Metamath Proof Explorer


Theorem setsxms

Description: The constructed metric space is a metric space iff the provided distance function is a metric. (Contributed by Mario Carneiro, 28-Aug-2015)

Ref Expression
Hypotheses setsms.x φ X = Base M
setsms.d φ D = dist M X × X
setsms.k φ K = M sSet TopSet ndx MetOpen D
setsms.m φ M V
Assertion setsxms φ K ∞MetSp D ∞Met X

Proof

Step Hyp Ref Expression
1 setsms.x φ X = Base M
2 setsms.d φ D = dist M X × X
3 setsms.k φ K = M sSet TopSet ndx MetOpen D
4 setsms.m φ M V
5 1 2 3 4 setsmstopn φ MetOpen D = TopOpen K
6 1 2 3 setsmsds φ dist M = dist K
7 1 2 3 setsmsbas φ X = Base K
8 7 sqxpeqd φ X × X = Base K × Base K
9 6 8 reseq12d φ dist M X × X = dist K Base K × Base K
10 2 9 eqtrd φ D = dist K Base K × Base K
11 10 fveq2d φ MetOpen D = MetOpen dist K Base K × Base K
12 5 11 eqtr3d φ TopOpen K = MetOpen dist K Base K × Base K
13 eqid TopOpen K = TopOpen K
14 eqid Base K = Base K
15 eqid dist K Base K × Base K = dist K Base K × Base K
16 13 14 15 isxms2 K ∞MetSp dist K Base K × Base K ∞Met Base K TopOpen K = MetOpen dist K Base K × Base K
17 16 rbaib TopOpen K = MetOpen dist K Base K × Base K K ∞MetSp dist K Base K × Base K ∞Met Base K
18 12 17 syl φ K ∞MetSp dist K Base K × Base K ∞Met Base K
19 7 fveq2d φ ∞Met X = ∞Met Base K
20 10 19 eleq12d φ D ∞Met X dist K Base K × Base K ∞Met Base K
21 18 20 bitr4d φ K ∞MetSp D ∞Met X