Metamath Proof Explorer


Theorem cncfcn

Description: Relate complex function continuity to topological continuity. (Contributed by Mario Carneiro, 17-Feb-2015)

Ref Expression
Hypotheses cncfcn.2 J = TopOpen fld
cncfcn.3 K = J 𝑡 A
cncfcn.4 L = J 𝑡 B
Assertion cncfcn A B A cn B = K Cn L

Proof

Step Hyp Ref Expression
1 cncfcn.2 J = TopOpen fld
2 cncfcn.3 K = J 𝑡 A
3 cncfcn.4 L = J 𝑡 B
4 eqid abs A × A = abs A × A
5 eqid abs B × B = abs B × B
6 eqid MetOpen abs A × A = MetOpen abs A × A
7 eqid MetOpen abs B × B = MetOpen abs B × B
8 4 5 6 7 cncfmet A B A cn B = MetOpen abs A × A Cn MetOpen abs B × B
9 cnxmet abs ∞Met
10 simpl A B A
11 1 cnfldtopn J = MetOpen abs
12 4 11 6 metrest abs ∞Met A J 𝑡 A = MetOpen abs A × A
13 9 10 12 sylancr A B J 𝑡 A = MetOpen abs A × A
14 2 13 syl5eq A B K = MetOpen abs A × A
15 simpr A B B
16 5 11 7 metrest abs ∞Met B J 𝑡 B = MetOpen abs B × B
17 9 15 16 sylancr A B J 𝑡 B = MetOpen abs B × B
18 3 17 syl5eq A B L = MetOpen abs B × B
19 14 18 oveq12d A B K Cn L = MetOpen abs A × A Cn MetOpen abs B × B
20 8 19 eqtr4d A B A cn B = K Cn L