Metamath Proof Explorer


Theorem liminf10ex

Description: The inferior limit of a function that alternates between two values. (Contributed by Glauco Siliprandi, 2-Jan-2022)

Ref Expression
Hypothesis liminf10ex.1 𝐹 = ( 𝑛 ∈ ℕ ↦ if ( 2 ∥ 𝑛 , 0 , 1 ) )
Assertion liminf10ex ( lim inf ‘ 𝐹 ) = 0

Proof

Step Hyp Ref Expression
1 liminf10ex.1 𝐹 = ( 𝑛 ∈ ℕ ↦ if ( 2 ∥ 𝑛 , 0 , 1 ) )
2 nftru 𝑘
3 nnex ℕ ∈ V
4 3 a1i ( ⊤ → ℕ ∈ V )
5 0xr 0 ∈ ℝ*
6 5 a1i ( 𝑛 ∈ ℕ → 0 ∈ ℝ* )
7 1xr 1 ∈ ℝ*
8 7 a1i ( 𝑛 ∈ ℕ → 1 ∈ ℝ* )
9 6 8 ifcld ( 𝑛 ∈ ℕ → if ( 2 ∥ 𝑛 , 0 , 1 ) ∈ ℝ* )
10 1 9 fmpti 𝐹 : ℕ ⟶ ℝ*
11 10 a1i ( ⊤ → 𝐹 : ℕ ⟶ ℝ* )
12 eqid ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) ) = ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) )
13 2 4 11 12 liminfval5 ( ⊤ → ( lim inf ‘ 𝐹 ) = sup ( ran ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) ) , ℝ* , < ) )
14 13 mptru ( lim inf ‘ 𝐹 ) = sup ( ran ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) ) , ℝ* , < )
15 id ( 𝑘 ∈ ℝ → 𝑘 ∈ ℝ )
16 1 15 limsup10exlem ( 𝑘 ∈ ℝ → ( 𝐹 “ ( 𝑘 [,) +∞ ) ) = { 0 , 1 } )
17 16 infeq1d ( 𝑘 ∈ ℝ → inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) = inf ( { 0 , 1 } , ℝ* , < ) )
18 xrltso < Or ℝ*
19 infpr ( ( < Or ℝ* ∧ 0 ∈ ℝ* ∧ 1 ∈ ℝ* ) → inf ( { 0 , 1 } , ℝ* , < ) = if ( 0 < 1 , 0 , 1 ) )
20 18 5 7 19 mp3an inf ( { 0 , 1 } , ℝ* , < ) = if ( 0 < 1 , 0 , 1 )
21 0lt1 0 < 1
22 21 iftruei if ( 0 < 1 , 0 , 1 ) = 0
23 20 22 eqtri inf ( { 0 , 1 } , ℝ* , < ) = 0
24 17 23 eqtrdi ( 𝑘 ∈ ℝ → inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) = 0 )
25 24 mpteq2ia ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) ) = ( 𝑘 ∈ ℝ ↦ 0 )
26 25 rneqi ran ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) ) = ran ( 𝑘 ∈ ℝ ↦ 0 )
27 eqid ( 𝑘 ∈ ℝ ↦ 0 ) = ( 𝑘 ∈ ℝ ↦ 0 )
28 ren0 ℝ ≠ ∅
29 28 a1i ( ⊤ → ℝ ≠ ∅ )
30 27 29 rnmptc ( ⊤ → ran ( 𝑘 ∈ ℝ ↦ 0 ) = { 0 } )
31 30 mptru ran ( 𝑘 ∈ ℝ ↦ 0 ) = { 0 }
32 26 31 eqtri ran ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) ) = { 0 }
33 32 supeq1i sup ( ran ( 𝑘 ∈ ℝ ↦ inf ( ( 𝐹 “ ( 𝑘 [,) +∞ ) ) , ℝ* , < ) ) , ℝ* , < ) = sup ( { 0 } , ℝ* , < )
34 supsn ( ( < Or ℝ* ∧ 0 ∈ ℝ* ) → sup ( { 0 } , ℝ* , < ) = 0 )
35 18 5 34 mp2an sup ( { 0 } , ℝ* , < ) = 0
36 14 33 35 3eqtri ( lim inf ‘ 𝐹 ) = 0