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Metamath Proof Explorer

Theorem clsss2

Description: If a subset is included in a closed set, so is the subset's closure. (Contributed by NM, 22-Feb-2007)

Ref Expression
Hypothesis clscld.1 
|- X = U. J
Assertion clsss2 
|- ( ( C e. ( Clsd ` J ) /\ S C_ C ) -> ( ( cls ` J ) ` S ) C_ C )

Proof

Step Hyp Ref Expression
1 clscld.1 
 |-  X = U. J
2 cldrcl 
 |-  ( C e. ( Clsd ` J ) -> J e. Top )
3 2 adantr 
 |-  ( ( C e. ( Clsd ` J ) /\ S C_ C ) -> J e. Top )
4 1 cldss 
 |-  ( C e. ( Clsd ` J ) -> C C_ X )
5 4 adantr 
 |-  ( ( C e. ( Clsd ` J ) /\ S C_ C ) -> C C_ X )
6 simpr 
 |-  ( ( C e. ( Clsd ` J ) /\ S C_ C ) -> S C_ C )
7 1 clsss 
 |-  ( ( J e. Top /\ C C_ X /\ S C_ C ) -> ( ( cls ` J ) ` S ) C_ ( ( cls ` J ) ` C ) )
8 3 5 6 7 syl3anc 
 |-  ( ( C e. ( Clsd ` J ) /\ S C_ C ) -> ( ( cls ` J ) ` S ) C_ ( ( cls ` J ) ` C ) )
9 cldcls 
 |-  ( C e. ( Clsd ` J ) -> ( ( cls ` J ) ` C ) = C )
10 9 adantr 
 |-  ( ( C e. ( Clsd ` J ) /\ S C_ C ) -> ( ( cls ` J ) ` C ) = C )
11 8 10 sseqtrd 
 |-  ( ( C e. ( Clsd ` J ) /\ S C_ C ) -> ( ( cls ` J ) ` S ) C_ C )