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It seems clear that for $A, B, C$ infinite cardinals with $A > B $ one could define an injection from $B^C \to A^C$ and so $A > B \Rightarrow A^C \ge B^C$, but is the inequality strict and what is the proof ?

After reading the counter example my question is extended and becomes "Are there clear circumstances in which $ A^C = B^C$ and in which $ A^C > B^C$ ?"

Tom Collinge
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1 Answers1

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Counterexample:

$$\aleph_0<2^{\aleph_0}\qquad\text{and}\qquad \aleph_0^{\aleph_0}=2^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}$$

Additionally, if $C$ is sufficiently large, then $A^C$ and $B^C$ are both equal to $2^C$.

Asaf Karagila
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  • Seems to answer:: $A = 2^{\aleph_0}, B = \aleph_0, C = \aleph_0$ – Tom Collinge Jul 18 '15 at 15:02
  • @Tom: Yes. And generally, taking $C\geq\max{A,B}$ would work too. – Asaf Karagila Jul 18 '15 at 15:03
  • I am interested in the answer in the context of the answer to http://math.stackexchange.com/questions/573378/u-otimes-v-versus-lu-v-for-infinite-dimensional-spaces . In the light of your answer I've extended this question. – Tom Collinge Jul 18 '15 at 16:14
  • I can't think of any reasonable and plausible rule in $\sf ZFC$ which will give a simple answer. This has probably much to do with the previous answer, and how much cardinal exponentiation is a wild card when you're not assuming additional hypotheses. – Asaf Karagila Jul 18 '15 at 16:34
  • Any chance an example of $A > B \Rightarrow A^C > B^C$ to demonstrate both possibilities ? – Tom Collinge Jul 18 '15 at 16:44
  • If $A>2^B>B=C$, then the inequality is preserved. – Asaf Karagila Jul 18 '15 at 16:50