Metamath Proof Explorer


Theorem 0psubclN

Description: The empty set is a closed projective subspace. (Contributed by NM, 25-Jan-2012) (New usage is discouraged.)

Ref Expression
Hypothesis 0psubcl.c
|- C = ( PSubCl ` K )
Assertion 0psubclN
|- ( K e. HL -> (/) e. C )

Proof

Step Hyp Ref Expression
1 0psubcl.c
 |-  C = ( PSubCl ` K )
2 0ss
 |-  (/) C_ ( Atoms ` K )
3 2 a1i
 |-  ( K e. HL -> (/) C_ ( Atoms ` K ) )
4 eqid
 |-  ( _|_P ` K ) = ( _|_P ` K )
5 4 2pol0N
 |-  ( K e. HL -> ( ( _|_P ` K ) ` ( ( _|_P ` K ) ` (/) ) ) = (/) )
6 eqid
 |-  ( Atoms ` K ) = ( Atoms ` K )
7 6 4 1 ispsubclN
 |-  ( K e. HL -> ( (/) e. C <-> ( (/) C_ ( Atoms ` K ) /\ ( ( _|_P ` K ) ` ( ( _|_P ` K ) ` (/) ) ) = (/) ) ) )
8 3 5 7 mpbir2and
 |-  ( K e. HL -> (/) e. C )