Metamath Proof Explorer


Theorem limsup0

Description: The superior limit of the empty set. (Contributed by Glauco Siliprandi, 23-Oct-2021)

Ref Expression
Assertion limsup0 ( lim sup ‘ ∅ ) = -∞

Proof

Step Hyp Ref Expression
1 0ex ∅ ∈ V
2 eqid ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) ) = ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) )
3 2 limsupval ( ∅ ∈ V → ( lim sup ‘ ∅ ) = inf ( ran ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) ) , ℝ* , < ) )
4 1 3 ax-mp ( lim sup ‘ ∅ ) = inf ( ran ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) ) , ℝ* , < )
5 0ima ( ∅ “ ( 𝑥 [,) +∞ ) ) = ∅
6 5 ineq1i ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) = ( ∅ ∩ ℝ* )
7 0in ( ∅ ∩ ℝ* ) = ∅
8 6 7 eqtri ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) = ∅
9 8 supeq1i sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) = sup ( ∅ , ℝ* , < )
10 xrsup0 sup ( ∅ , ℝ* , < ) = -∞
11 9 10 eqtri sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) = -∞
12 11 mpteq2i ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) ) = ( 𝑥 ∈ ℝ ↦ -∞ )
13 ren0 ℝ ≠ ∅
14 13 a1i ( ⊤ → ℝ ≠ ∅ )
15 12 14 rnmptc ( ⊤ → ran ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) ) = { -∞ } )
16 15 mptru ran ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) ) = { -∞ }
17 16 infeq1i inf ( ran ( 𝑥 ∈ ℝ ↦ sup ( ( ( ∅ “ ( 𝑥 [,) +∞ ) ) ∩ ℝ* ) , ℝ* , < ) ) , ℝ* , < ) = inf ( { -∞ } , ℝ* , < )
18 xrltso < Or ℝ*
19 mnfxr -∞ ∈ ℝ*
20 infsn ( ( < Or ℝ* ∧ -∞ ∈ ℝ* ) → inf ( { -∞ } , ℝ* , < ) = -∞ )
21 18 19 20 mp2an inf ( { -∞ } , ℝ* , < ) = -∞
22 4 17 21 3eqtri ( lim sup ‘ ∅ ) = -∞