Using cardinatility of sets in discrete mathematics the value of N is real numbers Currently using elements of discrete mathematics by Richard Hammack chapter 18 Let A be a collection of sets such that X in A if and only if X supset N text{and} |X| = n for some n in N. Prove that |A| = |N|.

Using cardinatility of sets in discrete mathematics the value of N is real numbers Currently using elements of discrete mathematics by Richard Hammack chapter 18 Let A be a collection of sets such that X in A if and only if X supset N text{and} |X| = n for some n in N. Prove that |A| = |N|.

Question
Discrete math
asked 2021-03-02
Using cardinatility of sets in discrete mathematics the value of N is real numbers Currently using elements of discrete mathematics by Richard Hammack chapter 18 Let A be a collection of sets such that X in A if and only if \(X \supset N\ \text{and} |X| = n\) for some n in N. Prove that \(|A| = |N|\).

Answers (1)

2021-03-03
Given that, A is collection of sets such that x in A if and only if \(X \subset N and |X|=n\) for some n in N. Here, it means that A is collection of finite subset of N. Now need to show \(|A|=|N|\). It is sufficient to show A is countable. Use induction method to show A is countable. The number of subsets of N with cardinality 1 is finite. It is trivially true. Assume that, the number of subsets of N with cardinality K is finite. Need to show the number of subsets of N with cardinality \(K +1\) is finite. The number of subsets of N with cardinality K +1 is exactly two times of number of subsets of cardinality K as each subset has two choices either \((K +1)^{th}\) term belongs to the set or not. Thus, the number of subsets of N with cardinality \(K +1\) is finite. Hence, for all n in N the number of subsets of N with cardinality n is finite. So, A, is equal to collection of subsets of N with cardinality n . Therefore, \(A=\bigcup_{n-1}^{\infty}A_{n}\). Tt is known that, countable union of countable set is countable. Here, A_n is countable for each n in N . So, A is countable Hence, \(|A|=|N|\) is proved.
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