I am trying to solve the following exercise. Let $X$ be a set of generators converging to 1 of a profinite group $G$. Then the topology on $X - \{1\}$ induced from G is the discrete topology. If $X$ is infinite, $\overline{X} = X \cup\{1\}$. If $1 \not\in X$ and $X$ is infinite, then $\overline{X}$ is the one-point compactification of $X$. But I am really stuck.
For the first part, the topology on $X - \{1\}$ induced from G is the discrete topology.
I am completely lost. I assume that I have to use the fact $X$ that converging to 1, this is, every open subgroup $U$ of $G$ contains all but a finite number of the elements in $X$, and that $X$ generates $G$ this is $\overline{\langle X\rangle}=\bigcup_{N\trianglelefteq_o G}\langle X\rangle N=G$. But I don't know how to combine this two facts to prove that $X - \{1\}$ has the discrete topology.
For the part: If $X$ is infinite, then $\overline{X} = X \cup\{1\}$.
I am able to show that $\overline{X} \supseteq X \cup\{1\}$. Clearly $X\subseteq \overline{X}$, and to show that $1\in \overline{X}$ it is enough to realize that any $N\trianglelefteq_o G$ will contain at least one element of $X$ and so its invese since every open subgroup $U$ of $G$ contains all but a finite number of the elements in $X$ and $X$ is infinite.This way $1\in \overline{X}=\bigcup_{N\trianglelefteq_o G}XN$.
But I am stuck proving that $\overline{X} \subseteq X \cup\{1\}$. I assume I have to prove that I have to use that the topology on $X - \{1\}$ induced from G is the discrete topology, but I am not sure on who to proceed.
Finally, if $1 \not\in X$ and $X$ is infinite, then $\overline{X}$ is the one-point compactification of $X$.
At this point we know that $X$ has the discrete topology and $\overline{X}=X\cup \{1\}$. And we need to show that the opens in $\overline{X}=X\cup \{1\}$ are either an open in $X$ (this is any subset of $X$) or a complement in \overline{X} of a compact subset in $X-\{1\}$(this is the complement in \overline{X} of a finite subset of $X − \{1\}$). I am trying to prove this as follows:
If $U$ is an open in $X\cup \{1\}$, then $U=V\cap (X\cup \{1\})$ where $V$ is open in $G$. If $1\not\in V$ then $U=V\cap (X\cup \{1\})=V\cap X$ wich is an open in $X$. But I don't know what happens when $1\in V$.
Any hint or idea would be very helpful.