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