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Let $I$ be an infinite set and each $C_i$ be a nonempty countable (finite or infinite) set. I would like to prove without ordinals/cardinal arithmetic that

$$\left|\bigcup_{i\in I}C_i\right|\leqslant |I|,$$

where $|\cdot|\leqslant|\cdot|$ is thought of only as an order relation based on existence of an injective function.

I have seen this fact justified with cardinal arithmetic, as at the end of this answer. However, I would like to be able to prove this with minimal set-theoretic background, if possible. All proofs I have found justifying cardinal arithmetic, such as the rule $\aleph_\alpha\cdot\aleph_\beta=\aleph_{\max(\alpha,\beta)}$, rely on ordinals. Thus I would rather avoid taking any rules of cardinal arithmetic as given and just show the existence of an injective function $f:\cup_{i\in I}C_i\to I$ directly.

It is fine if the proof uses axiom of choice or Zorn's lemma, as I imagine this is necessary.

Asaf Karagila
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WillG
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  • If you're willing to accept AC or ZL, you might as well bite the bullet and work with ordinals. Their existence is equivalent to AC (under the well-ordering principle). – nomen Apr 21 '20 at 18:40
  • @nomen My main motivation is (lack of) time and preparation. The ordinal route will require first proving basic theorems about ordinals and then that each well-ordered set is isomorphic to a unique ordinal, then that cardinals defined in terms of ordinals make sense, then establishing ordinal arithmetic and deriving cardinal arithmetic. This is several chapters of an intro set theory book, when the result I'm after might be obtainable much quicker! – WillG Apr 21 '20 at 18:45
  • OK, fair enough. I was just saying so, since "typically" the motivation to avoid ordinals is to avoid strong choice axioms. – nomen Apr 21 '20 at 21:31

1 Answers1

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Okay, we are going to do a sequence of reductions, and then I will send you elsewhere for the final details.

  1. It is enough to prove that $|I\times\Bbb N|=|I|$, if this is true, then we can choose for each $i$, an injection from $C_i$ into $\Bbb N$, and for each $x\in\bigcup C_i$, choose $i$ such that $x\in C_i$ (there may be several, you see), then combine these choices into an injection from $\bigcup C_i$ into $I\times\Bbb N$.

  2. It is enough, therefore, to show that there is some set $J$ such that $|I|=|J\times\Bbb N|$. Indeed, if this is true, then $$|I\times\Bbb N|=|(J\times\Bbb{N)\times N}|=|J\times(\Bbb{N\times N})|=|J\times\Bbb N|=|I|.$$

  3. Now we can use Zorn's lemma as in Prove that if $A$ is an infinite set then $A \times 2$ is equipotent to $A$ or Show that an infinite set $C$ is equipotent to its cartesian product $C\times C$ (the latter proves an even stronger statement, but goes through a similar step), or The cardinality of an uncountable set times $\mathbb N$..

Good luck.

Asaf Karagila
  • 393,674