Metamath Proof Explorer


Theorem 0cnf

Description: The empty set is a continuous function. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Assertion 0cnf
|- (/) e. ( { (/) } Cn { (/) } )

Proof

Step Hyp Ref Expression
1 f0
 |-  (/) : (/) --> (/)
2 cnv0
 |-  `' (/) = (/)
3 2 imaeq1i
 |-  ( `' (/) " x ) = ( (/) " x )
4 0ima
 |-  ( (/) " x ) = (/)
5 3 4 eqtri
 |-  ( `' (/) " x ) = (/)
6 0ex
 |-  (/) e. _V
7 6 snid
 |-  (/) e. { (/) }
8 5 7 eqeltri
 |-  ( `' (/) " x ) e. { (/) }
9 8 rgenw
 |-  A. x e. { (/) } ( `' (/) " x ) e. { (/) }
10 sn0topon
 |-  { (/) } e. ( TopOn ` (/) )
11 iscn
 |-  ( ( { (/) } e. ( TopOn ` (/) ) /\ { (/) } e. ( TopOn ` (/) ) ) -> ( (/) e. ( { (/) } Cn { (/) } ) <-> ( (/) : (/) --> (/) /\ A. x e. { (/) } ( `' (/) " x ) e. { (/) } ) ) )
12 10 10 11 mp2an
 |-  ( (/) e. ( { (/) } Cn { (/) } ) <-> ( (/) : (/) --> (/) /\ A. x e. { (/) } ( `' (/) " x ) e. { (/) } ) )
13 1 9 12 mpbir2an
 |-  (/) e. ( { (/) } Cn { (/) } )