Metamath Proof Explorer


Theorem smcn

Description: Scalar multiplication is jointly continuous in both arguments. (Contributed by NM, 16-Jun-2009) (Revised by Mario Carneiro, 5-May-2014) (New usage is discouraged.)

Ref Expression
Hypotheses smcn.c C = IndMet U
smcn.j J = MetOpen C
smcn.s S = 𝑠OLD U
smcn.k K = TopOpen fld
Assertion smcn U NrmCVec S K × t J Cn J

Proof

Step Hyp Ref Expression
1 smcn.c C = IndMet U
2 smcn.j J = MetOpen C
3 smcn.s S = 𝑠OLD U
4 smcn.k K = TopOpen fld
5 fveq2 U = if U NrmCVec U + × abs 𝑠OLD U = 𝑠OLD if U NrmCVec U + × abs
6 3 5 eqtrid U = if U NrmCVec U + × abs S = 𝑠OLD if U NrmCVec U + × abs
7 fveq2 U = if U NrmCVec U + × abs IndMet U = IndMet if U NrmCVec U + × abs
8 1 7 eqtrid U = if U NrmCVec U + × abs C = IndMet if U NrmCVec U + × abs
9 8 fveq2d U = if U NrmCVec U + × abs MetOpen C = MetOpen IndMet if U NrmCVec U + × abs
10 2 9 eqtrid U = if U NrmCVec U + × abs J = MetOpen IndMet if U NrmCVec U + × abs
11 10 oveq2d U = if U NrmCVec U + × abs K × t J = K × t MetOpen IndMet if U NrmCVec U + × abs
12 11 10 oveq12d U = if U NrmCVec U + × abs K × t J Cn J = K × t MetOpen IndMet if U NrmCVec U + × abs Cn MetOpen IndMet if U NrmCVec U + × abs
13 6 12 eleq12d U = if U NrmCVec U + × abs S K × t J Cn J 𝑠OLD if U NrmCVec U + × abs K × t MetOpen IndMet if U NrmCVec U + × abs Cn MetOpen IndMet if U NrmCVec U + × abs
14 eqid IndMet if U NrmCVec U + × abs = IndMet if U NrmCVec U + × abs
15 eqid MetOpen IndMet if U NrmCVec U + × abs = MetOpen IndMet if U NrmCVec U + × abs
16 eqid 𝑠OLD if U NrmCVec U + × abs = 𝑠OLD if U NrmCVec U + × abs
17 eqid BaseSet if U NrmCVec U + × abs = BaseSet if U NrmCVec U + × abs
18 eqid norm CV if U NrmCVec U + × abs = norm CV if U NrmCVec U + × abs
19 elimnvu if U NrmCVec U + × abs NrmCVec
20 eqid 1 1 + norm CV if U NrmCVec U + × abs y + x + 1 r = 1 1 + norm CV if U NrmCVec U + × abs y + x + 1 r
21 14 15 16 4 17 18 19 20 smcnlem 𝑠OLD if U NrmCVec U + × abs K × t MetOpen IndMet if U NrmCVec U + × abs Cn MetOpen IndMet if U NrmCVec U + × abs
22 13 21 dedth U NrmCVec S K × t J Cn J