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What is the cardinality of $A \times \mathbb N$ if the cardinality $\left \lvert A \right \rvert \geq \aleph_0\ $?

For $|A| = \aleph_0$ I can see that $A \times \mathbb N$ has the same cardinality as that of $A.$ So I intuitively feel that it is true for the case $|A| \geq \aleph_0$ as well but I can't conclude it analytically.

Could anyone help me out here?

Thanks!

Anacardium
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    "For $|A|\leq \aleph_0$ I can see that $A\times \mathbb{N}$ has the same cardinality as that of $A$." Really? Even when $A$ is finite? – Alex Kruckman May 23 '23 at 20:33
  • @AlexKruckman$:$ Sorry I made a typo. I meant to say this is true when $|A| = \aleph_0.$ Fixed it now. Thanks. – Anacardium May 23 '23 at 20:37
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    I have found an answer here given by Brian M. Scott $:$ https://math.stackexchange.com/q/89104/512080 – Anacardium May 23 '23 at 20:39
  • @ArturoMagidin$:$ You have given the same link as that of mine. – Anacardium May 23 '23 at 20:42
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    Arturo Magidin's comment is one that appears automatically when someone votes to close your question as a duplicate. Since you indicated that the linked answer satisfies you, I have also voted to close as a duplicate. Let me just point out that Brian M. Scott's answer didn't provide a proof. But it's a basic theorem of cardinal arithmetic that if $0 < \kappa \leq \lambda$ and $\lambda$ is infinite, then $\kappa\cdot \lambda = \lambda$. You can find a proof in just about any book on set theory. Let me also add (in case you care about such things), that the proof requires the Axiom of Choice. – Alex Kruckman May 23 '23 at 20:43
  • @AlexKruckman$:$ Hmm I will look into it. Although I am not a person of set theory. I have devised a proof of the fact that any two ONB of a Hilbert space have the same cardinality and I have used this fact there. I am just curious to know what's the bijection? – Anacardium May 23 '23 at 20:47
  • @AlexKruckman$:$ For the proof of the fact that "At most countable union of at most countable sets is at most countable" (e.g. this technique is used to show that $\mathbb Q$ is at most countable) we used Cantor's diagonal argument which is not available here. That's why I want to know the bijection between $A$ and $A \times \mathbb N$ when $|A| \gt \aleph_0.$ – Anacardium May 23 '23 at 20:49
  • It's not trivial. The key thing is to use transfinite induction on $\alpha$ to show that $\aleph_\alpha\times \aleph_\alpha = \aleph_\alpha$ for all $\alpha$, by well-ordering the cartesian product $\aleph_\alpha\times \aleph_\alpha$ in order type $\aleph_\alpha$. Then $|A| \leq |A\times \mathbb{N}|\leq |A\times A| = |A|$ and we're done by Cantor-Schroeder-Bernstein. For details, I suggest you look in a textbook. Hrabacek and Jech Introduction to Set Theory is a good choice. – Alex Kruckman May 23 '23 at 20:57
  • @AlexKruckman$:$ Thanks for your valuable suggestion. I will surely look into it and get back to you if I find anything there which is not understandable. Thanks again for your kind help. – Anacardium May 23 '23 at 21:01

1 Answers1

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The answer depends on the axiom of choice ($\mathsf{AC}$). If you assume $\mathsf{AC}$, then as mentioned in the answer you linked to in comments, $|A\times\mathbb N|=|A|$.

Suppose now $\mathsf{AC}$ fails (badly), so there is an infinite Dedekind finite set (I mean, the existence of such a set is relatively consistent with $\mathsf{ZF}$), say $B$, which we may assume disjoint from $\mathbb N$, and take $A=B\cup\mathbb N$. We have $|A|\ge|\mathbb N|$, but $$|A\times \mathbb N|=|B\times\mathbb N|+|\mathbb N|\ge|B\times\mathbb N|>|B|+|\mathbb N|=|A|.$$ To see the key inequality, note that $B\times\mathbb N=(B\times\{0\})\cup(B\times\{1\})\cup(B\times(\mathbb N\smallsetminus\{0,1\})$, so $$|A\times\mathbb N|\ge|B|+|B|+|\mathbb N|>|B|+|\mathbb N|$$ since $B$ is Dedekind finite.

In even more detail, if $f:B\to B\cup\mathbb N$ is injective, then $f[B]\cap \mathbb N$ is finite, so $f$ maps almost all of $B$, but a finite set, into $B$. Thus, if $B_0$ and $B_1$ are disjoint copies of $B$, and $f:B_0\cup B_1\cup\mathbb N\to B\cup\mathbb N$ is injective, then $f[B_0]\cap B$ and $f[B_1]\cap B$ are disjoint and infinite subsets of $B$, and $f[B_1]\smallsetminus B$ is finite, which means that $B$ contains a proper subset of size $|B|$ after all, a contradiction.

See here for more on infinite Dedekind finite sets.

Andrés E. Caicedo
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