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Is it possible to define this bijection?

  1. Apr 11, 2014 #1
    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 :(
     
    Last edited: Apr 11, 2014
  2. jcsd
  3. Apr 11, 2014 #2

    micromass

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    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).
     
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