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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.2011-05-23
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    @Apostolos: See my answer for that comment.2011-05-23

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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)?2011-05-23
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    in kunen, exponential cardinals define as a set of functions.2011-05-23
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    @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.2011-05-23
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    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?2011-05-23
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    @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.2011-05-23
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    yours minor addition is very useful for me.I am so sorry for basic question but i want to learn set theory. also thanks for books.2011-05-23
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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.2011-05-23
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    @Asaf: The question I found in Kunen's book asks to prove that $|H(\kappa)|=2^{<\kappa}$2011-05-23
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    for $\kappa>\omega$.2011-05-23
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    @Apostolos: p. 147, question (4). It requires $\kappa>\omega$.2011-05-23
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    @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?2011-05-23
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    @Apostolos: You are. I said two things: (i) I have added to the title $\kappa>\omega$; (ii) I address the issue of $\kappa$ being uncountable and inaccessible in my answer as well.2011-05-23
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    @Asaf:Oh, I see. Nevermind then, sorry for that.2011-05-23
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    if lambda less than kappa, is it imply that 2 to lambda is also less than kappa. still the set of 2 to kappa not clear for me?2011-05-23
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    Note that V(κ)=H(κ) if and only if κ=ω or κ is inaccessible. why you say it? k is not strong inaccessible!2011-05-23
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    @user10806: Inaccessible means strongly inaccessible, otherwise we distinguish by saying *weakly* inaccessible. However this is not 100% standard and I should have been more elaborate.2011-05-23
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    @user: It's not necessary that $2^\lambda$ is less than $\kappa$.2011-05-23