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|.

Dolly Robinson 2021-03-02 Answered
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 XN and|X|=n for some n in N. Prove that |A|=|N|.
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Nola Robson
Answered 2021-03-03 Author has 94 answers

Given that, A is collection of sets such that x in A if and only if XNand|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=n1An.

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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