0

Assuming the link: Infinite set as union of disjoint countable sets. implies the following proposition:

Prop: If $\{A_i:i\in \mathbb{N}\}$ is a collection of disjoint infinite sets of the same cardinality, then $$|\cup_i A_i|=|A_1|.$$ I would like to show it implies the next claim below. Given two sets $A$ and $B$, we define $$|A|\cdot |B|:=|A\times B|.$$ Claim: If $T$ is an infinite set, then $$\aleph_0\cdot |T|=|T|.$$

northcity4
  • 540
  • 1
    The linked answer implies the latter claim, although both are true – Rushabh Mehta Feb 28 '21 at 14:14
  • @DonThousand - I have edited my question since you mentioned both results are true. I am now more curious if the former latter result (my prop) can be used to imply my former first result (the claim). – northcity4 Feb 28 '21 at 20:29

1 Answers1

2

$\aleph_0 \times T$ is a countable collection of disjoint sets of the same cardinality, ( $A_i = \{i\} \times T$ ).

David Hartley
  • 1,823
  • Could you clarify? Also, did I misinterpret $\aleph_0\cdot |T|$? Should this be something like $|\mathbb{N}\times T|$? – northcity4 Mar 01 '21 at 02:55
  • 1
    @northcity4: Given your definition, $\aleph_0\cdot|T|=|\aleph_0\times T|$. Whether $\aleph_0=\Bbb N$ or merely has the same cardinality depends on your definitions; for me $\aleph_0=\Bbb N$, so $$\aleph_0\cdot|T|=|\Bbb N\times T|=\bigcup_{n\in\Bbb N}({n}\times T),,$$ where the sets ${n}\times T$ clearly all have cardinality $|T|$. – Brian M. Scott Mar 01 '21 at 04:36
  • @BrianM.Scott So would a solution look like this: Using $\mathbb{N}$ as our set of size $\aleph_0$ and $A_i$ as in David's answer, we have that $|\cup_i A_i|=|\mathbb{N}\times T|$ since the identity function is a bijection and $|\cup_i A_i|=|T|$ as a consequence of Zorn's lemma (proposition at top)? – northcity4 Mar 01 '21 at 05:19
  • @northcity4: I was simply using the prop. at the top, since I didn’t know how you proved it, but yes, that’s exactly what I was suggesting. – Brian M. Scott Mar 01 '21 at 05:25
  • @Brian: While formally not incorrect, it is very bad form to write $\aleph_0\times T$ when considering $\Bbb N\times T$, or $\omega\times T$. – Asaf Karagila Mar 02 '21 at 09:16