You cannot conclude that Because that is not true.
Can you show that
I need to prove the following:
Suppose is a metric space and . Then is totally bounded if and only if, for every , there is a finite -dense subset of .
: Let . Then . So there exist sets , with and . Now let . Let . So is a finite subset of . I claim that is -dense in . To see this, let . I must show that some element of so that . If , then for some . Since , , .
: Now suppose that for , has a finite -dense subset. I must prove that is totally bounded. Since has a finite -dense subset, so that . So l.u.b. so .
Now for the problem: I am not sure if I have included all the definition for totally bounded; that is, I think I still have another step, but am not sure. Thanks for your help.