The product of cardinals

Question

Let $\gamma_i$ be infinite cardinal numbers for $i=1,2,3$ such that $\gamma_i<\gamma_3$ for $i=1,2$.

Is it true that $\gamma_1.\gamma_2<\gamma_3$?

Answer 1

This is not true. For example let $\gamma_1=2,\gamma_2=3,\gamma_3=4$, then $\gamma_1\times \gamma_2>\gamma_3$.

Without loss of generality we assume $\gamma_1<\gamma_2$. I think what you want to show is for infinite cardinals, $\gamma_i\times \gamma_i=\gamma_i$. Then you have $\gamma_1\times \gamma_2=\gamma_2<\gamma_3$. But to show $\gamma_i\times \gamma_i=\gamma_i$, if $\gamma_i>\mathbb{N}$ may be tricky in general, even though it can be done for $\mathbb{N},c$ in a direct manner. The wikipage says it would be dependent on axiom of choice. I assume others at Math.SE will be more knowledgable on this matter, for I do not know how to show it even if given choice.

Answer 2

For finite cardinals this is certainly wrong. Consider $\gamma_1 = \gamma_2 = 2$ and $\gamma_3 = 3$.

For infinite cardinals it turns out to be correct, because then a theorem of Hessenberg (cf. also Question about proof of Hessenberg: $\kappa \cdot \lambda = \lambda$ ) states $\kappa \cdot \kappa = \kappa$ so that (w.l.o.g. assume $\gamma_1 \leq \gamma_2$) $\gamma_1 \cdot \gamma_2 \leq \gamma_2 \cdot \gamma_2 = \gamma_2 < \gamma_3$.