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Sep 22, 2022 at 19:27 vote accept Ibra
Sep 22, 2022 at 19:27 comment added Ibra Thank you very much for taking the time to explain to me every detail! It's cristal clear now.
Sep 22, 2022 at 18:57 comment added Dominik Kutek What I meant in the last paragraph, we can "add" all those sets finitelly generated by $B(x_k,\frac{1}{m})$ to our sequence $\phi_n$ (I've explained why there are only countable many of those) and then we would imediatelly get $\nu_1(U_n) = \nu_2(U_n)$ for any $n\in \mathbb N$, because $U_n$ is finitelly generated by $B(x_k,\frac{1}{m})$. Is this clear, now?
Sep 22, 2022 at 18:35 comment added Ibra Thank you very much for your detailed response. Unfortunately, I didn't understand well why we don't need to show the equality $\nu_{1}(U_{n}) = \nu_{2}(U_{n})$. I am unable to conclude.
Sep 22, 2022 at 18:05 history answered Dominik Kutek CC BY-SA 4.0