Find all finite sets A so that A x P(A) = P(A) x A, where P(A) is the power set of A.
$begingroup$
Now I'm a beginner at Discrete Math so I'm not sure how to tackle this problem. I was expecting that a set with elements would provided to prove this. My best hunch is that all sets should work since its the same on both sides of the equal sign. If you could help me find out how work through the question, instead of just an answer that would be great!
discrete-mathematics
asked 1 hour ago
BrownieBrownie
162
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Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Now I'm a beginner at Discrete Math so I'm not sure how to tackle this problem. I was expecting that a set with elements would provided to prove this. My best hunch is that all sets should work since its the same on both sides of the equal sign. If you could help me find out how work through the question, instead of just an answer that would be great!
discrete-mathematics
asked 1 hour ago
BrownieBrownie
162
New contributor
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
If by $times$ you mean the Cartesian Product, then it is not true that $A times mathcal{P}(A) = mathcal{P}(A) times A$. $mathcal{P}(A)$ is composed of sets of $A$ and $A$ is composed of elements of $A$, so I can't think of any scenario where this would be true.
$endgroup$
– Hyperion
1 hour ago
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In fact, $A times B neq B times A$ unless $A = B$, and because $mathcal{P}(A) neq A$, they can never be equal.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
@Hyperion $ emptyset times mbox{ anything } = mbox{ anything } times emptyset$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
@N.S. I didn't catch that; you're right.
$endgroup$
– Hyperion
1 hour ago
add a comment |
$begingroup$
Now I'm a beginner at Discrete Math so I'm not sure how to tackle this problem. I was expecting that a set with elements would provided to prove this. My best hunch is that all sets should work since its the same on both sides of the equal sign. If you could help me find out how work through the question, instead of just an answer that would be great!
discrete-mathematics
asked 1 hour ago
BrownieBrownie
162
New contributor
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Now I'm a beginner at Discrete Math so I'm not sure how to tackle this problem. I was expecting that a set with elements would provided to prove this. My best hunch is that all sets should work since its the same on both sides of the equal sign. If you could help me find out how work through the question, instead of just an answer that would be great!
discrete-mathematics
discrete-mathematics
asked 1 hour ago
BrownieBrownie
162
New contributor
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
BrownieBrownie
162
New contributor
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
BrownieBrownie
162
New contributor
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
BrownieBrownie
162
asked 1 hour ago
BrownieBrownie
162
162
New contributor
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Brownie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
If by $times$ you mean the Cartesian Product, then it is not true that $A times mathcal{P}(A) = mathcal{P}(A) times A$. $mathcal{P}(A)$ is composed of sets of $A$ and $A$ is composed of elements of $A$, so I can't think of any scenario where this would be true.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
In fact, $A times B neq B times A$ unless $A = B$, and because $mathcal{P}(A) neq A$, they can never be equal.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
@Hyperion $ emptyset times mbox{ anything } = mbox{ anything } times emptyset$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
@N.S. I didn't catch that; you're right.
$endgroup$
– Hyperion
1 hour ago
add a comment |
$begingroup$
If by $times$ you mean the Cartesian Product, then it is not true that $A times mathcal{P}(A) = mathcal{P}(A) times A$. $mathcal{P}(A)$ is composed of sets of $A$ and $A$ is composed of elements of $A$, so I can't think of any scenario where this would be true.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
In fact, $A times B neq B times A$ unless $A = B$, and because $mathcal{P}(A) neq A$, they can never be equal.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
@Hyperion $ emptyset times mbox{ anything } = mbox{ anything } times emptyset$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
@N.S. I didn't catch that; you're right.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
If by $times$ you mean the Cartesian Product, then it is not true that $A times mathcal{P}(A) = mathcal{P}(A) times A$. $mathcal{P}(A)$ is composed of sets of $A$ and $A$ is composed of elements of $A$, so I can't think of any scenario where this would be true.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
If by $times$ you mean the Cartesian Product, then it is not true that $A times mathcal{P}(A) = mathcal{P}(A) times A$. $mathcal{P}(A)$ is composed of sets of $A$ and $A$ is composed of elements of $A$, so I can't think of any scenario where this would be true.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
In fact, $A times B neq B times A$ unless $A = B$, and because $mathcal{P}(A) neq A$, they can never be equal.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
In fact, $A times B neq B times A$ unless $A = B$, and because $mathcal{P}(A) neq A$, they can never be equal.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
@Hyperion $ emptyset times mbox{ anything } = mbox{ anything } times emptyset$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
@Hyperion $ emptyset times mbox{ anything } = mbox{ anything } times emptyset$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
@N.S. I didn't catch that; you're right.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
@N.S. I didn't catch that; you're right.
