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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!

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    $\begingroup$ "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? $\endgroup$
    – Alex Kruckman
    May 23 at 20:33
  • $\begingroup$ @AlexKruckman$:$ Sorry I made a typo. I meant to say this is true when $|A| = \aleph_0.$ Fixed it now. Thanks. $\endgroup$
    – Anacardium
    May 23 at 20:37
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    $\begingroup$ I have found an answer here given by Brian M. Scott $:$ math.stackexchange.com/q/89104/512080 $\endgroup$
    – Anacardium
    May 23 at 20:39
  • $\begingroup$ @ArturoMagidin$:$ You have given the same link as that of mine. $\endgroup$
    – Anacardium
    May 23 at 20:42
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    $\begingroup$ 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. $\endgroup$
    – Alex Kruckman
    May 23 at 20:43

1 Answer 1

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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.

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