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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.

Asaf Karagila
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Sam Wong
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  • What do you mean by $|Y_1|<\infty$? and K(Y) trying to refer to some kind of group ring? – Pymamba Dec 25 '22 at 07:49
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    Did you mean that $Y_1$ was finite – Pymamba Dec 25 '22 at 07:55
  • @Pymamba Sorry for the ambiguity. I wanted to mean that $Y_1$ was finite, where $|Y_1|$ meant the cardinality of $Y_1$. – Sam Wong Dec 25 '22 at 08:14
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    No worries. The technically correct notation would be $|Y_1|<\aleph_0$ and as you probably know it is the cardinality of $\mathbb{N,Z}$, and $\mathbb{Q}$. – Pymamba Dec 25 '22 at 08:18

1 Answers1

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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 $$|K(Y)| = \sum_{n\in\mathbb{N}}|K_n(Y)| = |\mathbb{N}||Y|$$

Onur Oktay
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  • Thanks for the quick answer. I have a question though. ${a_1, ..., a_n}$ is a set which doesn't have an order actually, but $(a_1, ..., a_n)$ is a cartesian product which has a natural order. To make sense of the definition of $\phi_n$, do we need to identify all cartesian products whose coordinates are $a_1, \cdots, a_n$? i.e. We map $K_n(Y)$ to $(Y\times Y \cdots \times Y)/\sim$, where $\sim$ is the equivalence I just mentioned. Thanks. – Sam Wong Dec 25 '22 at 08:40
  • Alternatively, we can enumerate (or label) all elements for every $A_n\in K_n(Y)$ for all $n$. But then we have too many enumerations. I feel like in this situation the Axiom of Choice may get involved in at some points, is it? – Sam Wong Dec 25 '22 at 08:47
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    @SamWong I'm in favor of your equivalence class suggestion, which even makes $\phi_n$ bijection. Each equivalence class contains $2^n$ elements. Thus, these updated $\phi_n$'s also give us $|Y| = |K_n(Y)|$ as above. – Onur Oktay Dec 25 '22 at 08:58