$endgroup$
– Hyperion
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint 1: $mathcal P (A)$ always has more elements than $A$. In particular $A neq mathcal P(A)$.
Hint 2: If $A times B =B times A$ then you can prove that either $A=B$ or one of $A,B$ is empty.
Indeed, if you assume by contradiction that you have $A neq B$ and that neither $A$ nor $B$ is empty, then you can find an element $x$ which is in exactly one of the sets $A$ or $B$. Now, by using that neither $A$ nor $B$ is empty, you can pick some $y$ in the other set. Look at $(x,y)$.
Hint 3 Conclude that $A=emptyset$ is the only solution.
answered 1 hour ago
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
This helps out a lot, I understand why it works this way because of how a Cartesian products of sets are defined. If you don't mind me asking how would I continue to prove that they are not equal. Is it sufficient to state that because of how we define a Cartesian product , and because P(A) and A are not the same unless A is an empty set the two sides cannot be equal? Also this is my first post ever, and the lightning fast detailed response is really great, and super helpful. Thanks for helping people out
$endgroup$
– Brownie
1 hour ago
1
$begingroup$
@Viz-sa There are many ways to show that $mathcal P(A)$ is bigger. For example $f : A to mathcal{P}(A)$ defined by $f(x)= { x }$ is one to one but not onto, since it doesn't take the "value" ${ emptyset } in P(A)$.... Or you can show that if $A$ has $n$ elements, then $P(A)$ has $2^n$ and that $2^n >n$ for all $n geq 0$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
@Viz-sa Or you can use the Cantor theorem, which proves the claim even for infinite sets :)
$endgroup$
– N. S.
1 hour ago
add a comment |
$begingroup$
Assuming that "x" is the Cartesian product of two sets, we have the definition
For two sets $A$ and $B$, the cartesian product is given by
$$Atimes B={ (a,b):ain A, bin B}$$
Suppose that $Atimes P(A)$ for some set $A$.
Then, each element of $P(A)$ must be in $A$.
However, by Cantor's theorem, we have that $|A|<|P(A)|$ for any $A$, thus there is at least one element of $P(A)$ that is not in $A$.
Thus it cannot be true that $Atimes P(A)=P(A)times A$ for any set $A$.
answered 1 hour ago
高田航高田航
1,313418
$endgroup$
1
$begingroup$
"Then, each element of $P(A)$ must be in $A$" is not true.... You need the extra assumption that $A$ is not empty to deduce that.
$endgroup$
– N. S.
1 hour ago
3
$begingroup$
Note that $A=emptyset$ satisfies $A times P(A)=P(A) times A$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
Note that if $B times C subset D times E$, you need the extra assumption that $C neq emptyset$ to conclude that $B subset D$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Thank you for the correction, will edit.
$endgroup$
– 高田航
57 mins ago
add a comment |
Your Answer
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Hint 1: $mathcal P (A)$ always has more elements than $A$. In particular $A neq mathcal P(A)$.
Hint 2: If $A times B =B times A$ then you can prove that either $A=B$ or one of $A,B$ is empty.
