Metamath Proof Explorer


Theorem ressiooinf

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

Ref Expression
Hypotheses ressiooinf.a φA
ressiooinf.s S=supA*<
ressiooinf.n φ¬SA
ressiooinf.i I=S+∞
Assertion ressiooinf φAI

Proof

Step Hyp Ref Expression
1 ressiooinf.a φA
2 ressiooinf.s S=supA*<
3 ressiooinf.n φ¬SA
4 ressiooinf.i I=S+∞
5 ressxr *
6 5 a1i φ*
7 1 6 sstrd φA*
8 7 adantr φxAA*
9 8 infxrcld φxAsupA*<*
10 2 9 eqeltrid φxAS*
11 pnfxr +∞*
12 11 a1i φxA+∞*
13 1 adantr φxAA
14 simpr φxAxA
15 13 14 sseldd φxAx
16 7 sselda φxAx*
17 infxrlb A*xAsupA*<x
18 8 14 17 syl2anc φxAsupA*<x
19 2 18 eqbrtrid φxASx
20 id x=Sx=S
21 20 eqcomd x=SS=x
22 21 adantl xAx=SS=x
23 simpl xAx=SxA
24 22 23 eqeltrd xAx=SSA
25 24 adantll φxAx=SSA
26 3 ad2antrr φxAx=S¬SA
27 25 26 pm2.65da φxA¬x=S
28 27 neqned φxAxS
29 28 necomd φxASx
30 10 16 19 29 xrleneltd φxAS<x
31 15 ltpnfd φxAx<+∞
32 10 12 15 30 31 eliood φxAxS+∞
33 32 4 eleqtrrdi φxAxI
34 33 ralrimiva φxAxI
35 dfss3 AIxAxI
36 34 35 sylibr φAI