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I'd like to know if it is possible to define a bijection between the sets [itex][0,1]^{\mathbb{Z}}[/itex] and [itex][0,1]^{\mathbb{N}}[/itex]; [itex]\mathbb{N}^{\mathbb{N}}[/itex] and [itex]\mathbb{Z}^{\mathbb{Z}}[/itex].
I tried to define a bijection between [itex][0,1]^\mathbb{N}[/itex] and [itex][0,1]^\mathbb{Z}[/itex] as follows: Take the bijection [itex]g:\mathbb{Z}\to\mathbb{N}[/itex] defined by [itex]g(n)=\begin{cases}2n,&n\geq 0 \\ -(2n+1), & n<0 \end{cases}[/itex]. Now let [itex]F[/itex] be the function from [itex][0,1]^{\mathbb{Z}}[/itex] to [itex][0,1]^\mathbb{N}[/itex] given by [itex]F(f)=(f\circ g^{-1})[/itex].
Proving injectivity: Suppose [itex]F(f_1)=F(f_2)[/itex], I tried to show that [itex]f_1=f_2[/itex] as follows: [itex]F(f_1)=F(f_2)\implies (f_1\circ g^{-1})=(f_2 \circ g^{-1})[/itex] then for every [itex]x\in\mathbb{Z}[/itex] we have [itex]f_1(g^{-1}(x))=f_2(g^{-1}(x)) \forall x\in \mathbb{Z}[/itex], since [itex]g[/itex] defines a bijection then must be [itex]f_1=f_2[/itex].
I'm not sure how to prove surjectivy, so instead of do so, I defined [itex]h:\mathbb{Z}\to\mathbb{N}; h(n)=\begin{cases}-n/2,&\text{if n even} \\ (n+1)/2, &\text{otherwise} \end{cases}[/itex] and define [itex]D:[0,1]^{\mathbb{N}}\to [0,1]^{\mathbb{Z}}[/itex] by [itex]D(f)=(f\circ h^{-1}[/itex]. Now the same procedure as above would show that [itex]D[/itex] is an injection from [itex][0,1]^{\mathbb{N}}[/itex] to [itex][0,1]^{\mathbb{Z}}[/itex], and since the function [itex]F[/itex] defined an injection from [itex][0,1]^{\mathbb{N}}[/itex] to [itex][0,1]^\mathbb{Z}[/itex] is possible to define a bijection between the two sets.
Is this proof ok?.
Now for [itex]\mathbb{N}^{\mathbb{N}}[/itex] and [itex]\mathbb{Z}^{\mathbb{Z}}[/itex], I believe that the functions I defined above ([itex]F,D[/itex]) 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 :(
I tried to define a bijection between [itex][0,1]^\mathbb{N}[/itex] and [itex][0,1]^\mathbb{Z}[/itex] as follows: Take the bijection [itex]g:\mathbb{Z}\to\mathbb{N}[/itex] defined by [itex]g(n)=\begin{cases}2n,&n\geq 0 \\ -(2n+1), & n<0 \end{cases}[/itex]. Now let [itex]F[/itex] be the function from [itex][0,1]^{\mathbb{Z}}[/itex] to [itex][0,1]^\mathbb{N}[/itex] given by [itex]F(f)=(f\circ g^{-1})[/itex].
Proving injectivity: Suppose [itex]F(f_1)=F(f_2)[/itex], I tried to show that [itex]f_1=f_2[/itex] as follows: [itex]F(f_1)=F(f_2)\implies (f_1\circ g^{-1})=(f_2 \circ g^{-1})[/itex] then for every [itex]x\in\mathbb{Z}[/itex] we have [itex]f_1(g^{-1}(x))=f_2(g^{-1}(x)) \forall x\in \mathbb{Z}[/itex], since [itex]g[/itex] defines a bijection then must be [itex]f_1=f_2[/itex].
I'm not sure how to prove surjectivy, so instead of do so, I defined [itex]h:\mathbb{Z}\to\mathbb{N}; h(n)=\begin{cases}-n/2,&\text{if n even} \\ (n+1)/2, &\text{otherwise} \end{cases}[/itex] and define [itex]D:[0,1]^{\mathbb{N}}\to [0,1]^{\mathbb{Z}}[/itex] by [itex]D(f)=(f\circ h^{-1}[/itex]. Now the same procedure as above would show that [itex]D[/itex] is an injection from [itex][0,1]^{\mathbb{N}}[/itex] to [itex][0,1]^{\mathbb{Z}}[/itex], and since the function [itex]F[/itex] defined an injection from [itex][0,1]^{\mathbb{N}}[/itex] to [itex][0,1]^\mathbb{Z}[/itex] is possible to define a bijection between the two sets.
Is this proof ok?.
Now for [itex]\mathbb{N}^{\mathbb{N}}[/itex] and [itex]\mathbb{Z}^{\mathbb{Z}}[/itex], I believe that the functions I defined above ([itex]F,D[/itex]) 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 :(
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