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I could not see $|H(\kappa)| > 2^{<\kappa}$.

It is a question in Kunen book. The other part is answered, this part may be clear but I could not see.

Also, $2^{<\kappa}$ is not clear for me. Is it a set of functions which from $\kappa$ to 2?

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You probably mean $\geq$ rather than $>$ since the latter is not true. –  Apostolos May 23 '11 at 8:42
    
@Apostolos: See my answer for that comment. –  Asaf Karagila May 23 '11 at 8:44
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2 Answers 2

Firstly, $\displaystyle 2^{<\kappa}=\bigcup_{\lambda<\kappa}2^\lambda$.

Secondly, the set $H(\kappa)$ is the set of all elements whose transitive closure is of cardinality less than $\kappa$.

Note that $V(\kappa)=H(\kappa)$ if and only if $\kappa=\omega$ or $\kappa$ is inaccessible.

The claim, however, is not true in the countable case, as $H(\kappa)=V(\kappa)$ which is a countable set, and $2^{<\omega}$ is all the finite subsets of $\omega$, which is once again countable. (This is true in the strongly inaccessible case as well, however this claim is stronger than one can prove in ZFC, as the existence inaccessible cardinals cannot be proved in ZFC)

Note that if $A\in H(\kappa)$ then every subset of $A$ is also in $H(\kappa)$, i.e. $\mathcal P(A)\subseteq H(\kappa)$.

Therefore for every ordinal $\alpha<\kappa$ we have that $\mathcal P(\alpha)\subseteq H(\kappa)$

Therefore we have that $\displaystyle 2^{<\kappa} = \bigcup_{\alpha<\kappa}\mathcal P(\alpha)\subseteq H(\kappa)$, as needed.

A very minor addition:

The set $2^A$ is the same as $\mathcal P(A)$ by the function which takes $f\in 2^A$ to the set $\{x\in A\mid f(x)=1\}$, the fact that this is a bijection is left as an exercise.

Also Kunen uses the notation $R(\kappa)$ for the sets of rank $<\kappa$, where it is also standard to use $V_\kappa$ for the same set.

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how can you say 2 to less kappa = union of P(alpha)? –  user10806 May 23 '11 at 11:57
    
in kunen, exponential cardinals define as a set of functions. –  user10806 May 23 '11 at 11:58
    
@user10806: It seems that you are starting with set theory, as the homeomorphism between $\mathcal P(A)$ and $2^A$ is very basic and trivial, I will reiterate from your previous question: Kunen is not the best choice for entering set theory. –  Asaf Karagila May 23 '11 at 12:11
    
yes, it is the first time, but before I knew very little thing about set theory. What is your advise when i read kunen? which book is more useful? –  user10806 May 23 '11 at 12:21
    
@user10806: You can try Jech's Set Theory (the first part should be detailed enough), or Levy's Basic Set Theory, Halmos' Naive Set Theory, and so on. Kunen's book is very good for forcing, which is a more advanced technique, for the basics I would go by something else. –  Asaf Karagila May 23 '11 at 12:24
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You probably mean $\geq$ rather than $>$ and by "The other part is answered" I assume you mean this question.

First of all $2^{<\kappa}$ is used to denote the set $\bigcup_{\lambda<\kappa}2^\lambda$. Now the answer is fairly trivial since every ordinal less than $\kappa$ is in $H(\kappa)$ and thus each of its subsets is in $H(\kappa)$.

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Apostolos: The question in Kunen is for $\kappa>\omega$ as I have added to the title, and referenced in my answer. –  Asaf Karagila May 23 '11 at 9:27
    
@Asaf: The question I found in Kunen's book asks to prove that $|H(\kappa)|=2^{<\kappa}$ –  Apostolos May 23 '11 at 10:22
    
for $\kappa>\omega$. –  Apostolos May 23 '11 at 10:28
    
@Apostolos: p. 147, question (4). It requires $\kappa>\omega$. –  Asaf Karagila May 23 '11 at 10:29
    
@Asaf: Yes, I tried to edit it later on but the site said it was too late. Still, the strict inequality is false for every cardinal. Or am I misinterpreting your comment? –  Apostolos May 23 '11 at 10:44
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