Equivalents of Jensen's diamond principle Let $\kappa>\omega$ be regular. The principle $\diamondsuit_\kappa$ is as follows: There exists a sequence $\langle X_\alpha\mid \alpha<\kappa\rangle$ such that each $X_\alpha\subseteq\alpha$ and for any $A\subseteq\kappa$, the set $\{\alpha<\kappa\mid A\cap\alpha=X_\alpha\}$ is stationary in $\kappa$.
I'm trying to show that the above is equivalent to the following:
(1) For fixed $\gamma<\kappa$, we have a sequence $\langle Y_\alpha\mid \alpha<\kappa\rangle$ such that each $Y_\alpha\subseteq\alpha\times\gamma$ and for any $A\subseteq\kappa\times\gamma$, the set $\{\alpha<\kappa\mid A\cap(\alpha\times\gamma)=Y_\alpha\}$ is stationary in $\kappa$.
(2) We have a sequence $\langle Z_\alpha\mid \alpha<\kappa\rangle$ such that each $Z_\alpha\subseteq\alpha\times\alpha$ and for any $A\subseteq\kappa\times\kappa$, the set $\{\alpha<\kappa\mid A\cap(\alpha\times\alpha)=Z_\alpha\}$ is stationary in $\kappa$.
Not quite sure how to get started on this problem. I'm new to the subject and this is a bit overwhelming. Any help is appreciated
 A: For one direction of (2). First, let $<_L$ be the $<$-lexicographic order on $k\times k.$  For $(a1,b1), (a2,b2)$ in $k\times k$ define $$(a1,b1)<^* (a2,b2)\; \text {iff }$$ $$(i) \max (a1,b1)<\max (a2,b2),\;\text {OR}$$ $$\text { (ii) }  \max (a1,b1)=\max (a2,b2) \text { and }  (a1,b1)<_L (a2,b2).$$ Then $<^*$ is a well-order. And the cardinal of the set of $<^*$-predecessors of any $x\in k \times k$ is less than $k$. So $(k,<)$ and $(k\times k,<*)$ are order-isomorphic. 
Let $f:k \to k\times k$ be the order-isomorphism.
Lemma. The set $S=\{a\in k :f(a)=(a,a)\}$ is club in k.
Proof. (i).  $S\ne \phi$ because $0\in S.$ 
(ii).To show that $\sup S=k,$ let $a_0\in S.$ We will find $b\in S$ with $a_0\leq b.$
Observe that  for all $x\in k,$ we have $f(x)\leq^* (x,x) $ by transfinite induction. (Proof by contradiction.)
Now for $a_0\in k,$ take $\{a_n:0<n<\omega\} \subset k ,$  such that  (1)  if $f(a_n)=(a_n,a_n)$ then $a_n=a_j$ for  $n\leq j<\omega,$ and (2) if $f(a_n)<^* (a_n,a_n)$ then  $a_{n+1}$ satisfies  $f(a_{n+1})=(a_n,a_n).$ 
So if $f(a_n)=(a_n,a_n)$ for some $n>0$ then let $b=a_n$. But if $a_n\not \in S$ for all $n\in \omega $ then let $b=\sup \{a_n:n\in \omega\}$ ... (Because  $a_n<a_{n+1}$ and $f(a_n)<^*(a_n, a_n)=f(a_{n+1}) $,  hence $f(b)\leq^*(b,b) =$ $=\sup_{<^*}\{(a_n,a_n)):n\in \omega\}$ $ \leq^* \sup_{<^*}\{f(a_{n+1}):n\in \omega\}=f(b).$
(iii) It is a simple exercise to show that if $0<\sup l=l<k$ and if  $T=<a_y :y\in l>$ is an increasing $l$-sequence in $S,$ then $\sup T\in S.$ So $S$ is closed in $k.$ (End of proof of lemma.)
Suppose $<X_a>_{a\in k}$ is a $\diamond_k$ sequence. For any $A\subset k\times k,$ the set $T_A=S\cap \{a\in k: (f^{-1}A)\cap a= A_a\}$ is stationary in $k,$ being the intersection of a club set and a stationary set. And $a\in T_A $  $\implies A\cap (a\times a)=$ $=A\cap \{x\in k\times k:x<^*(a,a)\}=\{f(y):y\in A_a\}.$ 
So for $a\in k$ let $Z_a=\{f(y) : y\in A\}.$
