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Let $G$ be a non discrete Polish group. For every $n\ge 2$ equip $G^n$ with the product topology.

Saying that $F_n=\{(g_1,\dots,g_n)\in G^n: \{g_1,...,g_n\}$ freely generates a free subgroup of rank $n$} is comeager in $G^n$ is equivalent to say that $F'_n=\{(g_1,\dots,g_n)\in G^n: \{g_1,...,g_n\}$ freely generates a free subgroup of rank at most $n$} is comeager in $G^n$ ?

Since $F_n\subseteq F_n'$ we have that if $F_n$ is comeager that $F'_n$ is comeager too.

My problem is the other implication. My intuitive idea is the following: suppose $F'_n$ is comeager and take $(g_1,...,g_n)\in F'_n$ such that $\{g_1,...,g_n\}$ freely generates a free subgroup of rank$<n$, say $n-1$ for example. So there are two elements $g_i$ in $(g_1,...,g_n)$ which are equal. Suppose $g_1=g_2$. My hope would be that there is a tuple $(h_1,g_2,...,g_n)\in G^n$ such that $\{h_1,g_2,...,g_n\}$ freely generates a free subgroup of rank $n$. So this tuple would be in $F_n$ and maybe it might be helpful to deduce that also $F_n$ is comeager.

Observe that if $\{g_1,g_2,...,g_n\}$ freely generates a free subgroup of rank $n$ then $\{g_2,...,g_n\}$ freely generates a free subgroup of rank $n-1$.

Warning: before reading the answer, it might be helpful to look at this: If I say $\{g_1,\dots,g_n\}$ freely generates a subgroup of $G$, does that mean the elements $g_i$ are all distinct?

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Let $\Delta\subset G^n$ be the set of points whose coordinates are not all distinct. As you observe, $F'_n\setminus F_n\subseteq\Delta$. But $\Delta$ is meager: for each pair of distinct $i,j\in\{1,\dots,n\}$, the set of points $(g_1,\dots,g_n)$ with $g_i=g_j$ is clearly closed and has empty interior since $G$ is not discrete, and $\Delta$ is the union of these sets over the finitely many such pairs $(i,j)$. So if $F'_n$ is comeager, so is $F_n=F'_n\setminus\Delta$.

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  • $\begingroup$ sorry could you give me more details about the fact that the set of points $(g_1,...,g_n)$ with $g_i\not=g_j$ is closed and has empty interior ? thank you very much $\endgroup$ – Richard Feb 28 '16 at 20:03
  • $\begingroup$ Oops, I meant $g_i=g_j$. Hopefully that should make more sense now. $\endgroup$ – Eric Wofsey Feb 28 '16 at 20:08
  • $\begingroup$ The logic reasoning is correct now but actually I still can't see why that set is closed and has empty interior, in particular the latter... thank you $\endgroup$ – Richard Feb 28 '16 at 20:10
  • $\begingroup$ If you have a sequence of points for which $g_i=g_j$, then the limit also satisfies $g_i=g_j$. If you have a point with $g_i=g_j$ and any open neighborhood $U=\prod U_k$ of it, then since $G$ is not discrete, $U_i$ and $U_j$ have more than one point, so we can pick a point in $U$ whose $i$ coordinate and $j$ coordinates are not equal. $\endgroup$ – Eric Wofsey Feb 28 '16 at 20:13
  • $\begingroup$ got it, thank you very much. In your opinion, can we extend this result for $F_\infty$ and $F'_\infty$ where this time we have sequences $(g_i)_{i\in\mathbb N}$ such that $\{g_i\}_{i\in\mathbb N}$ freely generate a free subgroup of rank countably infinite/at most countably infinite? If you want I can edit the question adding this. $\endgroup$ – Richard Feb 28 '16 at 20:22

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