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For infinite cardinals $d$ and $e$, does $2^d \geq 2^e$ imply that $d \geq e$?

I would like a proof or a counterexample. Background so far: The cardinals form a chain under $\leq$; if $e$ is an infinite cardinal with $d \leq e$, then $de = e$ and $d + e = e$; for any cardinals we have $d^{e_1 + e_2} = d^{e_1} d^{e_2}$, $(d_1d_2)^e = d_1^e d_2^e, (d^e)^f = d^{ef}$.

Thank you in advance for any hints or answers.

Anu
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1 Answers1

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If you do not assume GCH, then it can happen that $2^{\aleph_0} = 2^{\aleph_1} = \aleph_2$, which gives you a counter-example $d = \aleph_0$ and $e = \aleph_1$.

Kenny Lau
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  • Although, if we were to change $\geq$ into $>$, or swap the direction of the implication, it would be true, right? – Arthur Jan 10 '18 at 13:27
  • Sure, as its contrapositive can be easily proved (which involves, indeed, swapping the implication). – Kenny Lau Jan 10 '18 at 13:28
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    There is a proof of "A surjects onto B implies P(B) injects into P(A)" that does not even use choice, so this proves $|A|\ge|B| \implies 2^{|A|}\ge2^{|B|}$ – Kenny Lau Jan 10 '18 at 13:30
  • The proof is by taking the preimage of the surjection. – Kenny Lau Jan 10 '18 at 13:30
  • Wasn't the continuum hypothesis established, though? What if we assume that $\aleph_1 = c$? – Anu Jan 10 '18 at 13:33
  • @Anu the continuum hypothesis has been proved to be independent of the ZFC axioms. Even if you assume CH, you can still deny GCH and come up with some counter-example. – Kenny Lau Jan 10 '18 at 13:34
  • As @Arthur asked, could you explain to me what happens if I swap $\geq$ with $>$? – Anu Jan 10 '18 at 13:43
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    Its contrapositive becomes $d \le e \implies 2^d \le 2^e$ which I just proved, so it becomes a true statement – Kenny Lau Jan 10 '18 at 13:43