Metamath Proof Explorer


Theorem flfcntr

Description: A continuous function's value is always in the trace of its filter limit. (Contributed by Thierry Arnoux, 30-Aug-2020)

Ref Expression
Hypotheses flfcntr.c C = J
flfcntr.b B = K
flfcntr.j φ J Top
flfcntr.a φ A C
flfcntr.1 φ F J 𝑡 A Cn K
flfcntr.y φ X A
Assertion flfcntr φ F X K fLimf nei J X 𝑡 A F

Proof

Step Hyp Ref Expression
1 flfcntr.c C = J
2 flfcntr.b B = K
3 flfcntr.j φ J Top
4 flfcntr.a φ A C
5 flfcntr.1 φ F J 𝑡 A Cn K
6 flfcntr.y φ X A
7 fveq2 x = X F x = F X
8 7 eleq1d x = X F x K fLimf nei J X 𝑡 A F F X K fLimf nei J X 𝑡 A F
9 oveq2 a = nei J X 𝑡 A J 𝑡 A fLim a = J 𝑡 A fLim nei J X 𝑡 A
10 oveq2 a = nei J X 𝑡 A K fLimf a = K fLimf nei J X 𝑡 A
11 10 fveq1d a = nei J X 𝑡 A K fLimf a F = K fLimf nei J X 𝑡 A F
12 11 eleq2d a = nei J X 𝑡 A F x K fLimf a F F x K fLimf nei J X 𝑡 A F
13 9 12 raleqbidv a = nei J X 𝑡 A x J 𝑡 A fLim a F x K fLimf a F x J 𝑡 A fLim nei J X 𝑡 A F x K fLimf nei J X 𝑡 A F
14 1 toptopon J Top J TopOn C
15 3 14 sylib φ J TopOn C
16 resttopon J TopOn C A C J 𝑡 A TopOn A
17 15 4 16 syl2anc φ J 𝑡 A TopOn A
18 cntop2 F J 𝑡 A Cn K K Top
19 5 18 syl φ K Top
20 2 toptopon K Top K TopOn B
21 19 20 sylib φ K TopOn B
22 cnflf J 𝑡 A TopOn A K TopOn B F J 𝑡 A Cn K F : A B a Fil A x J 𝑡 A fLim a F x K fLimf a F
23 17 21 22 syl2anc φ F J 𝑡 A Cn K F : A B a Fil A x J 𝑡 A fLim a F x K fLimf a F
24 5 23 mpbid φ F : A B a Fil A x J 𝑡 A fLim a F x K fLimf a F
25 24 simprd φ a Fil A x J 𝑡 A fLim a F x K fLimf a F
26 1 sscls J Top A C A cls J A
27 3 4 26 syl2anc φ A cls J A
28 27 6 sseldd φ X cls J A
29 4 6 sseldd φ X C
30 trnei J TopOn C A C X C X cls J A nei J X 𝑡 A Fil A
31 15 4 29 30 syl3anc φ X cls J A nei J X 𝑡 A Fil A
32 28 31 mpbid φ nei J X 𝑡 A Fil A
33 13 25 32 rspcdva φ x J 𝑡 A fLim nei J X 𝑡 A F x K fLimf nei J X 𝑡 A F
34 neiflim J 𝑡 A TopOn A X A X J 𝑡 A fLim nei J 𝑡 A X
35 17 6 34 syl2anc φ X J 𝑡 A fLim nei J 𝑡 A X
36 6 snssd φ X A
37 1 neitr J Top A C X A nei J 𝑡 A X = nei J X 𝑡 A
38 3 4 36 37 syl3anc φ nei J 𝑡 A X = nei J X 𝑡 A
39 38 oveq2d φ J 𝑡 A fLim nei J 𝑡 A X = J 𝑡 A fLim nei J X 𝑡 A
40 35 39 eleqtrd φ X J 𝑡 A fLim nei J X 𝑡 A
41 8 33 40 rspcdva φ F X K fLimf nei J X 𝑡 A F