Finding a bijective correspondence between $X^{\omega}$ and $\mathcal{P}(\mathbb{Z}_+)$

Question

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?

Answer 1

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 $\mathfrak c=\lvert P(\mathbb Z)\rvert$...

Answer 2

Hint:

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