Uncountable set Cantor's diagonal argument

this construction uses method devised cantor published in 1878. used construct bijection between closed interval [0, 1] , irrationals in open interval (0, 1). first removed countably infinite subset each of these sets there bijection between remaining uncountable sets. since there bijection between countably infinite subsets have been removed, combining 2 bijections produces bijection between original sets.

cantor s method can used modify function f 2(t) = 0.t2 produce bijection t (0, 1). because numbers have 2 binary expansions, f 2(t) not injective. example, f 2(1000…) = 0.1000…2 = 1/2 , f 2(0111…) = 0.0111…2 = 1/4 + 1/8 + 1/16 + … = 1/2, both 1000… , 0111… map same number, 1/2.

to modify f2 (t), observe bijection except countably infinite subset of (0, 1) , countably infinite subset of t. not bijection numbers in (0, 1) have 2 binary expansions. these numbers have form m / 2 m odd integer , n natural number. put these numbers in sequence: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, …). also, f2 (t) not bijection (0, 1) strings in t appearing after binary point in binary expansions of 0, 1, , numbers in sequence r. put these eventually-constant strings in sequence: s = (000…, 111…, 1000…, 0111…, 01000…, 00111…, 11000…, 10111…, ...). define bijection g(t) t (0, 1): if t n string in sequence s, let g(t) n number in sequence r ; otherwise, g(t) = 0.t2.

to construct bijection t r, start tangent function tan(x), bijection (−π/2, π/2) r (see figure shown on right). next observe linear function h(x) = πx – π/2 bijection (0, 1) (−π/2, π/2) (see figure shown on left). composite function tan(h(x)) = tan(πx – π/2) bijection (0, 1) r. composing function g(t) produces function tan(h(g(t))) = tan(πg(t) – π/2), bijection t r.