Let $S$ be any infinite set. Is it true that $S \times \mathbb{Z}$ has the same cardinality as $S$?

Asked Feb 28 '20 at 20:13

Active Feb 28 '20 at 20:13

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Consider any set $S$ which is not finite. Is it true that $S \times \mathbb{Z}$ has the same cardinality as $S$?

I thought the answer is "yes" and I thought it would be a trivial problem. But as I think about it, I have a nagging feeling it requires something like the Axiom of Choice and now, I'm not even sure the answer is "yes."

One cursory idea I had was to consider an injection of $\{s\} \times \mathbb{Z} \to S$ for each fixed $s \in S$. Somehow, if we can patch these together to form another injection $S \times \mathbb{Z}$, we would be done.

asked Feb 28 '20 at 20:13

inkievoyd

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Well, depending on your understanding of cardinal arithmetic, we know that $$|A \times B| = |A| \cdot |B|$$ which is $\max{|A|,|B|}$ for infinite $A$ or $B$ and both non-empty. This does assume the axiom of choice. – PrincessEev Feb 28 '20 at 20:18
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@Randall $S$ needn't be countable. – inkievoyd Feb 28 '20 at 20:21
@Eevee Trainer I see; I wasn't aware of this fact. Thanks! – inkievoyd Feb 28 '20 at 20:21
@inkievoyd woah,I totally missed that. – Randall Feb 28 '20 at 20:33
This question is a duplicate, but it's not a duplicate of the question that was linked in the closure. Instead this question or this one provide proofs of $|S\times \Bbb Z|=|S|$. – Vsotvep Feb 29 '20 at 11:39

Let $S$ be any infinite set. Is it true that $S \times \mathbb{Z}$ has the same cardinality as $S$?

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