- #1
issacnewton
- 1,035
- 37
Hi
I have to prove that \( ^{\mathbb{R}}\mathbb{R}\;\sim\;\mathcal{P}(\mathbb{R}) \).
My attempt is here. \( \mathbb{R}\;\sim\;\mathbb{R} \Rightarrow \mathbb{R}\;\precsim\;\mathbb{R}\). Since
\( \{0,1\}\subseteq \mathbb{R}\Rightarrow \{0,1\}\;\precsim\;\mathbb{R}\) . I am going to make use of the rule which I have proven.
if \(A\neq\varnothing\) and \( A\;\precsim\; B\) and \( C\;\precsim\; D \) then \( ^{A}C\;\precsim\; ^{B}D \). So we get
\( ^{\mathbb{R}}\{0,1\}\;\precsim\; ^{\mathbb{R}}\mathbb{R} \). Since \( \mathcal{P}(\mathbb{R})\;\sim\; ^{\mathbb{R}}\{0,1\} \), it
follows that \( \mathcal{P}(\mathbb{R})\;\precsim\; ^{\mathbb{R}}\{0,1\} \). So using transitivity of \( \precsim \) we get
\( \mathcal{P}(\mathbb{R})\;\precsim\;^{\mathbb{R}} \mathbb{R} \cdots (E1)\). Now
\[ ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\sim\;^{ \mathbb{R}}( ^{\mathbb{R}}\{0,1\})\;\sim\;^{(\mathbb{R}\times \mathbb{R})}\{0,1\} \]
But since \( \mathbb{R}\times\mathbb{R}\;\sim\; \mathbb{R} \) we have
\[ ^{(\mathbb{R}\times\mathbb{R})}\{0,1\}\;\sim\; ^{\mathbb{R}}\{0,1\}\;\sim\; \mathcal{P}(\mathbb{R}) \]
which implies, due to the transitivity of \( \sim \)
\[ ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\sim\; \mathcal{P}(\mathbb{R}) \]
\[ \Rightarrow ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\precsim\; \mathcal{P}(\mathbb{R}) \]
Now
\[ \mathbb{R}\;\precsim\;\mathcal{P}(\mathbb{R}); \; \mathbb{R}\;\precsim\; \mathbb{R} \]
since \( \mathbb{R}\neq \varnothing \) , we get
\[ ^{\mathbb{R}}\mathbb{R}\;\precsim\; ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\precsim\; \mathcal{P}(\mathbb{R}) \cdots (E2)\]
Using E1 and E2 , it follows from Cantor-Schroder-Bernstein theorem, that
\[ ^{\mathbb{R}}\mathbb{R}\;\sim\;\mathcal{P}(\mathbb{R}) \]
does it seem ok ? I have already proved all the identities I am using here...
I have to prove that \( ^{\mathbb{R}}\mathbb{R}\;\sim\;\mathcal{P}(\mathbb{R}) \).
My attempt is here. \( \mathbb{R}\;\sim\;\mathbb{R} \Rightarrow \mathbb{R}\;\precsim\;\mathbb{R}\). Since
\( \{0,1\}\subseteq \mathbb{R}\Rightarrow \{0,1\}\;\precsim\;\mathbb{R}\) . I am going to make use of the rule which I have proven.
if \(A\neq\varnothing\) and \( A\;\precsim\; B\) and \( C\;\precsim\; D \) then \( ^{A}C\;\precsim\; ^{B}D \). So we get
\( ^{\mathbb{R}}\{0,1\}\;\precsim\; ^{\mathbb{R}}\mathbb{R} \). Since \( \mathcal{P}(\mathbb{R})\;\sim\; ^{\mathbb{R}}\{0,1\} \), it
follows that \( \mathcal{P}(\mathbb{R})\;\precsim\; ^{\mathbb{R}}\{0,1\} \). So using transitivity of \( \precsim \) we get
\( \mathcal{P}(\mathbb{R})\;\precsim\;^{\mathbb{R}} \mathbb{R} \cdots (E1)\). Now
\[ ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\sim\;^{ \mathbb{R}}( ^{\mathbb{R}}\{0,1\})\;\sim\;^{(\mathbb{R}\times \mathbb{R})}\{0,1\} \]
But since \( \mathbb{R}\times\mathbb{R}\;\sim\; \mathbb{R} \) we have
\[ ^{(\mathbb{R}\times\mathbb{R})}\{0,1\}\;\sim\; ^{\mathbb{R}}\{0,1\}\;\sim\; \mathcal{P}(\mathbb{R}) \]
which implies, due to the transitivity of \( \sim \)
\[ ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\sim\; \mathcal{P}(\mathbb{R}) \]
\[ \Rightarrow ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\precsim\; \mathcal{P}(\mathbb{R}) \]
Now
\[ \mathbb{R}\;\precsim\;\mathcal{P}(\mathbb{R}); \; \mathbb{R}\;\precsim\; \mathbb{R} \]
since \( \mathbb{R}\neq \varnothing \) , we get
\[ ^{\mathbb{R}}\mathbb{R}\;\precsim\; ^{\mathbb{R}}\mathcal{P}(\mathbb{R})\;\precsim\; \mathcal{P}(\mathbb{R}) \cdots (E2)\]
Using E1 and E2 , it follows from Cantor-Schroder-Bernstein theorem, that
\[ ^{\mathbb{R}}\mathbb{R}\;\sim\;\mathcal{P}(\mathbb{R}) \]
does it seem ok ? I have already proved all the identities I am using here...
