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=x0Ax
Assertion cdivcncf AF:0cn

Proof

Step Hyp Ref Expression
1 cdivcncf.1 F=x0Ax
2 eqid TopOpenfld=TopOpenfld
3 2 cnfldtopon TopOpenfldTopOn
4 3 a1i ATopOpenfldTopOn
5 difss 0
6 resttopon TopOpenfldTopOn0TopOpenfld𝑡0TopOn0
7 4 5 6 sylancl ATopOpenfld𝑡0TopOn0
8 id AA
9 7 4 8 cnmptc Ax0ATopOpenfld𝑡0CnTopOpenfld
10 7 cnmptid Ax0xTopOpenfld𝑡0CnTopOpenfld𝑡0
11 eqid TopOpenfld𝑡0=TopOpenfld𝑡0
12 2 11 divcn ÷TopOpenfld×tTopOpenfld𝑡0CnTopOpenfld
13 12 a1i A÷TopOpenfld×tTopOpenfld𝑡0CnTopOpenfld
14 7 9 10 13 cnmpt12f Ax0AxTopOpenfld𝑡0CnTopOpenfld
15 ssid
16 3 toponrestid TopOpenfld=TopOpenfld𝑡
17 2 11 16 cncfcn 00cn=TopOpenfld𝑡0CnTopOpenfld
18 5 15 17 mp2an 0cn=TopOpenfld𝑡0CnTopOpenfld
19 14 1 18 3eltr4g AF:0cn