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The fact that for infinite sets $A,B$, $\text{card}(A\times B)=\max\{\text{card}(A),\text{card}(B)\}$ depends on the axiom of choice or Zorn's lemma, so presumably, it can be proven by transfinite induction. Here is my attempt to prove this by induction.

Note that this is obviously true for $|A|=|B|=\aleph_0$. Now suppose for some cardinal $\kappa$, $\text{card}(A\times B)=\max\{\text{card}(A),\text{card}(B)\}$ is true for all $|A|,|B|<\kappa$. Let $\alpha<\kappa$ be an ordinal. Then by induction assumption, we have $|\alpha\times\alpha|=|\alpha|$. So $$ |\alpha\times \kappa|=|\alpha\times \alpha\times \kappa|=\left|\bigcup_{\beta<\alpha} (\{\beta\}\times \alpha\times \kappa)\right|\\ \left|\alpha\times \kappa\times \bigcup_{\beta<\alpha} (\{\beta\})\right|=|\alpha\times\kappa||\alpha|. $$ But this is very far from the desired $\kappa.\kappa=\kappa$. How can I finish the proof?

Ma Joad
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1 Answers1

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You can write $$\kappa \times \kappa = \left( \bigcup_{\alpha < \kappa} \alpha \right) \times \left( \bigcup_{\beta < \kappa} \beta \right) = \bigcup_{\alpha,\beta < \kappa} \alpha \times \beta \le \bigcup_{\alpha,\beta < \kappa} (\alpha \cup \beta) \times (\alpha \cup \beta)$$ Now apply the induction hypothesis.

Clive Newstead
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  • Oh! Could you add more details about the last step, "apply the induction hypothesis"? The fact that $|(\alpha\cup\beta)\times(\alpha\cup\beta)|=|\alpha\cup\beta|$ doesn't seem to suggest that the union has cardinality $\kappa$. See here. Thank you! – Ma Joad Nov 18 '19 at 08:05