Metamath Proof Explorer


Theorem ipasslem8

Description: Lemma for ipassi . By ipasslem5 , F is 0 for all QQ ; since it is continuous and QQ is dense in RR by qdensere2 , we conclude F is 0 for all RR . (Contributed by NM, 24-Aug-2007) (Revised by Mario Carneiro, 6-May-2014) (New usage is discouraged.)

Ref Expression
Hypotheses ip1i.1 X=BaseSetU
ip1i.2 G=+vU
ip1i.4 S=𝑠OLDU
ip1i.7 P=𝑖OLDU
ip1i.9 UCPreHilOLD
ipasslem7.a AX
ipasslem7.b BX
ipasslem7.f F=wwSAPBwAPB
Assertion ipasslem8 F:0

Proof

Step Hyp Ref Expression
1 ip1i.1 X=BaseSetU
2 ip1i.2 G=+vU
3 ip1i.4 S=𝑠OLDU
4 ip1i.7 P=𝑖OLDU
5 ip1i.9 UCPreHilOLD
6 ipasslem7.a AX
7 ipasslem7.b BX
8 ipasslem7.f F=wwSAPBwAPB
9 0cn 0
10 qre xx
11 oveq1 w=xwSA=xSA
12 11 oveq1d w=xwSAPB=xSAPB
13 oveq1 w=xwAPB=xAPB
14 12 13 oveq12d w=xwSAPBwAPB=xSAPBxAPB
15 ovex xSAPBxAPBV
16 14 8 15 fvmpt xFx=xSAPBxAPB
17 10 16 syl xFx=xSAPBxAPB
18 qcn xx
19 5 phnvi UNrmCVec
20 1 3 nvscl UNrmCVecxAXxSAX
21 19 20 mp3an1 xAXxSAX
22 18 21 sylan xAXxSAX
23 1 4 dipcl UNrmCVecxSAXBXxSAPB
24 19 7 23 mp3an13 xSAXxSAPB
25 22 24 syl xAXxSAPB
26 1 2 3 4 5 7 ipasslem5 xAXxSAPB=xAPB
27 25 26 subeq0bd xAXxSAPBxAPB=0
28 6 27 mpan2 xxSAPBxAPB=0
29 17 28 eqtrd xFx=0
30 29 rgen xFx=0
31 8 funmpt2 FunF
32 qssre
33 ovex wSAPBwAPBV
34 33 8 dmmpti domF=
35 32 34 sseqtrri domF
36 funconstss FunFdomFxFx=0F-10
37 31 35 36 mp2an xFx=0F-10
38 30 37 mpbi F-10
39 qdensere clstopGenran.=
40 eqid TopOpenfld=TopOpenfld
41 40 cnfldhaus TopOpenfldHaus
42 haust1 TopOpenfldHausTopOpenfldFre
43 41 42 ax-mp TopOpenfldFre
44 eqid topGenran.=topGenran.
45 1 2 3 4 5 6 7 8 44 40 ipasslem7 FtopGenran.CnTopOpenfld
46 uniretop =topGenran.
47 40 cnfldtopon TopOpenfldTopOn
48 47 toponunii =TopOpenfld
49 46 48 dnsconst TopOpenfldFreFtopGenran.CnTopOpenfld0F-10clstopGenran.=F:0
50 43 45 49 mpanl12 0F-10clstopGenran.=F:0
51 9 38 39 50 mp3an F:0