Metamath Proof Explorer


Theorem heiborlem10

Description: Lemma for heibor . The last remaining piece of the proof is to find an element C such that C G 0 , i.e. C is an element of ( F0 ) that has no finite subcover, which is true by heiborlem1 , since ( F0 ) is a finite cover of X , which has no finite subcover. Thus, the rest of the proof follows to a contradiction, and thus there must be a finite subcover of U that covers X , i.e. X is compact. (Contributed by Jeff Madsen, 22-Jan-2014)

Ref Expression
Hypotheses heibor.1 J=MetOpenD
heibor.3 K=u|¬v𝒫UFinuv
heibor.4 G=yn|n0yFnyBnK
heibor.5 B=zX,m0zballD12m
heibor.6 φDCMetX
heibor.7 φF:0𝒫XFin
heibor.8 φn0X=yFnyBn
Assertion heiborlem10 φUJJ=Uv𝒫UFinJ=v

Proof

Step Hyp Ref Expression
1 heibor.1 J=MetOpenD
2 heibor.3 K=u|¬v𝒫UFinuv
3 heibor.4 G=yn|n0yFnyBnK
4 heibor.5 B=zX,m0zballD12m
5 heibor.6 φDCMetX
6 heibor.7 φF:0𝒫XFin
7 heibor.8 φn0X=yFnyBn
8 0nn0 00
9 inss2 𝒫XFinFin
10 ffvelcdm F:0𝒫XFin00F0𝒫XFin
11 9 10 sselid F:0𝒫XFin00F0Fin
12 6 8 11 sylancl φF0Fin
13 fveq2 n=0Fn=F0
14 oveq2 n=0yBn=yB0
15 13 14 iuneq12d n=0yFnyBn=yF0yB0
16 15 eqeq2d n=0X=yFnyBnX=yF0yB0
17 16 rspccva n0X=yFnyBn00X=yF0yB0
18 7 8 17 sylancl φX=yF0yB0
19 eqimss X=yF0yB0XyF0yB0
20 18 19 syl φXyF0yB0
21 ovex yB0V
22 1 2 21 heiborlem1 F0FinXyF0yB0XKyF0yB0K
23 oveq1 y=xyB0=xB0
24 23 eleq1d y=xyB0KxB0K
25 24 cbvrexvw yF0yB0KxF0xB0K
26 22 25 sylib F0FinXyF0yB0XKxF0xB0K
27 26 3expia F0FinXyF0yB0XKxF0xB0K
28 12 20 27 syl2anc φXKxF0xB0K
29 28 adantr φUJJ=UXKxF0xB0K
30 vex xV
31 c0ex 0V
32 1 2 3 30 31 heiborlem2 xG000xF0xB0K
33 1 2 3 4 5 6 7 heiborlem3 φgxGgxG2ndx+1BxgxB2ndx+1K
34 33 ad2antrr φUJJ=UxG0gxGgxG2ndx+1BxgxB2ndx+1K
35 5 ad2antrr φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1KDCMetX
36 6 ad2antrr φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1KF:0𝒫XFin
37 7 ad2antrr φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1Kn0X=yFnyBn
38 simprr φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1KxGgxG2ndx+1BxgxB2ndx+1K
39 fveq2 x=tgx=gt
40 fveq2 x=t2ndx=2ndt
41 40 oveq1d x=t2ndx+1=2ndt+1
42 39 41 breq12d x=tgxG2ndx+1gtG2ndt+1
43 fveq2 x=tBx=Bt
44 39 41 oveq12d x=tgxB2ndx+1=gtB2ndt+1
45 43 44 ineq12d x=tBxgxB2ndx+1=BtgtB2ndt+1
46 45 eleq1d x=tBxgxB2ndx+1KBtgtB2ndt+1K
47 42 46 anbi12d x=tgxG2ndx+1BxgxB2ndx+1KgtG2ndt+1BtgtB2ndt+1K
48 47 cbvralvw xGgxG2ndx+1BxgxB2ndx+1KtGgtG2ndt+1BtgtB2ndt+1K
49 38 48 sylib φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1KtGgtG2ndt+1BtgtB2ndt+1K
50 simprl φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1KxG0
51 eqeq1 g=mg=0m=0
52 oveq1 g=mg1=m1
53 51 52 ifbieq2d g=mifg=0xg1=ifm=0xm1
54 53 cbvmptv g0ifg=0xg1=m0ifm=0xm1
55 seqeq3 g0ifg=0xg1=m0ifm=0xm1seq0gg0ifg=0xg1=seq0gm0ifm=0xm1
56 54 55 ax-mp seq0gg0ifg=0xg1=seq0gm0ifm=0xm1
57 eqid nseq0gg0ifg=0xg1n32n=nseq0gg0ifg=0xg1n32n
58 simplrl φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1KUJ
59 cmetmet DCMetXDMetX
60 metxmet DMetXD∞MetX
61 1 mopnuni D∞MetXX=J
62 5 59 60 61 4syl φX=J
63 62 adantr φUJJ=UX=J
64 simprr φUJJ=UJ=U
65 63 64 eqtr2d φUJJ=UU=X
66 65 adantr φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1KU=X
67 1 2 3 4 35 36 37 49 50 56 57 58 66 heiborlem9 φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1K¬XK
68 67 expr φUJJ=UxG0xGgxG2ndx+1BxgxB2ndx+1K¬XK
69 68 exlimdv φUJJ=UxG0gxGgxG2ndx+1BxgxB2ndx+1K¬XK
70 34 69 mpd φUJJ=UxG0¬XK
71 32 70 sylan2br φUJJ=U00xF0xB0K¬XK
72 71 3exp2 φUJJ=U00xF0xB0K¬XK
73 8 72 mpi φUJJ=UxF0xB0K¬XK
74 73 rexlimdv φUJJ=UxF0xB0K¬XK
75 29 74 syld φUJJ=UXK¬XK
76 75 pm2.01d φUJJ=U¬XK
77 elfvdm DCMetXXdomCMet
78 sseq1 u=XuvXv
79 78 rexbidv u=Xv𝒫UFinuvv𝒫UFinXv
80 79 notbid u=X¬v𝒫UFinuv¬v𝒫UFinXv
81 80 2 elab2g XdomCMetXK¬v𝒫UFinXv
82 5 77 81 3syl φXK¬v𝒫UFinXv
83 82 adantr φUJJ=UXK¬v𝒫UFinXv
84 83 con2bid φUJJ=Uv𝒫UFinXv¬XK
85 76 84 mpbird φUJJ=Uv𝒫UFinXv
86 62 ad2antrr φUJJ=Uv𝒫UFinX=J
87 86 sseq1d φUJJ=Uv𝒫UFinXvJv
88 inss1 𝒫UFin𝒫U
89 88 sseli v𝒫UFinv𝒫U
90 89 elpwid v𝒫UFinvU
91 simprl φUJJ=UUJ
92 sstr vUUJvJ
93 92 unissd vUUJvJ
94 90 91 93 syl2anr φUJJ=Uv𝒫UFinvJ
95 94 biantrud φUJJ=Uv𝒫UFinJvJvvJ
96 eqss J=vJvvJ
97 95 96 bitr4di φUJJ=Uv𝒫UFinJvJ=v
98 87 97 bitrd φUJJ=Uv𝒫UFinXvJ=v
99 98 rexbidva φUJJ=Uv𝒫UFinXvv𝒫UFinJ=v
100 85 99 mpbid φUJJ=Uv𝒫UFinJ=v