Metamath Proof Explorer


Theorem totbndss

Description: A subset of a totally bounded metric space is totally bounded. (Contributed by Jeff Madsen, 2-Sep-2009) (Revised by Mario Carneiro, 12-Sep-2015)

Ref Expression
Assertion totbndss M TotBnd X S X M S × S TotBnd S

Proof

Step Hyp Ref Expression
1 istotbnd M TotBnd X M Met X d + v Fin v = X b v x X b = x ball M d
2 1 simprbi M TotBnd X d + v Fin v = X b v x X b = x ball M d
3 sseq2 v = X S v S X
4 3 biimprcd S X v = X S v
5 4 anim1d S X v = X b v x X b = x ball M d S v b v x X b = x ball M d
6 5 reximdv S X v Fin v = X b v x X b = x ball M d v Fin S v b v x X b = x ball M d
7 6 ralimdv S X d + v Fin v = X b v x X b = x ball M d d + v Fin S v b v x X b = x ball M d
8 2 7 mpan9 M TotBnd X S X d + v Fin S v b v x X b = x ball M d
9 totbndmet M TotBnd X M Met X
10 eqid M S × S = M S × S
11 10 sstotbnd M Met X S X M S × S TotBnd S d + v Fin S v b v x X b = x ball M d
12 9 11 sylan M TotBnd X S X M S × S TotBnd S d + v Fin S v b v x X b = x ball M d
13 8 12 mpbird M TotBnd X S X M S × S TotBnd S