Does $2^{\mathfrak m}=2^{\mathfrak n}$ imply $\mathfrak m=\mathfrak n$?

Question

Suppose $\mathfrak m$ and $\mathfrak n$ are infinite cardinals. Does $2^{\mathfrak m}=2^{\mathfrak n}$ imply $\mathfrak m=\mathfrak n$?

This is not a theorem of ZFC (unless ZFC happens to be inconsistent). — André Nicolas, Jul 22 '13 at 19:16
Do you bother searching the site? This has been asked at least twice before. — Asaf Karagila, Jul 22 '13 at 19:17
Also: http://math.stackexchange.com/questions/420400/sets-question-without-zorns-lemma/ and http://math.stackexchange.com/questions/74477/does-2x-cong-2y-imply-x-cong-y-without-assuming-the-axiom-of-choice — Asaf Karagila, Jul 22 '13 at 19:34
@AsafKaragila: Thanks for the links! I honestly tried to find it, but I worded it in a different way, so couldn't find any. — mathreader, Jul 22 '13 at 19:57

score 5 · Accepted Answer · answered Jul 22 '13 at 19:16

5

This is independent of ZFC. It is implied by GCH for example, but there exist models where (say) $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$.

answered Jul 22 '13 at 19:16

Chris Eagle

Thank you for the answer! Could you possibly direct me to any source that states this fact? – mathreader Jul 22 '13 at 19:28
Never mind, I got the links provided above. – mathreader Jul 22 '13 at 19:58

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