Metamath Proof Explorer


Theorem liminf10ex

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

Ref Expression
Hypothesis liminf10ex.1 F = n if 2 n 0 1
Assertion liminf10ex lim inf F = 0

Proof

Step Hyp Ref Expression
1 liminf10ex.1 F = n if 2 n 0 1
2 nftru k
3 nnex V
4 3 a1i V
5 0xr 0 *
6 5 a1i n 0 *
7 1xr 1 *
8 7 a1i n 1 *
9 6 8 ifcld n if 2 n 0 1 *
10 1 9 fmpti F : *
11 10 a1i F : *
12 eqid k sup F k +∞ * < = k sup F k +∞ * <
13 2 4 11 12 liminfval5 lim inf F = sup ran k sup F k +∞ * < * <
14 13 mptru lim inf F = sup ran k sup F k +∞ * < * <
15 id k k
16 1 15 limsup10exlem k F k +∞ = 0 1
17 16 infeq1d k sup F k +∞ * < = sup 0 1 * <
18 xrltso < Or *
19 infpr < Or * 0 * 1 * sup 0 1 * < = if 0 < 1 0 1
20 18 5 7 19 mp3an sup 0 1 * < = if 0 < 1 0 1
21 0lt1 0 < 1
22 21 iftruei if 0 < 1 0 1 = 0
23 20 22 eqtri sup 0 1 * < = 0
24 17 23 eqtrdi k sup F k +∞ * < = 0
25 24 mpteq2ia k sup F k +∞ * < = k 0
26 25 rneqi ran k sup F k +∞ * < = ran k 0
27 eqid k 0 = k 0
28 ren0
29 28 a1i
30 27 29 rnmptc ran k 0 = 0
31 30 mptru ran k 0 = 0
32 26 31 eqtri ran k sup F k +∞ * < = 0
33 32 supeq1i sup ran k sup F k +∞ * < * < = sup 0 * <
34 supsn < Or * 0 * sup 0 * < = 0
35 18 5 34 mp2an sup 0 * < = 0
36 14 33 35 3eqtri lim inf F = 0