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We know that, for $\alpha$, $\beta$, $\gamma$ cardinals, if $\alpha \leq \beta$, then $\alpha^\gamma \leq \beta^\gamma$, and, in general, the usual operations defined on cardinals are compatible with the order relation.

I have been given the task to determine if the reversed implication holds, i.e. $\alpha^\gamma \leq \beta^\gamma \Rightarrow \alpha \leq \beta$.

I think the answer is no, given that $4^{\aleph_0}=2^{\aleph_0}=\mathfrak{c}$ and $4>2$. But what happens if we have a strict inequality, does this implication hold?

$$\alpha^\gamma < \beta^\gamma \Rightarrow \alpha < \beta.$$

I proved that the above implication is true for $\gamma=$ finite.

P.S. I am asking this question, because I realize an interesting thing: if we replace the exponentiation operator with multiplication or addition, the implication does hold, that is, $$\alpha+\gamma < \beta+\gamma \Rightarrow \alpha < \beta.$$ $$\alpha\cdot\gamma < \beta\cdot\gamma \Rightarrow \alpha < \beta.$$

Also, there are counterexamples for division and subtraction, $\mathfrak{c}+\aleph_0=\mathfrak{c}\cdot\aleph_0=\mathfrak{c}+1=\mathfrak{c}\cdot1=\mathfrak{c}$, and $\aleph_0>1$.

Paul Rebenciuc
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    $\alpha^{\gamma} < \beta^{\gamma} \Rightarrow \alpha < \beta$ is the contraposition of (hence is equivalent to) $\alpha \geq \beta \Rightarrow \alpha^{\gamma} \geq \beta^{\gamma}$, which is true. – Aphelli Mar 20 '21 at 21:17

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Yes, it holds. Look at the contrapositive of your statement.

Jordan Mitchell Barrett
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