$\underline{\text{Definition:}}$ We say that a subset $X$ of a profinite group $G$ converges to 1 if every open (normal) subgroup $U$ of $G$ contains all but a finite number of the elements in $X$.
It is known that if $G$ is a profinite group, then there exists a set $X$ converging to 1 such that $\overline{\langle X \rangle}=G$ (where $\overline{\langle X \rangle}$ means the closure of $\langle X \rangle$). This is proven in Ribes and Zalesskii's book "Profinite Groups" Proposition 2.4.4 or in Fried and Jarden's book "field arithmetics" Proposition 17.1.1 ([Douady]).
I am trying to show (I do not know if it is true or not) that if $X$ generates a profinite group $G$, in the sense that $G=\overline{\langle X \rangle}$, then there exists $Y\subseteq X$ such that $G=\overline{\langle Y \rangle}$ and $Y$ converges to 1.
By mirroring the proof on Zalesskii's book "Profinite Groups" Proposition 2.4.4 I have the following:
$\underline{\text{Question:}}$Let $G$ be a profinite group and let $Y$ be a set of generators of $G$. Does it existst $\widetilde{Y}\subseteq Y$ such that $\widetilde{Y}$ is a set of generators converging to 1?
I have been able to show that the following set $\mathcal{P}$ has a maximal element. Where $\mathcal{P}$ is defined as follows:
Consider the set $\mathcal{P}$ of all the pairs $(N,X_N)$, where $N\triangleleft_c G$, $X_N\subseteq Y$ and $X_N\subseteq G-N$ such that
$\bullet$ for every open subgroup $U$ of $G$ containing $N$, $X_N-U$ is a finite set; and
$\bullet$ $G=\overline{\langle X_N,N\rangle}$.
The fact that $\mathcal{P}$ has a maximal element comes from Zorn's Lemma applied with this partial ordering: $(N, X_N)\preceq (M,X_M)$ if $N\geq M$, $X_N\subset X_M$ and $X_M-X_N\subseteq N$.
Let $(M,X)$ be the maximal element in $\mathcal{P}$.
I would like to have that $M=1$. But I am not able to get this. This is what is done in the book: Every profinite group $G$ admits a set of generators converging to 1:
They define $\mathcal{P}$ in a similar way, but without asking that $X_N\subseteq Y$. And so they get a maximal element in $\mathcal{P}$, namely $(M,X)$. They assume that $M\not=1$. Then, let $U\triangleleft_o G$ be such that $U\cup M$ is a proper subgroup of $M$. Then they choose a finite subset $T$ of $M-(U\cap M)$ such that $M=\langle T, U\cap M\rangle$. And then they have that $(U\cap M, X\cup T)\in \mathcal{P}$. But in our case, I can not ensure that $(U\cap M, X\cup T)\in \mathcal{P} $ since $T$ might not be a subset of $Y$.
Finally they show that $(M,X) \preceq (U\cap M, X\cup T)$ which is a contradiction. Thus $M=1$.
Is this even true, in my head it makes sense, but I am not sure? If it is not true in general, is it true for any class of group?
Any idea would be very appreciated.
This is not true. Let $G$ be a profinite group that is not finitely generated, and $H$ be a non-trivial finite group. Then $G\times H$ is profinite, and the set of generators $X=G\times (H\setminus \{e\})$ has no subset that converges to $1$. In fact, as $G\times\{e\}$ is open in $G\times H$ and $X\cap (G\times\{e\})=\emptyset$, any subset of $X$ that converges to $1$ must be finite, but $G$ and hence $G\times H$ is not finitely generated.
The technique can be used to show that the statement holds iff the profinite group is finitely generated.