Metamath Proof Explorer


Theorem ispnrm

Description: The property of being perfectly normal. (Contributed by Mario Carneiro, 26-Aug-2015)

Ref Expression
Assertion ispnrm ( 𝐽 ∈ PNrm ↔ ( 𝐽 ∈ Nrm ∧ ( Clsd ‘ 𝐽 ) ⊆ ran ( 𝑓 ∈ ( 𝐽m ℕ ) ↦ ran 𝑓 ) ) )

Proof

Step Hyp Ref Expression
1 fveq2 ( 𝑗 = 𝐽 → ( Clsd ‘ 𝑗 ) = ( Clsd ‘ 𝐽 ) )
2 oveq1 ( 𝑗 = 𝐽 → ( 𝑗m ℕ ) = ( 𝐽m ℕ ) )
3 2 mpteq1d ( 𝑗 = 𝐽 → ( 𝑓 ∈ ( 𝑗m ℕ ) ↦ ran 𝑓 ) = ( 𝑓 ∈ ( 𝐽m ℕ ) ↦ ran 𝑓 ) )
4 3 rneqd ( 𝑗 = 𝐽 → ran ( 𝑓 ∈ ( 𝑗m ℕ ) ↦ ran 𝑓 ) = ran ( 𝑓 ∈ ( 𝐽m ℕ ) ↦ ran 𝑓 ) )
5 1 4 sseq12d ( 𝑗 = 𝐽 → ( ( Clsd ‘ 𝑗 ) ⊆ ran ( 𝑓 ∈ ( 𝑗m ℕ ) ↦ ran 𝑓 ) ↔ ( Clsd ‘ 𝐽 ) ⊆ ran ( 𝑓 ∈ ( 𝐽m ℕ ) ↦ ran 𝑓 ) ) )
6 df-pnrm PNrm = { 𝑗 ∈ Nrm ∣ ( Clsd ‘ 𝑗 ) ⊆ ran ( 𝑓 ∈ ( 𝑗m ℕ ) ↦ ran 𝑓 ) }
7 5 6 elrab2 ( 𝐽 ∈ PNrm ↔ ( 𝐽 ∈ Nrm ∧ ( Clsd ‘ 𝐽 ) ⊆ ran ( 𝑓 ∈ ( 𝐽m ℕ ) ↦ ran 𝑓 ) ) )