Is $\left\{0,1,2\right\}^{\mathbb{Z}^2}=\left\{\left\{0,1,2\right\}^{\mathbb{Z}}\right\}^{\mathbb{Z}}$? I am asking myself if $\left\{0,1,2\right\}^{\mathbb{Z}^2}=\left\{\left\{0,1,2\right\}^{\mathbb{Z}}\right\}^{\mathbb{Z}}$?
Elements of $\left\{0,1,2\right\}^{\mathbb{Z}^2}$ are $0,1,2$-valued configurations on the 2d-lattice.
Edit
Sorry, I have to make my question more clear.
The background is the following:
Let $X=\left\{0,1,2\right\}^{\mathbb{Z}}$. And $T\colon X\to X$.
Now consider $Y\subset\left\{0,1,2\right\}^{\mathbb{Z}\times\mathbb{N}}$,
$$
Y:=\left\{(x,Tx,T^2x,\ldots): x\in X\right\}.
$$
Now consider the homeomorphism $f\colon Y\to X, (x,Tx,T^2x,...)\mapsto x$.
Is this a right form to write this? I mean that in the sense of a time-space diagram, i.e. that $x$ is a 0,1,2-configuration on the x-axis... $Tx$ is the 0,1,2-configuration standing in the "row" above the x-axis and so on.
Or how can I write $Y$ correctly?
 A: Not quite. But your intentions are probably correct.

*

*$\{x\}$ is a set with a single element. Therefore $\{x\}^\Bbb Z$ is the set of all functions from $\Bbb Z$ into that singleton, and there is only one function like that: $f(k)=x$ for all $k$.


*If, however, you mean $\left(\{0,1,2\}^\Bbb Z\right)^\Bbb Z$, then this is not equal to, but isomorphic to. You get that the elements of this sets are functions from $\Bbb Z$ into the set of functions from $\Bbb Z$ into $\{0,1,2\}$. So $f(k)=g$ for some $g\colon\Bbb Z\to\{0,1,2\}$.
Using Currying we can define a canonical bijection, $F(f)\colon\Bbb Z^2\to\{0,1,2\}$ is defined as $F(f)(n,k)=f(n)(k)$.
So to your question, no. But if you replace $\{\}$ by $()$ and equality by isomorphism, then the answer is yes.
A: It's not really correct to put an equal sign there. What would make more sense is to construct a bijection. Elements of $\{0,1,2\}^{\mathbb{Z}^2}$ are functions $f : \mathbb{Z}^2 \to \{0,1,2\}$. Elements of $(\{0,1,2\}^\mathbb{Z})^{\mathbb{Z}}$ are functions $g : \mathbb{Z} \to \{0,1,2\}^{\mathbb{Z}}$. The bijection is this: $f \leftrightarrow g$ if and only if, for all $m,n \in \mathbb{Z}$, $f(m,n) = g(n)(m)$.
Added to answer the OP's additional question: Yes, this bijection is a homeomorphism with respect to product topologies, assuming of course the discrete topology on $\{0,1,2\}$. In fact, both sets are homeomorphic to the Cantor set, although that is not of course enough to conclude that the bijection itself gives the homeomorphism. However, it is not too hard to prove that the bijection in one direction is continuous, and knowing that both spaces are compact and Hausdorff is then enough to conclude that it is a homeomorphism.
