Metamath Proof Explorer


Theorem infxrpnf2

Description: Removing plus infinity from a set does not affect its infimum. (Contributed by Glauco Siliprandi, 2-Jan-2022)

Ref Expression
Assertion infxrpnf2 ( 𝐴 ⊆ ℝ* → inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) = inf ( 𝐴 , ℝ* , < ) )

Proof

Step Hyp Ref Expression
1 ssdifss ( 𝐴 ⊆ ℝ* → ( 𝐴 ∖ { +∞ } ) ⊆ ℝ* )
2 infxrpnf ( ( 𝐴 ∖ { +∞ } ) ⊆ ℝ* → inf ( ( ( 𝐴 ∖ { +∞ } ) ∪ { +∞ } ) , ℝ* , < ) = inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) )
3 1 2 syl ( 𝐴 ⊆ ℝ* → inf ( ( ( 𝐴 ∖ { +∞ } ) ∪ { +∞ } ) , ℝ* , < ) = inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) )
4 3 adantr ( ( 𝐴 ⊆ ℝ* ∧ +∞ ∈ 𝐴 ) → inf ( ( ( 𝐴 ∖ { +∞ } ) ∪ { +∞ } ) , ℝ* , < ) = inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) )
5 difsnid ( +∞ ∈ 𝐴 → ( ( 𝐴 ∖ { +∞ } ) ∪ { +∞ } ) = 𝐴 )
6 5 infeq1d ( +∞ ∈ 𝐴 → inf ( ( ( 𝐴 ∖ { +∞ } ) ∪ { +∞ } ) , ℝ* , < ) = inf ( 𝐴 , ℝ* , < ) )
7 6 adantl ( ( 𝐴 ⊆ ℝ* ∧ +∞ ∈ 𝐴 ) → inf ( ( ( 𝐴 ∖ { +∞ } ) ∪ { +∞ } ) , ℝ* , < ) = inf ( 𝐴 , ℝ* , < ) )
8 4 7 eqtr3d ( ( 𝐴 ⊆ ℝ* ∧ +∞ ∈ 𝐴 ) → inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) = inf ( 𝐴 , ℝ* , < ) )
9 difsn ( ¬ +∞ ∈ 𝐴 → ( 𝐴 ∖ { +∞ } ) = 𝐴 )
10 9 infeq1d ( ¬ +∞ ∈ 𝐴 → inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) = inf ( 𝐴 , ℝ* , < ) )
11 10 adantl ( ( 𝐴 ⊆ ℝ* ∧ ¬ +∞ ∈ 𝐴 ) → inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) = inf ( 𝐴 , ℝ* , < ) )
12 8 11 pm2.61dan ( 𝐴 ⊆ ℝ* → inf ( ( 𝐴 ∖ { +∞ } ) , ℝ* , < ) = inf ( 𝐴 , ℝ* , < ) )