Indeed, if you assume by contradiction that you have $A neq B$ and that neither $A$ nor $B$ is empty, then you can find an element $x$ which is in exactly one of the sets $A$ or $B$. Now, by using that neither $A$ nor $B$ is empty, you can pick some $y$ in the other set. Look at $(x,y)$.
Hint 3 Conclude that $A=emptyset$ is the only solution.
answered 1 hour ago
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
This helps out a lot, I understand why it works this way because of how a Cartesian products of sets are defined. If you don't mind me asking how would I continue to prove that they are not equal. Is it sufficient to state that because of how we define a Cartesian product , and because P(A) and A are not the same unless A is an empty set the two sides cannot be equal? Also this is my first post ever, and the lightning fast detailed response is really great, and super helpful. Thanks for helping people out
$endgroup$
– Brownie
1 hour ago
1
$begingroup$
@Viz-sa There are many ways to show that $mathcal P(A)$ is bigger. For example $f : A to mathcal{P}(A)$ defined by $f(x)= { x }$ is one to one but not onto, since it doesn't take the "value" ${ emptyset } in P(A)$.... Or you can show that if $A$ has $n$ elements, then $P(A)$ has $2^n$ and that $2^n >n$ for all $n geq 0$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
@Viz-sa Or you can use the Cantor theorem, which proves the claim even for infinite sets :)
$endgroup$
– N. S.
1 hour ago
add a comment |
$begingroup$
Hint 1: $mathcal P (A)$ always has more elements than $A$. In particular $A neq mathcal P(A)$.
Hint 2: If $A times B =B times A$ then you can prove that either $A=B$ or one of $A,B$ is empty.
Indeed, if you assume by contradiction that you have $A neq B$ and that neither $A$ nor $B$ is empty, then you can find an element $x$ which is in exactly one of the sets $A$ or $B$. Now, by using that neither $A$ nor $B$ is empty, you can pick some $y$ in the other set. Look at $(x,y)$.
Hint 3 Conclude that $A=emptyset$ is the only solution.
answered 1 hour ago
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
This helps out a lot, I understand why it works this way because of how a Cartesian products of sets are defined. If you don't mind me asking how would I continue to prove that they are not equal. Is it sufficient to state that because of how we define a Cartesian product , and because P(A) and A are not the same unless A is an empty set the two sides cannot be equal? Also this is my first post ever, and the lightning fast detailed response is really great, and super helpful. Thanks for helping people out
$endgroup$
– Brownie
1 hour ago
1
$begingroup$
@Viz-sa There are many ways to show that $mathcal P(A)$ is bigger. For example $f : A to mathcal{P}(A)$ defined by $f(x)= { x }$ is one to one but not onto, since it doesn't take the "value" ${ emptyset } in P(A)$.... Or you can show that if $A$ has $n$ elements, then $P(A)$ has $2^n$ and that $2^n >n$ for all $n geq 0$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
@Viz-sa Or you can use the Cantor theorem, which proves the claim even for infinite sets :)
$endgroup$
– N. S.
1 hour ago
add a comment |
$begingroup$
Hint 1: $mathcal P (A)$ always has more elements than $A$. In particular $A neq mathcal P(A)$.
Hint 2: If $A times B =B times A$ then you can prove that either $A=B$ or one of $A,B$ is empty.
Indeed, if you assume by contradiction that you have $A neq B$ and that neither $A$ nor $B$ is empty, then you can find an element $x$ which is in exactly one of the sets $A$ or $B$. Now, by using that neither $A$ nor $B$ is empty, you can pick some $y$ in the other set. Look at $(x,y)$.
Hint 3 Conclude that $A=emptyset$ is the only solution.
answered 1 hour ago
N. S.N. S.
103k6111208
$endgroup$
Hint 1: $mathcal P (A)$ always has more elements than $A$. In particular $A neq mathcal P(A)$.
