This question has a proof of that fact, and I was thinking if it is necessary to impose $1/2^n$ as the shrinking radius of the balls. Why not just $1/n$?
Asked
Active
Viewed 35 times
1
-
That's just a choice, it is not important. It seems to me that, in this case, any sequence that tends to 0 will do. – Giuseppe Negro Jan 22 '19 at 10:57