Removing all balls which do not contain points of and rearranging the subindices if necessary: .
For all there exists one point . Then, .
In fact, if we have:
So, hence, is totally bounded.
I try to prove that every subset of a totally bounded set is totally bounded, but I am not really sure with my argument here. Could someone help me check it?
Suppose A is a subset of a totally bounded set X. Since X is totally bounded, there exists an -net that covers X, i.e . Since A is a subset of X, A is also a subset of the union above. So, for some . Let . Then covers A since this ball contains every ball with radius above. Something does not sound right here since I am not sure if the new ball with radius contains all the balls with radius