Hint 2: If $A times B =B times A$ then you can prove that either $A=B$ or one of $A,B$ is empty.
Indeed, if you assume by contradiction that you have $A neq B$ and that neither $A$ nor $B$ is empty, then you can find an element $x$ which is in exactly one of the sets $A$ or $B$. Now, by using that neither $A$ nor $B$ is empty, you can pick some $y$ in the other set. Look at $(x,y)$.
Hint 3 Conclude that $A=emptyset$ is the only solution.
answered 1 hour ago
N. S.N. S.
103k6111208
edited 1 hour ago
answered 1 hour ago
N. S.N. S.
103k6111208
answered 1 hour ago
N. S.N. S.
103k6111208
answered 1 hour ago
N. S.N. S.
103k6111208
103k6111208
$begingroup$
This helps out a lot, I understand why it works this way because of how a Cartesian products of sets are defined. If you don't mind me asking how would I continue to prove that they are not equal. Is it sufficient to state that because of how we define a Cartesian product , and because P(A) and A are not the same unless A is an empty set the two sides cannot be equal? Also this is my first post ever, and the lightning fast detailed response is really great, and super helpful. Thanks for helping people out
$endgroup$
– Brownie
1 hour ago
1
$begingroup$
@Viz-sa There are many ways to show that $mathcal P(A)$ is bigger. For example $f : A to mathcal{P}(A)$ defined by $f(x)= { x }$ is one to one but not onto, since it doesn't take the "value" ${ emptyset } in P(A)$.... Or you can show that if $A$ has $n$ elements, then $P(A)$ has $2^n$ and that $2^n >n$ for all $n geq 0$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
@Viz-sa Or you can use the Cantor theorem, which proves the claim even for infinite sets :)
$endgroup$
– N. S.
1 hour ago
add a comment |
$begingroup$
This helps out a lot, I understand why it works this way because of how a Cartesian products of sets are defined. If you don't mind me asking how would I continue to prove that they are not equal. Is it sufficient to state that because of how we define a Cartesian product , and because P(A) and A are not the same unless A is an empty set the two sides cannot be equal? Also this is my first post ever, and the lightning fast detailed response is really great, and super helpful. Thanks for helping people out
$endgroup$
– Brownie
1 hour ago
1
$begingroup$
@Viz-sa There are many ways to show that $mathcal P(A)$ is bigger. For example $f : A to mathcal{P}(A)$ defined by $f(x)= { x }$ is one to one but not onto, since it doesn't take the "value" ${ emptyset } in P(A)$.... Or you can show that if $A$ has $n$ elements, then $P(A)$ has $2^n$ and that $2^n >n$ for all $n geq 0$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
@Viz-sa Or you can use the Cantor theorem, which proves the claim even for infinite sets :)
$endgroup$
– N. S.
1 hour ago
$begingroup$
This helps out a lot, I understand why it works this way because of how a Cartesian products of sets are defined. If you don't mind me asking how would I continue to prove that they are not equal. Is it sufficient to state that because of how we define a Cartesian product , and because P(A) and A are not the same unless A is an empty set the two sides cannot be equal? Also this is my first post ever, and the lightning fast detailed response is really great, and super helpful. Thanks for helping people out
$endgroup$
– Brownie
1 hour ago
$begingroup$
This helps out a lot, I understand why it works this way because of how a Cartesian products of sets are defined. If you don't mind me asking how would I continue to prove that they are not equal. Is it sufficient to state that because of how we define a Cartesian product , and because P(A) and A are not the same unless A is an empty set the two sides cannot be equal? Also this is my first post ever, and the lightning fast detailed response is really great, and super helpful. Thanks for helping people out
$endgroup$
– Brownie
1 hour ago
1
1
$begingroup$
@Viz-sa There are many ways to show that $mathcal P(A)$ is bigger. For example $f : A to mathcal{P}(A)$ defined by $f(x)= { x }$ is one to one but not onto, since it doesn't take the "value" ${ emptyset } in P(A)$.... Or you can show that if $A$ has $n$ elements, then $P(A)$ has $2^n$ and that $2^n >n$ for all $n geq 0$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
@Viz-sa There are many ways to show that $mathcal P(A)$ is bigger. For example $f : A to mathcal{P}(A)$ defined by $f(x)= { x }$ is one to one but not onto, since it doesn't take the "value" ${ emptyset } in P(A)$.... Or you can show that if $A$ has $n$ elements, then $P(A)$ has $2^n$ and that $2^n >n$ for all $n geq 0$.
