Homework Help: Is it possible to define this bijection?

1. Apr 11, 2014

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 :(

Last edited: Apr 11, 2014
2. Apr 11, 2014

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).