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The power set of N is uncountable

The power set of N is uncountable. Prof. Ting-Lu Huang Introduction to Formal Languages National Chiao Tung University. sets. A set is a collection of distinct entities. ex. Let N={0,1,2,3,4,5,…} then N is also equal to {1,0,3,2,5,4,…}

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The power set of N is uncountable

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  1. The power set of N is uncountable Prof. Ting-Lu Huang Introduction to Formal Languages National Chiao Tung University

  2. sets • A set is a collection of distinct entities. ex. Let N={0,1,2,3,4,5,…} then N is also equal to {1,0,3,2,5,4,…} ex. { } is a set containing nothing, called empty set. ex. { } is not equal to {{ }}, because the latter contains { } while the former contains nothing. ex. {1,2,3} = {2,1,3} , and {1,2,3}  {2,3,4}

  3. Membership in a set To say that 2 is in the set N, we write 2  N. We also say that 2 is an element of the set N. Example: { }  {{ }} is true, while { }  { } is false

  4. Subsets • Given two sets M and K. If K contains every element in M, we say M is a subset of K, written as M  K. • Note that M is a subset of itself. • If M  K and M  K, then we say M is a proper subset of K.

  5. Comparing sizes of sets • Let M = {0,2,4,6,8}, K = {0,1,2,3,9} • Then M and K have the same size since both contain the same number of elements. • It is easy to compare the sizes of M and K because they are both finite sets. • Challenging question: How to compare the sizes of the following two infinite sets? • N={0,1,2,3,4,5,6,7,…} • E={0,2,4,6,8,10,12,14,…}

  6. Cardinality of sets • Given set N and set E, if there exists a perfect match between elements in the two sets, then we say N and E have the same cardinality, written as N  E. • In mathematics, such a perfect match is a function mapping N to E, so that each element in N matches one and only one element in E, and vice versa. Such a function is called a 1-to-1 correspondence,. • If N and a subset of E have the same cardinality, we write N  E. • Bernstein Theorem: If N  E and E  N, then N  E.

  7. Cardinality: examples • N: natural numbers • E: even numbers • N={0,1,2,3,4,5,6,7,…} • E={0,2,4,6,8,10,12,14,…} • Both can be put in a linear order. • Pairing up by the position in the linear order is a perfect match between N and E. Therefore, they have the same cardinality. • It is counter-intuitive since E is a proper subset of N but both E and N have the same cardinality .

  8. Countable sets • Fact: Among all infinite sets, N is one of those that have the smallest cardinality. • Definition: If a set is finite or has the same cardinality as N, it is called a countable set. Otherwise, it is called an uncountable set. • Exercise: Prove or disprove that the set of all rational numbers is countable.

  9. Power sets • Given a set K, the set consisting of all possible subsets of K is called the power set of K, written as P(K). • Ex. Let M={3,4}. Then P(M)={{},{3},{4},{3,4}}.

  10. A challenging question Let N={0,1,2,3,…}. Then P(N) denotesthe set of all subsets of N. Can we put all subsets of N in a linear order? Try this: {{},{0},{1},{2},{3},…,{0,1},{0,2},{0,3},…} Try that: Try some more: It turns out that it is impossible to put them in a linear order. To say it more formally: P(N) is uncountable. We need a proof for this impossibility.

  11. Theorem: P(N) is uncountable. (Cantor, 1891)Proof: • The proof method is one instance of the general scheme called proof by contradiction. (Assuming the assertion is false, then we derive a contradiction. Hence, the assertion is true.) Assume that there exists a way to put every element of P(N) in a linear order: S1,S2,S3,…where Si is a subset of N.

  12. Constructing a subset to obtain a contradiction • Observe the bit string in the diagonal line from upper-left to lower-right: TTFF… • Let Sd be the set associated with the negated bit string FFTT… • Then, Sdis a legitimate subset of N. By the assumption that a linear order exists, Sd is equal to some Si. Unfortunately, it differs from every Si in at least one bit. So, Sd is equal to no Si . This is a contradiction. • Therefore, there exists no such liner order. QED

  13. Comparing Cardinalities of N and P(N) • It is easy to see that N  P(N). • From the diagonalization proof we know that the assertion N  P(N) does not hold. • So, the cardinality of N is strictly smaller than that of P(N), written as N < P(N).

  14. Wikipedia entry

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