Metamath Proof Explorer


Theorem fprodadd2cncf

Description: F is continuous. (Contributed by Glauco Siliprandi, 8-Apr-2021)

Ref Expression
Hypotheses fprodadd2cncf.k kφ
fprodadd2cncf.a φAFin
fprodadd2cncf.b φkAB
fprodadd2cncf.f F=xkAB+x
Assertion fprodadd2cncf φF:cn

Proof

Step Hyp Ref Expression
1 fprodadd2cncf.k kφ
2 fprodadd2cncf.a φAFin
3 fprodadd2cncf.b φkAB
4 fprodadd2cncf.f F=xkAB+x
5 4 a1i φF=xkAB+x
6 eqid TopOpenfld=TopOpenfld
7 6 cnfldtopon TopOpenfldTopOn
8 7 a1i φTopOpenfldTopOn
9 eqid xB+x=xB+x
10 3 9 add2cncf φkAxB+x:cn
11 6 cncfcn1 cn=TopOpenfldCnTopOpenfld
12 11 a1i φkAcn=TopOpenfldCnTopOpenfld
13 10 12 eleqtrd φkAxB+xTopOpenfldCnTopOpenfld
14 1 6 8 2 13 fprodcn φxkAB+xTopOpenfldCnTopOpenfld
15 5 14 eqeltrd φFTopOpenfldCnTopOpenfld
16 11 a1i φcn=TopOpenfldCnTopOpenfld
17 16 eqcomd φTopOpenfldCnTopOpenfld=cn
18 15 17 eleqtrd φF:cn