 # 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 for some n in N. Prove that $|A|=|N|$.
You can still ask an expert for help

## Want to know more about Discrete math?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Nola Robson

Given that, A is collection of sets such that x in A if and only if $X\subset Nand|X|=n$ 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 $\left(K+1{\right)}^{th}$ 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, $A=\bigcup _{n-1}^{\mathrm{\infty }}{A}_{n}$.

$⇑$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.