Metamath Proof Explorer


Theorem ressiocsup

Description: If the supremum belongs to a set of reals, the set is a subset of the unbounded below, right-closed interval, with upper bound equal to the supremum. (Contributed by Glauco Siliprandi, 26-Jun-2021)

Ref Expression
Hypotheses ressiocsup.a ⊢ φ → A ⊆ ℝ
ressiocsup.s ⊢ S = sup A ℝ * <
ressiocsup.e ⊢ φ → S ∈ A
ressiocsup.5 ⊢ I = −∞ S
Assertion ressiocsup ⊢ φ → A ⊆ I

Proof

Step Hyp Ref Expression
1 ressiocsup.a ⊢ φ → A ⊆ ℝ
2 ressiocsup.s ⊢ S = sup A ℝ * <
3 ressiocsup.e ⊢ φ → S ∈ A
4 ressiocsup.5 ⊢ 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 9 sselda ⊢ φ ∧ x ∈ A → x ∈ ℝ *
14 1 adantr ⊢ φ ∧ x ∈ A → A ⊆ ℝ
15 simpr ⊢ φ ∧ x ∈ A → x ∈ A
16 14 15 sseldd ⊢ φ ∧ x ∈ A → x ∈ ℝ
17 16 mnfltd ⊢ φ ∧ x ∈ A → −∞ < x
18 supxrub ⊢ A ⊆ ℝ * ∧ x ∈ A → x ≤ sup A ℝ * <
19 10 15 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 6 12 13 17 22 eliocd ⊢ φ ∧ x ∈ A → x ∈ −∞ S
24 23 4 eleqtrrdi ⊢ φ ∧ x ∈ A → x ∈ I
25 24 ralrimiva ⊢ φ → ∀ x ∈ A x ∈ I
26 dfss3 ⊢ A ⊆ I ↔ ∀ x ∈ A x ∈ I
27 25 26 sylibr ⊢ φ → A ⊆ I