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Let $X = \{ 0,1 \}$ and let $\mathcal{P} (\mathbb{Z}_+) $. Find a bijective correspondence between $\mathcal{P} (\mathbb{Z}_+) $ and the cartesian product $X^{\omega} $ or ${\bf show}$ there isn't one

Attempt to solution:

I claim we can find one bijection. Here is my idea. Notice that the elements of $X^{\omega}$ are sequences $(a_n)$ where $a_n $ is either $1$ or $0$

Now, let $A \subset \mathbb{Z}_+$, then $0 \leq |A| \leq \infty $ and let $n = |A|$. Now, we define $f: \mathcal{P} (\mathbb{Z}_+) \to X^{\omega}$ as :

If $n=0$, then define $f(A) = (0,0,.....) $

if $n=1$, then define $f(A_k) = (a_k)$ where $a_k = 1$ in the kth position and $0$ eveyrwhere else.

if $n=2$, then this approach becomes more complicated.

Is this a good way to start the construction? Is it possible to find a closed form function?

Community
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Theoneandonly
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    define $f:X^\omega\to\mathcal P(\mathbb Z_+)$ as $(a_n)\mapsto{n\in \mathbb Z_+|a_n=0}$ – J. W. Tanner May 25 '20 at 05:05
  • gee how didnt I see this – Theoneandonly May 25 '20 at 05:08
  • that's why $\mathcal P(X)$ is sometimes written $2^X$ – J. W. Tanner May 25 '20 at 05:10
  • @Theoneandonly: Or, if you wanted to follow your own approach more closely, turn J. W. Tanner’s definition ‘upside down’ and send $\langle a_n:n\in\Bbb Z^+\rangle$ to ${n\in\Bbb Z^+:a_n=1}$. – Brian M. Scott May 25 '20 at 05:10
  • so, to see injectivity: if $f(a_n) \neq f(b_n)$ then there is an element $i \in Z_+$ such that $i \in { n : a_n =0 } - { n : b_n =0 }$ and thus $a_n$ contains a $0$ in the ith position and $b_n$ does not contain a $0$ thus $a_n \neq b_n$. Now, for surjectivity: take any ${ 1,2,3,...}$ of size $k$ so that $a_k = 0$, then the element $(0,0,0,.....,0,1,1,1,...)$ would get map to that set. So function if bijective. Is this correct? – Theoneandonly May 25 '20 at 05:15

2 Answers2

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Hint:

Define $f:X^\omega\to\mathcal P(\mathbb Z_+)$ as $(a_n)\mapsto\{n\in \mathbb Z_+|a_n=1\}$.

J. W. Tanner
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  • so, to see injectivity: if $f(a_n) \neq f(b_n)$ then there is an element $i \in Z_+$ such that $i \in { n : a_n =0 } - { n : b_n =0 }$ and thus $a_n$ contains a $0$ in the ith position and $b_n$ does not contain a $0$ thus $a_n \neq b_n$. Now, for surjectivity: take any ${ 1,2,3,...}$ of size $k$ so that $a_k = 0$, then the element $(0,0,0,.....,0,1,1,1,...)$ would get map to that set. So function if bijective. Is this correct? – Theoneandonly May 25 '20 at 05:15
  • for injective, you want to show if $(a_n)\ne(b_n)$ then $f((a_n))\ne f(b_n))$; anyway, it may be simpler to show there's an inverse map – J. W. Tanner May 25 '20 at 05:17
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First note that $Y^X$ is the notation in functional analysis, or just in general, for the space of all functions $f$ from $X$ to $Y$.

There is one, and the key to achieving it is to note that a function $f$ from $X$ to a two element set, say $\{0,1\}$, corresponds to the element of the power set, $S\in P(X)$, defined by $S=\{x\in X: f(x)=1\}$.

See my Easy proof that $\mathfrak c=\lvert P(\mathbb Z)\rvert$...