$endgroup$
– N. S.
1 hour ago
1
1
$begingroup$
@Viz-sa Or you can use the Cantor theorem, which proves the claim even for infinite sets :)
$endgroup$
– N. S.
1 hour ago
$begingroup$
@Viz-sa Or you can use the Cantor theorem, which proves the claim even for infinite sets :)
$endgroup$
– N. S.
1 hour ago
add a comment |
$begingroup$
Assuming that "x" is the Cartesian product of two sets, we have the definition
For two sets $A$ and $B$, the cartesian product is given by
$$Atimes B={ (a,b):ain A, bin B}$$
Suppose that $Atimes P(A)$ for some set $A$.
Then, each element of $P(A)$ must be in $A$.
However, by Cantor's theorem, we have that $|A|<|P(A)|$ for any $A$, thus there is at least one element of $P(A)$ that is not in $A$.
Thus it cannot be true that $Atimes P(A)=P(A)times A$ for any set $A$.
answered 1 hour ago
高田航高田航
1,313418
$endgroup$
1
$begingroup$
"Then, each element of $P(A)$ must be in $A$" is not true.... You need the extra assumption that $A$ is not empty to deduce that.
$endgroup$
– N. S.
1 hour ago
3
$begingroup$
Note that $A=emptyset$ satisfies $A times P(A)=P(A) times A$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
Note that if $B times C subset D times E$, you need the extra assumption that $C neq emptyset$ to conclude that $B subset D$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Thank you for the correction, will edit.
$endgroup$
– 高田航
57 mins ago
add a comment |
$begingroup$
Assuming that "x" is the Cartesian product of two sets, we have the definition
For two sets $A$ and $B$, the cartesian product is given by
$$Atimes B={ (a,b):ain A, bin B}$$
Suppose that $Atimes P(A)$ for some set $A$.
Then, each element of $P(A)$ must be in $A$.
However, by Cantor's theorem, we have that $|A|<|P(A)|$ for any $A$, thus there is at least one element of $P(A)$ that is not in $A$.
Thus it cannot be true that $Atimes P(A)=P(A)times A$ for any set $A$.
answered 1 hour ago
高田航高田航
1,313418
$endgroup$
1
$begingroup$
"Then, each element of $P(A)$ must be in $A$" is not true.... You need the extra assumption that $A$ is not empty to deduce that.
$endgroup$
– N. S.
1 hour ago
3
$begingroup$
Note that $A=emptyset$ satisfies $A times P(A)=P(A) times A$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
Note that if $B times C subset D times E$, you need the extra assumption that $C neq emptyset$ to conclude that $B subset D$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Thank you for the correction, will edit.
$endgroup$
– 高田航
57 mins ago
add a comment |
$begingroup$
Assuming that "x" is the Cartesian product of two sets, we have the definition
For two sets $A$ and $B$, the cartesian product is given by
$$Atimes B={ (a,b):ain A, bin B}$$
Suppose that $Atimes P(A)$ for some set $A$.
Then, each element of $P(A)$ must be in $A$.
However, by Cantor's theorem, we have that $|A|<|P(A)|$ for any $A$, thus there is at least one element of $P(A)$ that is not in $A$.
