empty set = {} = \0

integers = Z = {...,-2,-1,0,1,2,...}

natural numbers = N = {1, 2, 3, 4, ...} = { x \in Z: x > 0}

rational numbers = Q = {m/n \in R: m,n \in Z, n \neq 0}

irrational = \Q = {x \in R: x \notin Q} = Q^c uncountable

has infinitely many points

[0,1]

[0,1/3][2/3,3/3]

[9,1/9][2/9,1/3][2/3,7/9][8/9],1]

[0,1]

[0,.1=.02xxx][.2,1]

[0,0.01 = .002xxx][.02,.02xxx][.2,.202xxx][.22,1]

length of each closed interval is (1/3)^(n-1)

# of intervals = 2^(n-1)

-> 2^(n-1)(1/3(^(n-1) = (2/3)^(n-1)

the total length of C is 0 (so technically there are no intervals)

2^(n-1) = 1 interval in C_1, 2 in C_2, C_3 is 4, C_4 is 8

a perfect set, because it is closed and it contains only limit points (no isolated points)

dimension log2/log3