Нить над натуральным рядом
Пусть N0={ }, N1= N={1,2,…,n,…}, N*={0} ÈN, d0=0, d1=1.
Sk,m={ k+mn: nÎN*}={ k, k+m, k+m2,…,k+mn,… }, k £ m,
bk,m= 2m-k/(2m-1)= - число, двоичное представление которого является индикатором подмножества Sk,m Ì N.
N01=S2,2,
d01=b2,2= 1/3 =0.01010101…(2),
N0ÌN01ÌN1, 0<d01<1.
N001=S4,4,
d001=b4,4= 1/15 =0.00010001…(2),
N0ÌN001ÌN01ÌN1, 0<d001<d01<1.
N011= N01ÈS3,4,
d011=d01+b3,4= 7/15 =0.01110111…(2),
N0ÌN001ÌN01ÌN011ÌN1, 0<d001<d01<d011<1.
N0001=S1,8,
d0001=b1,8= 1/225 =0.0000000100000001…(2),
N0ÌN0001ÌN001ÌN01ÌN011ÌN1.
N0011= N001ÈS6,8
d0011=d001+b6,8= 81/255 =0.0001010100010101…(2),
N0ÌN0001ÌN001ÌN0011ÌN01ÌN011ÌN1.
N0101= N01ÈS7,8,
d0101=d01+b7,8= 87/255 =0.0101011101010111…(2),
N0ÌN0001ÌN001ÌN0011ÌN01ÌN0101ÌN011ÌN1.
N0111= N011ÈS5,8,
d0111=d011+b5,8= 127/255 =0.0111111101111111…(2),
N0ÌN0001ÌN001ÌN0011ÌN01ÌN0101ÌN011ÌN0111ÌN1.
0,0 0 00000000 0
8,8 0001 00000001 1
4,4 001 00010001 17
6,8 0011 00010101 81
2,2 01 01010101 85
7,8 0101 01010111 87
3,4 011 01110111 119
5,8 0111 01111111 127
1,1 1 11111111 255