- #1
issacnewton
- 1,041
- 37
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)
\[ ^{\mathbb{Z^+}}\mathcal{P}(\mathbb{Z^+})\;\sim\; \mathcal{P}(\mathbb{Z^+}) \]
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)