Is it possible to define this bijection?

[trash]

I'd like to know if it is possible to define a bijection between the sets [0,1]^{\mathbb{Z}} and [0,1]^{\mathbb{N}}; \mathbb{N}^{\mathbb{N}} and \mathbb{Z}^{\mathbb{Z}}.

I tried to define a bijection between [0,1]^\mathbb{N} and [0,1]^\mathbb{Z} as follows: Take the bijection g:\mathbb{Z}\to\mathbb{N} defined by g(n)=\begin{cases}2n,&n\geq 0 \\ -(2n+1), & n<0 \end{cases}. Now let F be the function from [0,1]^{\mathbb{Z}} to [0,1]^\mathbb{N} given by F(f)=(f\circ g^{-1}).
Proving injectivity: Suppose F(f_1)=F(f_2), I tried to show that f_1=f_2 as follows: F(f_1)=F(f_2)\implies (f_1\circ g^{-1})=(f_2 \circ g^{-1}) then for every x\in\mathbb{Z} we have f_1(g^{-1}(x))=f_2(g^{-1}(x)) \forall x\in \mathbb{Z}, since g defines a bijection then must be f_1=f_2.

I'm not sure how to prove surjectivy, so instead of do so, I defined h:\mathbb{Z}\to\mathbb{N}; h(n)=\begin{cases}-n/2,&\text{if n even} \\ (n+1)/2, &\text{otherwise} \end{cases} and define D:[0,1]^{\mathbb{N}}\to [0,1]^{\mathbb{Z}} by D(f)=(f\circ h^{-1}. Now the same procedure as above would show that D is an injection from [0,1]^{\mathbb{N}} to [0,1]^{\mathbb{Z}}, and since the function F defined an injection from [0,1]^{\mathbb{N}} to [0,1]^\mathbb{Z} is possible to define a bijection between the two sets.
Is this proof ok?.

Now for \mathbb{N}^{\mathbb{N}} and \mathbb{Z}^{\mathbb{Z}}, I believe that the functions I defined above (F,D) would work as well changing the domain and codomain of the functions, am I right?.

Edit: Can someone move this cuestion to the homework forum?. I posted in the wrong forum :(

 

[micromass]

Yes. This is perfectly ok.

It is not necessary, but you might still want to be interested in showing $F$ to be bijection. To do so, take $h\in [0,1]^\mathbb{N}$. You wish to find a function $f\in [0,1]^\mathbb{Z}$ such that $F(f) = h$. This means that $f\circ g^{-1} = h$. So choose $f = h\circ g$, this will work (show it).