# Velleman problem 5(d) section 7.2

#### issacnewton

##### Member
Hi I have to prove

$^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\;\sim\; \mathcal{P}(\mathbb{Z^+})$

here is my attempt. I have proven that $$\mathcal{P}(\mathbb{Z^+})\;\sim\; ^{\mathbb{Z^+}}\{0,1\}$$. Also I am going to use the fact that
if $$A\;\sim B$$ and $$C\;\sim D$$ then $$^{A}C\;\sim ^{B}D$$. So we get

$^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\;\sim\; ^{\mathbb{Z^+}}(^{\mathbb{Z^+}} \{0,1\} )$

Also I have proven that for any sets A,B,C we have $$^{(A\times B)}C\;\sim\; ^{A}( ^{B}C)$$. So

$^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\;\sim\; ^{(\mathbb{Z^+}\times \mathbb{Z^+} )} \{0,1\}$

Since $$\mathbb{Z^+}\times \mathbb{Z^+}\;\sim \mathbb{Z^+}$$ and $$\{0,1\}\;\sim \{0,1\}$$ , we have

$^{(\mathbb{Z^+}\times \mathbb{Z^+} )} \{0,1\}\;\sim\; ^{\mathbb{Z^+}} \{0,1\}$

So it follows that
$^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\;\sim\; ^{\mathbb{Z^+}} \{0,1\}$

since $$\mathcal{P}(\mathbb{Z^+})\;\sim\; ^{\mathbb{Z^+}}\{0,1\}$$ , we get

$^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\;\sim\; \mathcal{P}(\mathbb{Z^+})$

Is it ok ?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Yes, I think this is fine.