Thus it cannot be true that $Atimes P(A)=P(A)times A$ for any set $A$.
answered 1 hour ago
高田航高田航
1,313418
$endgroup$
Assuming that "x" is the Cartesian product of two sets, we have the definition
For two sets $A$ and $B$, the cartesian product is given by
$$Atimes B={ (a,b):ain A, bin B}$$
Suppose that $Atimes P(A)$ for some set $A$.
Then, each element of $P(A)$ must be in $A$.
However, by Cantor's theorem, we have that $|A|<|P(A)|$ for any $A$, thus there is at least one element of $P(A)$ that is not in $A$.
Thus it cannot be true that $Atimes P(A)=P(A)times A$ for any set $A$.
answered 1 hour ago
高田航高田航
1,313418
answered 1 hour ago
高田航高田航
1,313418
answered 1 hour ago
高田航高田航
1,313418
answered 1 hour ago
高田航高田航
1,313418
1,313418
1
$begingroup$
"Then, each element of $P(A)$ must be in $A$" is not true.... You need the extra assumption that $A$ is not empty to deduce that.
$endgroup$
– N. S.
1 hour ago
3
$begingroup$
Note that $A=emptyset$ satisfies $A times P(A)=P(A) times A$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
Note that if $B times C subset D times E$, you need the extra assumption that $C neq emptyset$ to conclude that $B subset D$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Thank you for the correction, will edit.
$endgroup$
– 高田航
57 mins ago
add a comment |
1
$begingroup$
"Then, each element of $P(A)$ must be in $A$" is not true.... You need the extra assumption that $A$ is not empty to deduce that.
$endgroup$
– N. S.
1 hour ago
3
$begingroup$
Note that $A=emptyset$ satisfies $A times P(A)=P(A) times A$.
$endgroup$
– N. S.
1 hour ago
1
$begingroup$
Note that if $B times C subset D times E$, you need the extra assumption that $C neq emptyset$ to conclude that $B subset D$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Thank you for the correction, will edit.
$endgroup$
– 高田航
57 mins ago
1
1
$begingroup$
"Then, each element of $P(A)$ must be in $A$" is not true.... You need the extra assumption that $A$ is not empty to deduce that.
$endgroup$
– N. S.
1 hour ago
$begingroup$
"Then, each element of $P(A)$ must be in $A$" is not true.... You need the extra assumption that $A$ is not empty to deduce that.
$endgroup$
– N. S.
1 hour ago
3
3
$begingroup$
Note that $A=emptyset$ satisfies $A times P(A)=P(A) times A$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Note that $A=emptyset$ satisfies $A times P(A)=P(A) times A$.
$endgroup$
– N. S.
1 hour ago
1
1
$begingroup$
Note that if $B times C subset D times E$, you need the extra assumption that $C neq emptyset$ to conclude that $B subset D$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Note that if $B times C subset D times E$, you need the extra assumption that $C neq emptyset$ to conclude that $B subset D$.
$endgroup$
– N. S.
1 hour ago
$begingroup$
Thank you for the correction, will edit.
$endgroup$
– 高田航
57 mins ago
$begingroup$
Thank you for the correction, will edit.
$endgroup$
– 高田航
57 mins ago
add a comment |
Brownie is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
If by $times$ you mean the Cartesian Product, then it is not true that $A times mathcal{P}(A) = mathcal{P}(A) times A$. $mathcal{P}(A)$ is composed of sets of $A$ and $A$ is composed of elements of $A$, so I can't think of any scenario where this would be true.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
In fact, $A times B neq B times A$ unless $A = B$, and because $mathcal{P}(A) neq A$, they can never be equal.
$endgroup$
– Hyperion
1 hour ago
$begingroup$
@Hyperion $ emptyset times mbox{ anything } = mbox{ anything } times emptyset$.
$endgroup$
– N. S.
1 hour ago
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@N.S. I didn't catch that; you're right.
$endgroup$
– Hyperion
1 hour ago