Metamath Proof Explorer


Theorem cmsss

Description: The restriction of a complete metric space is complete iff it is closed. (Contributed by Mario Carneiro, 15-Oct-2015)

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

Proof

Step Hyp Ref Expression
1 cmsss.h K=M𝑠A
2 cmsss.x X=BaseM
3 cmsss.j J=TopOpenM
4 simpr MCMetSpAXAX
5 xpss12 AXAXA×AX×X
6 4 5 sylancom MCMetSpAXA×AX×X
7 6 resabs1d MCMetSpAXdistMX×XA×A=distMA×A
8 2 fvexi XV
9 8 ssex AXAV
10 9 adantl MCMetSpAXAV
11 eqid distM=distM
12 1 11 ressds AVdistM=distK
13 10 12 syl MCMetSpAXdistM=distK
14 1 2 ressbas2 AXA=BaseK
15 14 adantl MCMetSpAXA=BaseK
16 15 sqxpeqd MCMetSpAXA×A=BaseK×BaseK
17 13 16 reseq12d MCMetSpAXdistMA×A=distKBaseK×BaseK
18 7 17 eqtrd MCMetSpAXdistMX×XA×A=distKBaseK×BaseK
19 15 fveq2d MCMetSpAXCMetA=CMetBaseK
20 18 19 eleq12d MCMetSpAXdistMX×XA×ACMetAdistKBaseK×BaseKCMetBaseK
21 eqid distMX×X=distMX×X
22 2 21 cmscmet MCMetSpdistMX×XCMetX
23 22 adantr MCMetSpAXdistMX×XCMetX
24 eqid MetOpendistMX×X=MetOpendistMX×X
25 24 cmetss distMX×XCMetXdistMX×XA×ACMetAAClsdMetOpendistMX×X
26 23 25 syl MCMetSpAXdistMX×XA×ACMetAAClsdMetOpendistMX×X
27 20 26 bitr3d MCMetSpAXdistKBaseK×BaseKCMetBaseKAClsdMetOpendistMX×X
28 cmsms MCMetSpMMetSp
29 ressms MMetSpAVM𝑠AMetSp
30 1 29 eqeltrid MMetSpAVKMetSp
31 28 9 30 syl2an MCMetSpAXKMetSp
32 eqid BaseK=BaseK
33 eqid distKBaseK×BaseK=distKBaseK×BaseK
34 32 33 iscms KCMetSpKMetSpdistKBaseK×BaseKCMetBaseK
35 34 baib KMetSpKCMetSpdistKBaseK×BaseKCMetBaseK
36 31 35 syl MCMetSpAXKCMetSpdistKBaseK×BaseKCMetBaseK
37 28 adantr MCMetSpAXMMetSp
38 3 2 21 mstopn MMetSpJ=MetOpendistMX×X
39 37 38 syl MCMetSpAXJ=MetOpendistMX×X
40 39 fveq2d MCMetSpAXClsdJ=ClsdMetOpendistMX×X
41 40 eleq2d MCMetSpAXAClsdJAClsdMetOpendistMX×X
42 27 36 41 3bitr4d MCMetSpAXKCMetSpAClsdJ