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For $\kappa \geq \omega$, which is a cardinal, then how to show $\kappa^{cf(\kappa)}>\kappa$?

My idea: When $\kappa=\aleph_\omega$, then $cf(\kappa)=\omega$, is $\kappa^\omega>\kappa$? It seems $\kappa^{cf(\kappa)}>\kappa$ is wrong.

Could you help me?

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1 Answer 1

up vote 6 down vote accepted

This is known as Koenig's theorem.

In this thread there are several answers about the proof that if $A_i<B_i$ for all $i\in I$ then $|\bigcup A_i|<|\prod B_i|$.

Apply this result to the case where $I=\operatorname{cf}(\kappa)$, $\langle A_i\mid i<\kappa\rangle$ is a partition of $\kappa$ into sets of cardinality $<\kappa$, and $B_i=\kappa$ for all $i$. We have, if so:

$$\kappa=\left|\bigcup_{i\in I}A_i\right|<\left|\prod_{i\in I}B_i\right|=\kappa^{\operatorname{cf}(\kappa)}$$

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Could you tell me what is wrong about my idea? –  Simple Mar 12 '13 at 9:10
1  
@Simple: You didn't really write your idea. You just stated that for $\kappa=\aleph_\omega$ it seems wrong that $\aleph_\omega^\omega>\aleph_\omega$. –  Asaf Karagila Mar 12 '13 at 9:15
    
My puzzeling is: $\aleph_\omega^\omega=\aleph_\omega$ could witness that Koenig's theorem seems not right. –  Simple Mar 12 '13 at 9:18
    
@Simple Note that the arithmetic of cardinals acts different from it of ordinals. E.g. $2^{\aleph_0} > \sup\{2^n|n<\aleph_0\}=\aleph_0$. –  Popopo Mar 12 '13 at 9:43
    
@Simple: In mathematics we often have natural intuitions which are almost usually turn out to be wrong when it comes to infinite objects. It is best to stick by the definitions and follow proofs closely. Note that ordinal arithmetics does not coincide with cardinal arithmetics on infinite cardinals, and Popopo points out the definition of exponentiation is especially incompatible between the two notions. $\aleph_{\aleph_0}^\omega>\aleph_\omega$ in cardinal exponentiation, whereas $\omega_\omega^\omega=\omega_\omega$ for ordinal exponentiation. (That's a lot of $\omega$ right there!) –  Asaf Karagila Mar 12 '13 at 9:47

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