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Velleman problem 5(d) section 7.2

issacnewton

Member
Jan 30, 2012
61
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 ?

(Emo)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
Yes, I think this is fine.