I need to prove, without assuming the Axiom of Choice, that for cardinals $a,b,b'$, if $a\ge 2$ and $b<b'$, then $a^b <a^{b'}$.

I have already proved that for cardinals $c,c',d,d'$, if $c\neq0$, $c\leq c'$ and $d\leq d'$, then $c^d\leq c'^{d'}$. So I think I need to assume for a contradiction that $a^b=a^{b'}$ and find a contradiction.

Of course, I couldn't come up with a counterexample, so I guess it is true.

  • $\begingroup$ Why do you think this is true? Also, $a'$ does not appear anywhere in your question except in $a\leq a'$. $\endgroup$ – Asaf Karagila Apr 14 at 9:25
  • 2
    $\begingroup$ $2<3$, but $\aleph_0^2=\aleph_0^3$. $\endgroup$ – TonyK Apr 14 at 9:27
  • $\begingroup$ I couldn't come up with a counterexample. I meant that in the case that $a=a'$, the result I have already proved implies the inequality I need, but weakly; hence I thought to proceed by assuming for a contradiction that $a^b=a^{b'}$. I'll clarify the notation. $\endgroup$ – Davide Apr 14 at 9:29
  • $\begingroup$ Ooh thanks @TonyK, that was more straightforward than I expected $\endgroup$ – Davide Apr 14 at 9:34
  • $\begingroup$ @bof: How do you prove this without choice, though? :) $\endgroup$ – Asaf Karagila Apr 14 at 9:46

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