# 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|.

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Discrete math
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 $$|A| = |N|$$.

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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