Metamath Proof Explorer


Theorem setsms

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 φX=BaseM
setsms.d φD=distMX×X
setsms.k φK=MsSetTopSetndxMetOpenD
setsms.m φMV
Assertion setsms φKMetSpDMetX

Proof

Step Hyp Ref Expression
1 setsms.x φX=BaseM
2 setsms.d φD=distMX×X
3 setsms.k φK=MsSetTopSetndxMetOpenD
4 setsms.m φMV
5 1 2 3 4 setsxms φK∞MetSpD∞MetX
6 1 2 3 setsmsds φdistM=distK
7 1 2 3 setsmsbas φX=BaseK
8 7 sqxpeqd φX×X=BaseK×BaseK
9 6 8 reseq12d φdistMX×X=distKBaseK×BaseK
10 2 9 eqtr2d φdistKBaseK×BaseK=D
11 7 fveq2d φMetX=MetBaseK
12 11 eqcomd φMetBaseK=MetX
13 10 12 eleq12d φdistKBaseK×BaseKMetBaseKDMetX
14 5 13 anbi12d φK∞MetSpdistKBaseK×BaseKMetBaseKD∞MetXDMetX
15 eqid TopOpenK=TopOpenK
16 eqid BaseK=BaseK
17 eqid distKBaseK×BaseK=distKBaseK×BaseK
18 15 16 17 isms KMetSpK∞MetSpdistKBaseK×BaseKMetBaseK
19 metxmet DMetXD∞MetX
20 19 pm4.71ri DMetXD∞MetXDMetX
21 14 18 20 3bitr4g φKMetSpDMetX