from sets $|A| = |B|$ prove $A$ and $B$'s uncountable subsets also have same cardinality?
$begingroup$
For two uncountable sets $A$ and $B$ such that $|A| = |B|$, let Z(A) denote the set of uncountable subsets of $A$. How can we prove the cardinality of the uncountable subsets of $A$ and $B$ also equal ? that is $|Z(A)| = |Z(B)|$ ?
elementary-set-theory
edited Dec 7 '18 at 23:56
Andrews
3831317
$endgroup$
add a comment |
$begingroup$
For two uncountable sets $A$ and $B$ such that $|A| = |B|$, let Z(A) denote the set of uncountable subsets of $A$. How can we prove the cardinality of the uncountable subsets of $A$ and $B$ also equal ? that is $|Z(A)| = |Z(B)|$ ?
elementary-set-theory
edited Dec 7 '18 at 23:56
Andrews
3831317
$endgroup$
add a comment |
$begingroup$
For two uncountable sets $A$ and $B$ such that $|A| = |B|$, let Z(A) denote the set of uncountable subsets of $A$. How can we prove the cardinality of the uncountable subsets of $A$ and $B$ also equal ? that is $|Z(A)| = |Z(B)|$ ?
elementary-set-theory
edited Dec 7 '18 at 23:56
Andrews
3831317
$endgroup$
For two uncountable sets $A$ and $B$ such that $|A| = |B|$, let Z(A) denote the set of uncountable subsets of $A$. How can we prove the cardinality of the uncountable subsets of $A$ and $B$ also equal ? that is $|Z(A)| = |Z(B)|$ ?
elementary-set-theory
elementary-set-theory
edited Dec 7 '18 at 23:56
Andrews
3831317
edited Dec 7 '18 at 23:56
Andrews
3831317
edited Dec 7 '18 at 23:56
Andrews
3831317
edited Dec 7 '18 at 23:56
Andrews
3831317
edited Dec 7 '18 at 23:56
Andrews
3831317
3831317
asked Dec 7 '18 at 17:34
user623970
add a comment |
add a comment |
1 Answer
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$begingroup$
If you have a bijection $phi:Ato B$, you also have a bijection $mathcal P(A)tomathcal P(B)$. Can you find it?
Then just restrict this new bijection to $Z(A)$.
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
$endgroup$
$begingroup$
Note to the OP that not every bijection will work here - you need its restriction to behave appropriately - but the natural bijection will work.
$endgroup$
– Noah Schweber
Dec 7 '18 at 17:47
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
If you have a bijection $phi:Ato B$, you also have a bijection $mathcal P(A)tomathcal P(B)$. Can you find it?
Then just restrict this new bijection to $Z(A)$.
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
$endgroup$
$begingroup$
Note to the OP that not every bijection will work here - you need its restriction to behave appropriately - but the natural bijection will work.
$endgroup$
– Noah Schweber
Dec 7 '18 at 17:47
add a comment |
$begingroup$
If you have a bijection $phi:Ato B$, you also have a bijection $mathcal P(A)tomathcal P(B)$. Can you find it?
Then just restrict this new bijection to $Z(A)$.
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
$endgroup$
$begingroup$
Note to the OP that not every bijection will work here - you need its restriction to behave appropriately - but the natural bijection will work.
$endgroup$
– Noah Schweber
Dec 7 '18 at 17:47
add a comment |
$begingroup$
If you have a bijection $phi:Ato B$, you also have a bijection $mathcal P(A)tomathcal P(B)$. Can you find it?
Then just restrict this new bijection to $Z(A)$.
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
$endgroup$
If you have a bijection $phi:Ato B$, you also have a bijection $mathcal P(A)tomathcal P(B)$. Can you find it?
Then just restrict this new bijection to $Z(A)$.
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
answered Dec 7 '18 at 17:38
FedericoFederico
4,919514
4,919514
$begingroup$
Note to the OP that not every bijection will work here - you need its restriction to behave appropriately - but the natural bijection will work.
$endgroup$
– Noah Schweber
Dec 7 '18 at 17:47
add a comment |
$begingroup$
Note to the OP that not every bijection will work here - you need its restriction to behave appropriately - but the natural bijection will work.
$endgroup$
– Noah Schweber
Dec 7 '18 at 17:47
$begingroup$
Note to the OP that not every bijection will work here - you need its restriction to behave appropriately - but the natural bijection will work.
$endgroup$
– Noah Schweber
Dec 7 '18 at 17:47
$begingroup$
Note to the OP that not every bijection will work here - you need its restriction to behave appropriately - but the natural bijection will work.
$endgroup$
– Noah Schweber
Dec 7 '18 at 17:47
add a comment |
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