For each $ n \in \mathbb{N}$, ($\mathbb{N}$ being the natural numbers set), let $ C_{n} $ be the set of counting numbers from $ 1 $ to $ n $, so that $ C_{n} = \{1, 2, …, n\}, $ and let $ P(A) $ be the power set of of a set $ A $. My question is whether $ \bigcup_{n=1}^{\infty}{P(C_{n})} = P(\mathbb{N}) $ ?
I believe this seems true because on the LHS above, we have a union of an infinite number of power sets of the counting number sets, $ C_{1}, C_{2}, …$, so this should give us all of the subsets of the natural numbers $\mathbb{N}$, which by definition is $P(\mathbb{N})$ on the RHS.
However in the book they argue that $\bigcup_{n=1}^{\infty}{P(C_{n})}$ consists of all finite sets of counting numbers, while $P(\mathbb{N})$ contains all subsets of $\mathbb{N}$, including infinite ones, therefore they are not equal. However I don't understand why they are saying that $\bigcup_{n=1}^{\infty}{P(C_{n})}$ consists of only finite sets?
Best Answer
Note that if $A\in\bigcup_{n=1}^\infty\mathcal P(C_n)$, then there exists some $n$ such that $A\in\mathcal P(C_n)$. Therefore $A\subseteq\{1,\ldots,n\}$ and so $A$ must be finite.
So indeed $\Bbb N$ itself is not in this union, despite the fact that the union is infinite, since $\Bbb N$ is not a finite set for itself.