Suppose $kappa$ is a cardinal, such that $2 kappa = kappa$. Can it be proven that $kappa ! = 2^kappa$ without...
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The original question is given in an exercise. The first part of the problem is to show that if $left|Aright| = left|Aright| +left|Aright|$ for some set $A$, then there exists a partition of $A$, with every cell in the partition having cardinality $2$ (i.e. a partition of $A$ into pairs). The second part of the problem is to show that the cardinality of the set of permutations of $A$ is $2^{left|Aright|}$.
Using the first part of the problem, it's not hard to show that that cardinality of set of permutations of $A$ is bounded between $2^{left|Aright|}$ and $left|Aright|^{left|Aright|}$. But I couldn't find a way to proceed from here without assuming that $left|Aright| = left|Aright|cdotleft|Aright|$.
Is it possible to complete the proof without choice?
elementary-set-theory cardinals
asked Nov 15 at 9:15
j3M
604516
add a comment |
up vote
1
down vote
favorite
The original question is given in an exercise. The first part of the problem is to show that if $left|Aright| = left|Aright| +left|Aright|$ for some set $A$, then there exists a partition of $A$, with every cell in the partition having cardinality $2$ (i.e. a partition of $A$ into pairs). The second part of the problem is to show that the cardinality of the set of permutations of $A$ is $2^{left|Aright|}$.
Using the first part of the problem, it's not hard to show that that cardinality of set of permutations of $A$ is bounded between $2^{left|Aright|}$ and $left|Aright|^{left|Aright|}$. But I couldn't find a way to proceed from here without assuming that $left|Aright| = left|Aright|cdotleft|Aright|$.
Is it possible to complete the proof without choice?
elementary-set-theory cardinals
asked Nov 15 at 9:15
j3M
604516
Is that an exercise in a book?
– bof
Nov 15 at 10:39
@bof No, it's from an exercise sheet given in a course. The exercise doesn't ask to avoid choice explicitly, but I thought it is implicit because the only assumption is that $left|Aright| + left|Aright| = left|Aright|$, and because the next problem goes "In the previous problem we saw that $left|Aright|+left|Aright|=left|Aright|$ implies that $A$ can be partitioned to pairs. What partition can we find if $left|Aright|cdotleft|Aright|=left|Aright|$?".
– j3M
Nov 15 at 11:05
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The original question is given in an exercise. The first part of the problem is to show that if $left|Aright| = left|Aright| +left|Aright|$ for some set $A$, then there exists a partition of $A$, with every cell in the partition having cardinality $2$ (i.e. a partition of $A$ into pairs). The second part of the problem is to show that the cardinality of the set of permutations of $A$ is $2^{left|Aright|}$.
Using the first part of the problem, it's not hard to show that that cardinality of set of permutations of $A$ is bounded between $2^{left|Aright|}$ and $left|Aright|^{left|Aright|}$. But I couldn't find a way to proceed from here without assuming that $left|Aright| = left|Aright|cdotleft|Aright|$.
Is it possible to complete the proof without choice?
elementary-set-theory cardinals
asked Nov 15 at 9:15
j3M
604516
The original question is given in an exercise. The first part of the problem is to show that if $left|Aright| = left|Aright| +left|Aright|$ for some set $A$, then there exists a partition of $A$, with every cell in the partition having cardinality $2$ (i.e. a partition of $A$ into pairs). The second part of the problem is to show that the cardinality of the set of permutations of $A$ is $2^{left|Aright|}$.
Using the first part of the problem, it's not hard to show that that cardinality of set of permutations of $A$ is bounded between $2^{left|Aright|}$ and $left|Aright|^{left|Aright|}$. But I couldn't find a way to proceed from here without assuming that $left|Aright| = left|Aright|cdotleft|Aright|$.
Is it possible to complete the proof without choice?
elementary-set-theory cardinals
elementary-set-theory cardinals
asked Nov 15 at 9:15
j3M
604516
asked Nov 15 at 9:15
j3M
604516
asked Nov 15 at 9:15
j3M
604516
asked Nov 15 at 9:15
j3M
604516
asked Nov 15 at 9:15
j3M
604516
604516
Is that an exercise in a book?
– bof
Nov 15 at 10:39
@bof No, it's from an exercise sheet given in a course. The exercise doesn't ask to avoid choice explicitly, but I thought it is implicit because the only assumption is that $left|Aright| + left|Aright| = left|Aright|$, and because the next problem goes "In the previous problem we saw that $left|Aright|+left|Aright|=left|Aright|$ implies that $A$ can be partitioned to pairs. What partition can we find if $left|Aright|cdotleft|Aright|=left|Aright|$?".
– j3M
Nov 15 at 11:05
add a comment |
Is that an exercise in a book?
– bof
Nov 15 at 10:39
@bof No, it's from an exercise sheet given in a course. The exercise doesn't ask to avoid choice explicitly, but I thought it is implicit because the only assumption is that $left|Aright| + left|Aright| = left|Aright|$, and because the next problem goes "In the previous problem we saw that $left|Aright|+left|Aright|=left|Aright|$ implies that $A$ can be partitioned to pairs. What partition can we find if $left|Aright|cdotleft|Aright|=left|Aright|$?".
– j3M
Nov 15 at 11:05
Is that an exercise in a book?
– bof
Nov 15 at 10:39
Is that an exercise in a book?
– bof
Nov 15 at 10:39
@bof No, it's from an exercise sheet given in a course. The exercise doesn't ask to avoid choice explicitly, but I thought it is implicit because the only assumption is that $left|Aright| + left|Aright| = left|Aright|$, and because the next problem goes "In the previous problem we saw that $left|Aright|+left|Aright|=left|Aright|$ implies that $A$ can be partitioned to pairs. What partition can we find if $left|Aright|cdotleft|Aright|=left|Aright|$?".
– j3M
Nov 15 at 11:05
@bof No, it's from an exercise sheet given in a course. The exercise doesn't ask to avoid choice explicitly, but I thought it is implicit because the only assumption is that $left|Aright| + left|Aright| = left|Aright|$, and because the next problem goes "In the previous problem we saw that $left|Aright|+left|Aright|=left|Aright|$ implies that $A$ can be partitioned to pairs. What partition can we find if $left|Aright|cdotleft|Aright|=left|Aright|$?".
– j3M
Nov 15 at 11:05
add a comment |
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Is that an exercise in a book?
– bof
Nov 15 at 10:39
@bof No, it's from an exercise sheet given in a course. The exercise doesn't ask to avoid choice explicitly, but I thought it is implicit because the only assumption is that $left|Aright| + left|Aright| = left|Aright|$, and because the next problem goes "In the previous problem we saw that $left|Aright|+left|Aright|=left|Aright|$ implies that $A$ can be partitioned to pairs. What partition can we find if $left|Aright|cdotleft|Aright|=left|Aright|$?".
– j3M
Nov 15 at 11:05