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Assume $x_n$ converges to $x$ and $x$ has well ordered nbhd basis $U_m$. Since $U_m$ for every $m$ meets almost all $x_n$, we can apply a similar arguemtn as in here.
Maybe there is a better way to phrase this theorem.
1 new trait.
Assume$x_n$ converges to $x$ and $x$ has well ordered nbhd basis $U_m$ . Since $U_m$ for every $m$ meets almost all $x_n$ , we can apply a similar arguemtn as in here.
Maybe there is a better way to phrase this theorem.