# Is $\kappa^\lambda=2^\lambda$($2 \le \kappa<\lambda$,$\lambda$ infinite) valid in set models of ZF?

Let $2 \le \kappa<\lambda$(both cardinal numbers), in which $\lambda$ is infinite. Then these formula as follows hold where in ZFC:

1. $\lambda+\kappa=\lambda$

2. $\lambda\cdot\kappa=\lambda$

3. $\kappa^\lambda=2^\lambda$

However, if $\lambda$ is not Dedekind-infinite, then 1,2 fail.

But for 3, it's not quite clear.

To prove it. Obviously $2^\lambda\le\kappa^\lambda$; for the other direction, I only got $\kappa^\lambda\le 2^{\kappa \cdot \lambda}$, but $2^{\kappa \cdot \lambda}=2^{\lambda}$ seems not valid.

So my question: Is 3 also valid in all set models of ZF?

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Note that in ZFC $\lambda^\kappa\neq\lambda$ for many pairs of $\kappa<\lambda$. The third, if so, does not need AC to fail in order to fail. It just fails a lot. One example would be $\kappa=\aleph_0$ and $\lambda=\aleph_\omega$. Another would be the failure of CH and taking $\aleph_1^{\aleph_0}$. –  Asaf Karagila Dec 2 '12 at 14:04
@AsafKaragila Okay, I'm going to correct it. –  Popopo Dec 2 '12 at 14:13
@P.., a dozen edits in a matter of minutes floods the front page with old questions, and is generally seen as a bad idea. Please, take it easy. –  Gerry Myerson Mar 21 '13 at 12:22
@GerryMyerson: Sorry about that. I didn't think it through. –  P.. Mar 21 '13 at 14:04

Assuming that $\lambda\cdot\kappa=\lambda$, yes -- as the proof follows through immediately. In particular when the two are ordinals then it is true.
However for general cardinals this may be false. For example if $\lambda$ is the cardinal of an amorphous set then $2^\lambda$ is Dedekind-finite. It follows that $3^\lambda$ is strictly larger than $2^\lambda$, otherwise we could have omitted some of the functions and retain the same cardinality, which would imply that $2^\lambda$ is Dedekind-infinite.
@Popopo: There are several ways. The simplest in this case is the direct method. Show that if there was a countable collection of subsets then you may assume they are all finite (note that the finite and infinite subsets of an amorphous sets stand in bijection) and therefore the union is infinite. Without loss of generality you can therefore assume that the union is the entire amorphous set; fix an enumeration of these sets, $X_n$. Let $A_e$ be the sets which appear for the first time in the in an even-index set, and $A_o$ for odd-index sets. Show that these are disjoint infinite sets. –  Asaf Karagila Dec 2 '12 at 15:46
Why $A_e$ and $A_o$ are disjoint? –  Popopo Dec 2 '12 at 15:55