Cardinality of a collection consisting of subsets of an infinite set

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Sorry I didn't learn rigorous set theory and my question comes from a proof in the theory of modules, and so if you can avoid using notations only appearing in set theory, I will be so grateful.


Suppose we have an infinite set $Y$. And we define $K(Y)=\{Y_1\subseteq Y;|Y_1|<\infty\}$.

How do we prove $|K(Y)|\leq|Y||\mathbb Z|$ and $|Y||\mathbb Z|\leq|Y|?$

I only know $|\mathbb Z|=\aleph_0$ and pretty much nothing else...

Thanks for help.

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I understand that $K(Y)$ is the set of finite subsets of $Y$. Let $K_n(Y)$ be the set of subsets of $Y$ with exactly $n$-elements. Then $\displaystyle K(Y) = \bigcup_{n\in\mathbb{N}} K_n(Y)$.

Second, let $\phi_n:K_n(Y)\to Y\times Y\times\cdots\times Y$ (n-fold cartesian product) be the map defined by $$\forall A=\{a_1,\dots,a_n\} \in K_n(Y) \hspace{9mm} \phi_n(A) = (a_1,\dots,a_n)$$ It is up to the reader to show that $\phi_n$ is well-defined and one-to-one. Thus, $|Y|\leq |K_n(Y)|\leq |Y\times Y\times\cdots\times Y| = |Y|$ so $|K_n(Y)| = |Y|$.

Finally, $\displaystyle K(Y) = \bigcup_{n\in\mathbb{N}} K_n(Y)$ as a disjoint union implies that $$|K(Y)| = \sum_{n\in\mathbb{N}}|K_n(Y)| = |\mathbb{N}||Y|$$