- #1
issacnewton
- 1,000
- 29
Hi
Here is the problem I am doing. Prove that
[tex]^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\sim \mathcal{P}(\mathbb{Z^+})[/tex]
Where [itex]^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})[/itex] is the set of functions
from [itex]\mathbb{Z^+}[/itex] to [itex]\mathcal{P}(\mathbb{Z^+})[/itex].
To prove this I will need to come up with some bijection from [itex]^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})[/itex] to [itex]\mathcal{P}(\mathbb{Z^+})[/itex]
I am having hard time with this bijection. How do you map a subset of [itex]\mathbb{Z^+}[/itex] to some function from [itex]\mathbb{Z^+}[/itex] to
[itex]\mathcal{P}(\mathbb{Z^+})[/itex]
I have been able to get some insight into the problem so far. I have proven that
[itex]\mathcal{P}(\mathbb{Z^+}) [/itex] is uncountable. So it means that any function
from [itex]\mathbb{Z^+}[/itex] to [itex]\mathcal{P}(\mathbb{Z^+}) [/itex] can not be onto.
I would appreciate further insight to attack the problem.
thanks
Here is the problem I am doing. Prove that
[tex]^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\sim \mathcal{P}(\mathbb{Z^+})[/tex]
Where [itex]^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})[/itex] is the set of functions
from [itex]\mathbb{Z^+}[/itex] to [itex]\mathcal{P}(\mathbb{Z^+})[/itex].
To prove this I will need to come up with some bijection from [itex]^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})[/itex] to [itex]\mathcal{P}(\mathbb{Z^+})[/itex]
I am having hard time with this bijection. How do you map a subset of [itex]\mathbb{Z^+}[/itex] to some function from [itex]\mathbb{Z^+}[/itex] to
[itex]\mathcal{P}(\mathbb{Z^+})[/itex]
I have been able to get some insight into the problem so far. I have proven that
[itex]\mathcal{P}(\mathbb{Z^+}) [/itex] is uncountable. So it means that any function
from [itex]\mathbb{Z^+}[/itex] to [itex]\mathcal{P}(\mathbb{Z^+}) [/itex] can not be onto.
I would appreciate further insight to attack the problem.
thanks