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 ⊢ φ → A ⊆ ℝ
ressioosup.s ⊢ S = sup A ℝ * <
ressioosup.n ⊢ φ → ¬ S ∈ A
ressioosup.i ⊢ I = −∞ S
Assertion ressioosup ⊢ φ → A ⊆ I

Proof

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