The answers are as follows: these arguments are correct, the indicated function is not needed.

Indeed, your argument might as well prove that for all $H_1$,…,$H_n$, and $H$ such that $c_F(H_1,…,H_n)≤c$, you have $∥P_n⋯P_2P_1−P_0∥≤g_n(c)$. Quantifying over a collection of sets is perfectly legitimate even if this collection of sets forms a proper class. There is no need to mention the set $A_n(c)$ at all, it only creates further confusion.

The answers are as follows: these arguments are correct, the indicated function is not needed.

Indeed, your argument might as well prove that for all $H_1,\dotsc,H_n$, and $H$ such that $c_F(H_1,\dotsc,H_n)\le c$, you have $\lVert P_n\dotsm P_2P_1−P_0\rVert \le g_n(c)$. Quantifying over a collection of sets is perfectly legitimate even if this collection of sets forms a proper class. There is no need to mention the set $A_n(c)$ at all, it only creates further confusion.

The answers are as follows: these arguments are correct, the indicated function is not needed.

Indeed, your argument might as well prove that for all H_1,…,H_n, and H such that c_F(H_1,…,H_n)≤c, you have ∥P_n⋯P_2P_1−P_0∥≤g_n(c). Quantifying over a collection of sets is perfectly legitimate even if this collection of sets forms a proper class. There is no need to mention the set A_n(c) at all, it only creates further confusion.

The answers are as follows: these arguments are correct, the indicated function is not needed.

Indeed, your argument might as well prove that for all $H_1$,…,$H_n$, and $H$ such that $c_F(H_1,…,H_n)≤c$, you have $∥P_n⋯P_2P_1−P_0∥≤g_n(c)$. Quantifying over a collection of sets is perfectly legitimate even if this collection of sets forms a proper class. There is no need to mention the set $A_n(c)$ at all, it only creates further confusion.