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 φX=BaseM
setsms.d φD=distMX×X
setsms.k φK=MsSetTopSetndxMetOpenD
setsms.m φMV
Assertion setsxms φK∞MetSpD∞MetX

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 setsmstopn φMetOpenD=TopOpenK
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 eqtrd φD=distKBaseK×BaseK
11 10 fveq2d φMetOpenD=MetOpendistKBaseK×BaseK
12 5 11 eqtr3d φTopOpenK=MetOpendistKBaseK×BaseK
13 eqid TopOpenK=TopOpenK
14 eqid BaseK=BaseK
15 eqid distKBaseK×BaseK=distKBaseK×BaseK
16 13 14 15 isxms2 K∞MetSpdistKBaseK×BaseK∞MetBaseKTopOpenK=MetOpendistKBaseK×BaseK
17 16 rbaib TopOpenK=MetOpendistKBaseK×BaseKK∞MetSpdistKBaseK×BaseK∞MetBaseK
18 12 17 syl φK∞MetSpdistKBaseK×BaseK∞MetBaseK
19 7 fveq2d φ∞MetX=∞MetBaseK
20 10 19 eleq12d φD∞MetXdistKBaseK×BaseK∞MetBaseK
21 18 20 bitr4d φK∞MetSpD∞MetX