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

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!

• "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? May 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. May 23 at 20:37
• I have found an answer here given by Brian M. Scott $:$ math.stackexchange.com/q/89104/512080 May 23 at 20:39
• @ArturoMagidin$:$ You have given the same link as that of mine. May 23 at 20:42
• 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. May 23 at 20:43

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.