Back and forth games Consider two countably infinite structures $\mathcal{A}$ and $\mathcal{B}$ in the language of a single binary relation $R$. Assume that the relation $R$ is an equivalence relation on both $\mathcal{A}$ and $\mathcal{B}$, with:
$\mathcal{A}$ contains one block of size $n$ for each natural number $n$, and no other blocks aside from these;
$\mathcal{B}$ contains one block of size $n$ for each natural number $n$ but alos one infinite block. 
(i) Show that $\mathcal{A}$ and $\mathcal{B}$ are not back forth equivalent, by giving a strtegy for $\forall$berlard to play that will defeat $\exists$loise in the $\omega$ back and forth game. \
(ii) Show that for any $k \in \mathbb{N}$, $\exists$loise does have a strategy for winning the $k$-round back and forth game. Thus $\mathcal{A}$ and $\mathcal{B}$ are elementarily equivalent. \
(iii) Give an example of an embedding of $\mathcal{A}$ into $\mathcal{B}$ that is not an elementary embedding. \
(iv) Let $T$ be the theory of $\mathcal{A}$. Show that $T$ does not admit quantifier elimination, by showing that $T$ is not model complete.  
I have a solid idea no how to approach part (i) and (ii) by just using the definitions of a back and forth game but don't know how to formalize it. For part (iii) and (iv), I am pretty lost, any help is appreciated. 
 A: I will answer i) and ii) since it seems (from the comments) that you know the others now.
To "formalize" a back and forth game we have to describe a strategy and show that no matter how $\forall$berland or $\exists$loise plays the other will allways win. So just stating that there is an infinite block is not enough, but rather we have to state a strategy which uses the infinite block.
i) Strategy: $\forall$berland chooses only elements in the inifnite block in $\mathcal{B}$.
Why will $\forall$berland win? In step 1 $\exists$loise will have to choose an element somewhere in $\mathcal{A}$. In the following steps, in order to keep the back and forth equivalence she will have to choose elements in the same block. If there are $t$ elements in the block where she choose the first element she will fail the back and forth equivalence after $t+1$ steps where $\forall$berland chooses elements in the same class still in $\mathcal B$, but she can not in $\mathcal A$. Thus $\forall$berland win.
ii) It is very important here that $k$ is chosen before they start play the game. Note that it will be impossible in $k$ steps to see the difference between a class containing $k+1, k^2$ or $\infty$ amount of elements.
Strategy: As long as $\forall$berland does not chose any element in the infnite class, $\exists$loise just chooses an element in the corresponding fininte class in the other structure. If $\forall$berland chooses an element in the infinite class, then $\exists$loise chooses an element in the first nonchoosen class of size at least $k+1$. If $\forall$berland after this choose the class of size $k+1$ in $\mathcal B$ then $\exists$loise will have to move one step higher up and continue to push up the nonused classes.
Why Will $\exists$loise Win? Note that can $\forall$berland choose $k$ elements in the infinite class, and  $\exists$loise will be able to respond with the same amount in the finite choose class. If $\forall$berland on the other hand chooses elements in a finite class $\exists$loise can respond with the same class (or a bigger if size greater than k).
Now any game like this can allways be described even more detailed an explicit with even more elaborately proof of why it works. 
Good luck.
