as follows. If $\alpha$ is a limit ordinal, let $C_\alpha=\bigcap_{\beta<\alpha}C_\beta$ (for $\alpha=0$ this mean $C_0=G$). If $\alpha=\beta+1$ is a successor ordinal, define $C_\alpha$ as follows. If $1$ is isolated in $C_\beta$, let $C_\alpha=C_\beta$ if $1$ is isolated in $C_\beta$. Otherwise, let $C_\alpha$ be a closed neighborhood of $1$ in $C_\beta$ that is not all of $C_\beta$ (so in particular $1$ is not isolated in $C_\alpha$ either).

This descending sequence must eventually stabilize on some set $C_\infty$ in which $1$ is isolated; let $\alpha$ be the least ordinal such that $C_\alpha=C_\infty$ (which must be a limit ordinal). Let $N\subseteq G$ be a closed neighborhood of $1$ such that $N\cap C_\infty=\{1\}$ and let $D_\beta=C_\beta\cap N$. Then $(D_\beta)_{\beta<\alpha}$ is a descending sequence of closed subsets of $G$ such that $\bigcap_{\beta<\alpha}D_\beta=\{1\}$ but $D_\beta\neq\{1\}$ for each $\beta<\alpha$.

Now we are finally ready to choose the $V_i$'s. For each $i\in I$, note that since $\bigcap_{\beta<\alpha}D_\beta=\{1\}\subseteq U_i$ and $G\setminus U_i$ is compact, there must exist some $\beta_i<\alpha$ such that $D_{\beta_i}\subseteq U_i$. Pick a point $x\in D_{\beta_i}\setminus\{1\}$ and let $V_i\subseteq U_i$ be a normal open subgroup that does not contain $x$.

Now I claim that for any $i_1,\dots,i_n\in I$, $\bigcap_{k=1}^n V_{i_k}$ is a proper subgroup of $\bigcap_{k=1}^nU_{i_k}$. Indeed, let $\beta$ be the maximum of the ordinals $\beta_{i_1},\dots,\beta_{i_n}$. Then $D_{\beta}\subseteq D_{\beta_{i_k}}\subseteq U_{i_k}$ for each $k$, so $D_\beta\subseteq \bigcap_{k=1}^n U_{i_k}$. However, if $k$ is such that $\beta_{i_k}=\beta$, then $D_\beta\not\subseteq V_{i_k}$ by construction, and so $D_\beta\not\subseteq \bigcap_{k=1}^n V_{i_k}$.

(To generalize a bit, the same argument should work with $G$ replaced by any compact Hausdorff space, $1$ replaced by some non-isolated point of $G$, and "normal open subgroups" replaced by any neighborhood base at $1$.)