Metamath Proof Explorer


Theorem limsup10ex

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

Ref Expression
Hypothesis limsup10ex.1 F=nif2n01
Assertion limsup10ex lim supF=1

Proof

Step Hyp Ref Expression
1 limsup10ex.1 F=nif2n01
2 nftru k
3 nnex V
4 3 a1i V
5 0xr 0*
6 5 a1i n0*
7 1xr 1*
8 7 a1i n1*
9 6 8 ifcld nif2n01*
10 1 9 fmpti F:*
11 10 a1i F:*
12 eqid ksupFk+∞*<=ksupFk+∞*<
13 2 4 11 12 limsupval3 lim supF=supranksupFk+∞*<*<
14 13 mptru lim supF=supranksupFk+∞*<*<
15 id kk
16 1 15 limsup10exlem kFk+∞=01
17 16 supeq1d ksupFk+∞*<=sup01*<
18 xrltso <Or*
19 suppr <Or*0*1*sup01*<=if1<001
20 18 5 7 19 mp3an sup01*<=if1<001
21 0le1 01
22 0re 0
23 1re 1
24 22 23 lenlti 01¬1<0
25 21 24 mpbi ¬1<0
26 25 iffalsei if1<001=1
27 20 26 eqtri sup01*<=1
28 17 27 eqtrdi ksupFk+∞*<=1
29 28 mpteq2ia ksupFk+∞*<=k1
30 29 rneqi ranksupFk+∞*<=rank1
31 eqid k1=k1
32 ren0
33 32 a1i
34 31 33 rnmptc rank1=1
35 34 mptru rank1=1
36 30 35 eqtri ranksupFk+∞*<=1
37 36 infeq1i supranksupFk+∞*<*<=sup1*<
38 infsn <Or*1*sup1*<=1
39 18 7 38 mp2an sup1*<=1
40 14 37 39 3eqtri lim supF=1