Metamath Proof Explorer


Theorem cdivcncf

Description: Division with a constant numerator is continuous. (Contributed by Mario Carneiro, 28-Dec-2016)

Ref Expression
Hypothesis cdivcncf.1 F = x 0 A x
Assertion cdivcncf A F : 0 cn

Proof

Step Hyp Ref Expression
1 cdivcncf.1 F = x 0 A x
2 eqid TopOpen fld = TopOpen fld
3 2 cnfldtopon TopOpen fld TopOn
4 3 a1i A TopOpen fld TopOn
5 difss 0
6 resttopon TopOpen fld TopOn 0 TopOpen fld 𝑡 0 TopOn 0
7 4 5 6 sylancl A TopOpen fld 𝑡 0 TopOn 0
8 id A A
9 7 4 8 cnmptc A x 0 A TopOpen fld 𝑡 0 Cn TopOpen fld
10 7 cnmptid A x 0 x TopOpen fld 𝑡 0 Cn TopOpen fld 𝑡 0
11 eqid TopOpen fld 𝑡 0 = TopOpen fld 𝑡 0
12 2 11 divcn ÷ TopOpen fld × t TopOpen fld 𝑡 0 Cn TopOpen fld
13 12 a1i A ÷ TopOpen fld × t TopOpen fld 𝑡 0 Cn TopOpen fld
14 7 9 10 13 cnmpt12f A x 0 A x TopOpen fld 𝑡 0 Cn TopOpen fld
15 ssid
16 3 toponrestid TopOpen fld = TopOpen fld 𝑡
17 2 11 16 cncfcn 0 0 cn = TopOpen fld 𝑡 0 Cn TopOpen fld
18 5 15 17 mp2an 0 cn = TopOpen fld 𝑡 0 Cn TopOpen fld
19 14 1 18 3eltr4g A F : 0 cn