14 Uncountable Sets(Denial of Self Recursion)
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1 14 Uncountable Sets(Denial of Self Recursion) 14.1 Countable sets Def. 1 Two sets A and B are isomorphic with respect to the bijective function f, iff a bijection f: A B, written A f B or for shorta B. Or Atwo sets A and B has same cardinality, written A = B or A = f B. Lem. 1 The set A can be reproduced with the set B and the function f 1, and The set B also can be reproduced with A and f. f: A B f 1 : B A f: A B A = {f 1 (b) A b B} = f 1 (B) A f 1 B B = {f(a) B a A} = f(a) B f A. 5/17/16 Kwang-Moo Choe 1
2 Def. 1 Let the set = {0, 1, 2, } be a set of natural numbers. Def. 2 A set S is said to be countable, if has same cardinality with a subset of. Uncountable, otherwise. Let a set S be countable. Then S is either finite or countabley infinite. If the set S is countably infinite, we write S = (aleph). Fact 1-1 Let N n = {1, 2,, n} and A = {a 1, a 2,, a n }. Then N n = f A = n <. n. 1 i n: f(i) = a i and f 1 (a i ) = i. N n. N n is (countable and) finite of size n. Fact 1-2 Let N 0,n = {0, 1,, n} and B = {b 0, b 1,, b n }. Then N n = f A = n < N 0,n = B = n+1. n+1. 0 i n: f(i) = a i and f 1 (b i ) = i. N 0,n is finite of size n+1. N 0,n N n and N n = A = n < N 0,n = B = n+1 <<. 5/17/16 Kwang-Moo Choe 2
3 Def. 3 Let 1 = {1, 2, 3, } = {k+1 k } be a set of natural numbers starting from 1. Then 1 but 1 f, i 1 : f(i) = i 1 and n : f 1 (n) = n+1. 1 is countable and finite(countably infinite). Fact 2-1 Let E = {0, 2,, 2n, } = {2k k } be a set of even numbers. Then E but E f, e E : f(e) = e/2 and n : f 1 (n) = 2 n. E = is countably infinite. Fact 2-2 Let O = {1, 3,, 2n+1, } = {2k+1 k } be a set of odd numbers. Then O but O f, o O : f(o)= (o 1)/2 and n : f 1 (n)=2 n+1. O = is countably infinite. 5/17/16 Kwang-Moo Choe 3
4 Def. 4 Let Z = {, 2, 1, 0, 1, 2, } f {0, 1, 1, 2, 2, } = {k Z k 1 } {0} { k Z k 1 } be a set of integers. Then Z but Z f where f: Z, i Z: if i 0 f(i) = 2 i i 0 f(i) = (2 i+1) fi. f 1 : Z, n : if n=2 k f(n) = n/2 n=2 k+1 f(n) = n/2 fi. Z = is countably infinite. Def. 5 Let Z p n = {k Z k mod n = p or [k] n = p} be a set integers of congruence n of modulus p. Then Z p n Z but Z p n f Z. Z p n = Z = is countably infinite. f: Z n p Z, m Z n p : f(m) = m div n. f 1 : Z Z n p, i Z, f 1 (i) = i n + p. {, p 2n, p n, p, p+n,, p+2n, } f {, 2, 1, 0, 1, 2, }. 5/17/16 Kwang-Moo Choe 4
5 Thm. 1 = {(i, j) i, j } is countably infinite. (0, 0), (0, 1), (0, 2),, (0, n), (1, 0), (1, 1), (1, 2),, (1, n), (n, 0), (n, 1), (n, 2),, (n, n), f:, i, j : f(i, j) = (i+j) (i+j+1)/2 + j. f 1 :, n : f 1 (n) =? = 2 =. Lem. 2 Let Q be a set of rational numbers. Since Q, Q =. Lem. 3 Let k. Then = k is countably infinite. = k =. = 1 = E = O = Z = Z p n = = Q = k =. 5/17/16 Kwang-Moo Choe 5
6 Let V be a set of symbols, called vocabulary(alphabet). We define a set of strings over V of length n(n 0). V 0 = B { } : epmty string of lengh 0, identity for concatenation( ) V n+1 = R V n V x V n is a string over V whose length x = n. V n is a set of string over V and V n = V n. We define V * = i 0 V i = { } V V 2 as an universe of strings over V. Let x = b 1 b 2 b n V n (n 0), 1 i n: b i V. Then f(x) = f(b 1 b 2 b n ) = b 1 V n + b 2 V n b n-1 V + b n. V * = is countably infinite. 5/17/16 Kwang-Moo Choe 6
7 14-2 Uncountable(Denial of Self Recursion) Cantor s Diagonal Argument(1874, 1892) Consider f: {0, 1} and 2 = P( ) infinite binary string Power set of natural numbers Assme 2 is countable. Then b n 2 can be enumerated. b 0 = (b 00, b 01,, b 0n, ) 2. b 1 = (b 10, b 11,, b 1n, ) 2. b n = (b n0, b n1,, b nn, ) 2. i, j : b ij {0, 1} Consider diagonal b d = (b 00, b 11,, b nn, ) 2. Consider complement of b d, b d = (b 00, b 11,, b nn, ) b ii = b ii b d is an infinite binary string(b d 2 ) but b d 2. Assumption 2 is countable. is a contradiction. 2 is uncountable. 5/17/16 Kwang-Moo Choe 7
8 Russel s Paradox(1901, 1911) Let R = {x x R} in x = R. Then R = {R R R}. R R R R Gödel s Incompleteness Theorem(1931) Assume proof(theorem) {proved, proved} exists. Hilbert s dream(1929) Let Gödel(proof, theorem) = if proof(theorem) proved proof(theorem) proved fi in proof = theorem = Gödel then Gödel(Gödel, Gödel) proved Gödel(Gödel) proved Gödel(Gödel, Gödel) proved Gödel(Gödel) proved 5/17/16 Kwang-Moo Choe 8
9 Halting problem Let Halting(program, data) = if program(data) stop program(data) stop fi in program = data = Halting then Halting(Halting, Halting) stop Halting(Halting) Halting(Halting, Halting) stop Halting(Halting). Denial of self recursion! A Barber problem A barber cuts everyone who cannot cut himself. Shall the barber cut himself? 5/17/16 Kwang-Moo Choe 9
10 A memo written by a liar This memo is written by me who is a liar. Is the memo written by the liar or not? Heterlogical adverb( 이종형용사 ) monosyllabic( 단음절 ) heterlogical polysyllabic( 다음절 ) not heterlogical heterlogical heterlogical or not heterlogical 5/17/16 Kwang-Moo Choe 10
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