Metamath Proof Explorer


Theorem divcn

Description: Complex number division is a continuous function, when the second argument is nonzero. (Contributed by Mario Carneiro, 12-Aug-2014) Avoid ax-mulf . (Revised by GG, 16-Mar-2025)

Ref Expression
Hypotheses mpomulcn.j J=TopOpenfld
divcn.k K=J𝑡0
Assertion divcn ÷J×tKCnJ

Proof

Step Hyp Ref Expression
1 mpomulcn.j J=TopOpenfld
2 divcn.k K=J𝑡0
3 df-div ÷=x,y0ιz|yz=x
4 eldifsn y0yy0
5 divval xyy0xy=ιz|yz=x
6 divrec xyy0xy=x1y
7 5 6 eqtr3d xyy0ιz|yz=x=x1y
8 7 3expb xyy0ιz|yz=x=x1y
9 4 8 sylan2b xy0ιz|yz=x=x1y
10 9 mpoeq3ia x,y0ιz|yz=x=x,y0x1y
11 3 10 eqtri ÷=x,y0x1y
12 1 cnfldtopon JTopOn
13 12 a1i JTopOn
14 difss 0
15 resttopon JTopOn0J𝑡0TopOn0
16 13 14 15 sylancl J𝑡0TopOn0
17 2 16 eqeltrid KTopOn0
18 13 17 cnmpt1st x,y0xJ×tKCnJ
19 13 17 cnmpt2nd x,y0yJ×tKCnK
20 eqid z01z=z01z
21 eldifi z0z
22 eldifsni z0z0
23 21 22 reccld z01z
24 20 23 fmpti z01z:0
25 eqid if1xw1xwx2=if1xw1xwx2
26 25 reccn2 x0w+a+y0yx<a1y1x<w
27 ovres x0y0xabs0×0y=xabsy
28 eldifi x0x
29 eldifi y0y
30 eqid abs=abs
31 30 cnmetdval xyxabsy=xy
32 abssub xyxy=yx
33 31 32 eqtrd xyxabsy=yx
34 28 29 33 syl2an x0y0xabsy=yx
35 27 34 eqtrd x0y0xabs0×0y=yx
36 35 breq1d x0y0xabs0×0y<ayx<a
37 oveq2 z=x1z=1x
38 ovex 1xV
39 37 20 38 fvmpt x0z01zx=1x
40 oveq2 z=y1z=1y
41 ovex 1yV
42 40 20 41 fvmpt y0z01zy=1y
43 39 42 oveqan12d x0y0z01zxabsz01zy=1xabs1y
44 eldifsni x0x0
45 28 44 reccld x01x
46 eldifsni y0y0
47 29 46 reccld y01y
48 30 cnmetdval 1x1y1xabs1y=1x1y
49 abssub 1x1y1x1y=1y1x
50 48 49 eqtrd 1x1y1xabs1y=1y1x
51 45 47 50 syl2an x0y01xabs1y=1y1x
52 43 51 eqtrd x0y0z01zxabsz01zy=1y1x
53 52 breq1d x0y0z01zxabsz01zy<w1y1x<w
54 36 53 imbi12d x0y0xabs0×0y<az01zxabsz01zy<wyx<a1y1x<w
55 54 ralbidva x0y0xabs0×0y<az01zxabsz01zy<wy0yx<a1y1x<w
56 55 rexbidv x0a+y0xabs0×0y<az01zxabsz01zy<wa+y0yx<a1y1x<w
57 56 adantr x0w+a+y0xabs0×0y<az01zxabsz01zy<wa+y0yx<a1y1x<w
58 26 57 mpbird x0w+a+y0xabs0×0y<az01zxabsz01zy<w
59 58 rgen2 x0w+a+y0xabs0×0y<az01zxabsz01zy<w
60 cnxmet abs∞Met
61 xmetres2 abs∞Met0abs0×0∞Met0
62 60 14 61 mp2an abs0×0∞Met0
63 eqid abs0×0=abs0×0
64 1 cnfldtopn J=MetOpenabs
65 eqid MetOpenabs0×0=MetOpenabs0×0
66 63 64 65 metrest abs∞Met0J𝑡0=MetOpenabs0×0
67 60 14 66 mp2an J𝑡0=MetOpenabs0×0
68 2 67 eqtri K=MetOpenabs0×0
69 68 64 metcn abs0×0∞Met0abs∞Metz01zKCnJz01z:0x0w+a+y0xabs0×0y<az01zxabsz01zy<w
70 62 60 69 mp2an z01zKCnJz01z:0x0w+a+y0xabs0×0y<az01zxabsz01zy<w
71 24 59 70 mpbir2an z01zKCnJ
72 71 a1i z01zKCnJ
73 13 17 19 17 72 40 cnmpt21 x,y01yJ×tKCnJ
74 1 mpomulcn u,vuvJ×tJCnJ
75 74 a1i u,vuvJ×tJCnJ
76 oveq12 u=xv=1yuv=x1y
77 13 17 18 73 13 13 75 76 cnmpt22 x,y0x1yJ×tKCnJ
78 77 mptru x,y0x1yJ×tKCnJ
79 11 78 eqeltri ÷J×tKCnJ