How to prove that if $\kappa>\omega$ then $|H(\kappa)|>2^{<\kappa}$ 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? 
 A: 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.
A: 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 $H(\kappa)$.
