HiGot the solution...
I am going to use the following identities in set algebra which deal
with equinumerous sets. I had asked the help for the proofs on
many identities at MHF and eventually posted the solutions there.
But now MHF is not there, all that work is lost... So here are the
identities I am going to use.. I have proved these before...1) For any set A \[ A\;\precsim\mathcal{P}(A) \]
2)For any sets A,B,C \[ ^{A}(B\times C)\;\sim\; ^{A}B\times\; ^{A}C \]
3) \[ C\subseteq D \Rightarrow C\precsim D \]
4)For any sets A,B,C \[ ^{(A\times B)}C\;\sim\; ^{A}( ^{B}C) \]
5)For any set A, \[ \mathcal{P}(A)\;\sim\; ^{A}\{1,0\} \]
here we can replace 1 and 0 by any other elements.
6)if \( A\;\precsim B\) and \( C\;\precsim D \) then
\[ A\times C\;\precsim B\times D \]
7)if \( A\;\precsim B\) then \[ \mathcal{P}(A)\precsim \mathcal{P}(B) \]
8)if \(A\neq\varnothing\) and \( A\;\precsim B\) and \( C\;\precsim D \)
then \[ ^{A}C\;\precsim\; ^{B}D \]
9)if \(A\cap B=\varnothing \) then
\[ \mathcal{P}(A\cup B)\;\sim \mathcal{P}(A)\times \mathcal{P}(B) \]so here is the solution I got. since A has at least two elements, let
them be a and b . Now define, \( B=A\setminus \{a,b\} \). Using rule 9, we
have \[ \mathcal{P}(B\cup \{a,b\})\;\sim \mathcal{P}(B)\times \mathcal{P}(\{a,b\}) \]\[ \therefore \mathcal{P}(A)\;\sim \mathcal{P}(B)\times \mathcal{P}(\{a,b\}) \]which means
\[ \mathcal{P}(A)\;\precsim \mathcal{P}(B)\times \mathcal{P}(\{a,b\}) \]
Now \( B\subseteq A\) and \( \{a,b\}\subseteq A\), using rule 3 we have\[ B\;\precsim A\;;\;\{a,b\}\precsim A \]so using rule 7 we have\[ \mathcal{P}(B)\;\precsim \mathcal{P}(A) \;;\;\mathcal{P}(\{a,b\})\;\precsim \mathcal{P}(A)\]using rule 6, we get\[ \mathcal{P}(B)\times \mathcal{P}(\{a,b\})\;\precsim \mathcal{P}(A)\times \mathcal{P}(A) \]since \( \precsim \) is transitive , we have \[ \mathcal{P}(A)\;\precsim \mathcal{P}(A)\times \mathcal{P}(A) \cdots (E1)\]
Now \( \mathcal{P}(A)\;\sim\; ^{A}\{a,b\} \) so using rule 6 (but for equinumerous sets), we have\[ \mathcal{P}(A)\times \mathcal{P}(A)\;\sim\; ^{A}\{a,b\}\times ^{A}\{a,b\} \cdots (E2)\]using rule 2 we get\[ ^{A}\{a,b\}\times ^{A}\{a,b\}\;\sim\; ^{A}(\{a,b\}\times \{a,b\}) \cdots (E3)\]since \( \{a,b\}\;\precsim A \) , using rule 6 we get\[ \{a,b\}\times \{a,b\}\;\precsim A\times A \]but we have been given \( A\times A\;\sim A \), which implies that
\( A\times A\;\precsim A \), so using transitivity of \(\precsim\) we get\[ \{a,b\}\times \{a,b\}\;\precsim A \] Now we notice that for set A we have \( A\;\precsim A\) , and since
\( A\neq\varnothing \), using rule 8 we have\[ ^{A}(\{a,b\}\times \{a,b\})\;\precsim\; ^{A}A \cdots (E4)\] but using equations E2 ,E3 and E4 , we get \[ \mathcal{P}(A)\times \mathcal{P}(A)\;\precsim ^{A}A \cdots (E5)\]Now, \( A\;\precsim \mathcal{P}(A) \) using rule 1 and also we have
\( A\;\precsim A \). Since \( A\neq\varnothing \), using rule 8 we have
\( ^{A}A\;\precsim ^{A}\mathcal{P}(A) \). So using E5 we get\[ \mathcal{P}(A)\times \mathcal{P}(A)\;\precsim\; ^{A}\mathcal{P}(A) \cdots (E6)\]Now \( A\;\sim A\) and \( \mathcal{P}(A)\;\sim ^{A}\{a,b\} \) and since
, using rule 8 (but for equinumerous sets), we have\[ ^{A}\mathcal{P}(A)\;\sim\; ^{A}( ^{A}\{a,b\}) \cdots (E7)\]using rule 4 we get\[ ^{A}( ^{A}\{a,b\})\;\sim\; ^{(A\times A)}\{a,b\} \cdots (E8)\] But \( ^{(A\times A)}\{a,b\}\;\sim \mathcal{P}(A\times A) \) using rule 5, So
using E7 and E8 and transitivity of \(\sim\) we get
\[ ^{A}\mathcal{P}(A)\;\sim \mathcal{P}(A\times A) \] Now using rule 7 (but for equinumerous sets), we have\[ \because A\times A\;\sim A \Rightarrow \mathcal{P}(A\times A)\;\sim \mathcal{P}(A) \]\[ \therefore ^{A}\mathcal{P}(A)\;\sim \mathcal{P}(A) \]\[ \therefore ^{A}\mathcal{P}(A)\;\precsim \mathcal{P}(A) \] so using E6 we have\[ \mathcal{P}(A)\times \mathcal{P}(A)\;\precsim \mathcal{P}(A) \cdots (E9)\]Using E1 and E9 , it follows from Cantor-Schroder-Bernstein theorem that\[ \mathcal{P}(A)\times \mathcal{P}(A)\;\sim \mathcal{P}(A) \]It seems too long for me. Is there a shorter way to get this ? its a lot of typing...(Whew)Thanks
