Metamath Proof Explorer


Theorem cmssmscld

Description: The restriction of a metric space is closed if it is complete. (Contributed by AV, 9-Oct-2022)

Ref Expression
Hypotheses cmsss.h K=M𝑠A
cmsss.x X=BaseM
cmsss.j J=TopOpenM
Assertion cmssmscld MMetSpAXKCMetSpAClsdJ

Proof

Step Hyp Ref Expression
1 cmsss.h K=M𝑠A
2 cmsss.x X=BaseM
3 cmsss.j J=TopOpenM
4 eqid distMX×X=distMX×X
5 2 4 msmet MMetSpdistMX×XMetX
6 5 3ad2ant1 MMetSpAXKCMetSpdistMX×XMetX
7 xpss12 AXAXA×AX×X
8 7 anidms AXA×AX×X
9 8 3ad2ant2 MMetSpAXKCMetSpA×AX×X
10 9 resabs1d MMetSpAXKCMetSpdistMX×XA×A=distMA×A
11 2 sseq2i AXABaseM
12 fvex BaseMV
13 12 ssex ABaseMAV
14 11 13 sylbi AXAV
15 14 3ad2ant2 MMetSpAXKCMetSpAV
16 eqid distM=distM
17 1 16 ressds AVdistM=distK
18 15 17 syl MMetSpAXKCMetSpdistM=distK
19 18 reseq1d MMetSpAXKCMetSpdistMA×A=distKA×A
20 10 19 eqtrd MMetSpAXKCMetSpdistMX×XA×A=distKA×A
21 eqid BaseK=BaseK
22 eqid distKBaseK×BaseK=distKBaseK×BaseK
23 21 22 iscms KCMetSpKMetSpdistKBaseK×BaseKCMetBaseK
24 1 2 ressbas2 AXA=BaseK
25 24 adantr AXKMetSpA=BaseK
26 25 eqcomd AXKMetSpBaseK=A
27 26 sqxpeqd AXKMetSpBaseK×BaseK=A×A
28 27 reseq2d AXKMetSpdistKBaseK×BaseK=distKA×A
29 26 fveq2d AXKMetSpCMetBaseK=CMetA
30 28 29 eleq12d AXKMetSpdistKBaseK×BaseKCMetBaseKdistKA×ACMetA
31 30 biimpd AXKMetSpdistKBaseK×BaseKCMetBaseKdistKA×ACMetA
32 31 expimpd AXKMetSpdistKBaseK×BaseKCMetBaseKdistKA×ACMetA
33 23 32 biimtrid AXKCMetSpdistKA×ACMetA
34 33 imp AXKCMetSpdistKA×ACMetA
35 34 3adant1 MMetSpAXKCMetSpdistKA×ACMetA
36 20 35 eqeltrd MMetSpAXKCMetSpdistMX×XA×ACMetA
37 eqid MetOpendistMX×X=MetOpendistMX×X
38 37 metsscmetcld distMX×XMetXdistMX×XA×ACMetAAClsdMetOpendistMX×X
39 6 36 38 syl2anc MMetSpAXKCMetSpAClsdMetOpendistMX×X
40 3 2 4 mstopn MMetSpJ=MetOpendistMX×X
41 40 3ad2ant1 MMetSpAXKCMetSpJ=MetOpendistMX×X
42 41 fveq2d MMetSpAXKCMetSpClsdJ=ClsdMetOpendistMX×X
43 39 42 eleqtrrd MMetSpAXKCMetSpAClsdJ