Metamath Proof Explorer


Theorem ressioosup

Description: If the supremum does not belong to a set of reals, the set is a subset of the unbounded below, right-open interval, with upper bound equal to the supremum. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypotheses ressioosup.a ( 𝜑𝐴 ⊆ ℝ )
ressioosup.s 𝑆 = sup ( 𝐴 , ℝ* , < )
ressioosup.n ( 𝜑 → ¬ 𝑆𝐴 )
ressioosup.i 𝐼 = ( -∞ (,) 𝑆 )
Assertion ressioosup ( 𝜑𝐴𝐼 )

Proof

Step Hyp Ref Expression
1 ressioosup.a ( 𝜑𝐴 ⊆ ℝ )
2 ressioosup.s 𝑆 = sup ( 𝐴 , ℝ* , < )
3 ressioosup.n ( 𝜑 → ¬ 𝑆𝐴 )
4 ressioosup.i 𝐼 = ( -∞ (,) 𝑆 )
5 mnfxr -∞ ∈ ℝ*
6 5 a1i ( ( 𝜑𝑥𝐴 ) → -∞ ∈ ℝ* )
7 ressxr ℝ ⊆ ℝ*
8 7 a1i ( 𝜑 → ℝ ⊆ ℝ* )
9 1 8 sstrd ( 𝜑𝐴 ⊆ ℝ* )
10 9 adantr ( ( 𝜑𝑥𝐴 ) → 𝐴 ⊆ ℝ* )
11 10 supxrcld ( ( 𝜑𝑥𝐴 ) → sup ( 𝐴 , ℝ* , < ) ∈ ℝ* )
12 2 11 eqeltrid ( ( 𝜑𝑥𝐴 ) → 𝑆 ∈ ℝ* )
13 1 adantr ( ( 𝜑𝑥𝐴 ) → 𝐴 ⊆ ℝ )
14 simpr ( ( 𝜑𝑥𝐴 ) → 𝑥𝐴 )
15 13 14 sseldd ( ( 𝜑𝑥𝐴 ) → 𝑥 ∈ ℝ )
16 15 mnfltd ( ( 𝜑𝑥𝐴 ) → -∞ < 𝑥 )
17 9 sselda ( ( 𝜑𝑥𝐴 ) → 𝑥 ∈ ℝ* )
18 supxrub ( ( 𝐴 ⊆ ℝ*𝑥𝐴 ) → 𝑥 ≤ sup ( 𝐴 , ℝ* , < ) )
19 10 14 18 syl2anc ( ( 𝜑𝑥𝐴 ) → 𝑥 ≤ sup ( 𝐴 , ℝ* , < ) )
20 2 a1i ( ( 𝜑𝑥𝐴 ) → 𝑆 = sup ( 𝐴 , ℝ* , < ) )
21 20 eqcomd ( ( 𝜑𝑥𝐴 ) → sup ( 𝐴 , ℝ* , < ) = 𝑆 )
22 19 21 breqtrd ( ( 𝜑𝑥𝐴 ) → 𝑥𝑆 )
23 id ( 𝑥 = 𝑆𝑥 = 𝑆 )
24 23 eqcomd ( 𝑥 = 𝑆𝑆 = 𝑥 )
25 24 adantl ( ( 𝑥𝐴𝑥 = 𝑆 ) → 𝑆 = 𝑥 )
26 simpl ( ( 𝑥𝐴𝑥 = 𝑆 ) → 𝑥𝐴 )
27 25 26 eqeltrd ( ( 𝑥𝐴𝑥 = 𝑆 ) → 𝑆𝐴 )
28 27 adantll ( ( ( 𝜑𝑥𝐴 ) ∧ 𝑥 = 𝑆 ) → 𝑆𝐴 )
29 3 ad2antrr ( ( ( 𝜑𝑥𝐴 ) ∧ 𝑥 = 𝑆 ) → ¬ 𝑆𝐴 )
30 28 29 pm2.65da ( ( 𝜑𝑥𝐴 ) → ¬ 𝑥 = 𝑆 )
31 30 neqned ( ( 𝜑𝑥𝐴 ) → 𝑥𝑆 )
32 17 12 22 31 xrleneltd ( ( 𝜑𝑥𝐴 ) → 𝑥 < 𝑆 )
33 6 12 15 16 32 eliood ( ( 𝜑𝑥𝐴 ) → 𝑥 ∈ ( -∞ (,) 𝑆 ) )
34 33 4 eleqtrrdi ( ( 𝜑𝑥𝐴 ) → 𝑥𝐼 )
35 34 ralrimiva ( 𝜑 → ∀ 𝑥𝐴 𝑥𝐼 )
36 dfss3 ( 𝐴𝐼 ↔ ∀ 𝑥𝐴 𝑥𝐼 )
37 35 36 sylibr ( 𝜑𝐴𝐼 )