If $x$ is a set then $cap x$ is a set
$begingroup$
Consider the following problem:
(*) If $A$ is a non-empty set, then $cap A$ is a set.
But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?
What did I do wrong?
elementary-set-theory logic
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
asked Dec 6 '18 at 7:15
bbwbbw
50039
$endgroup$
add a comment |
$begingroup$
Consider the following problem:
(*) If $A$ is a non-empty set, then $cap A$ is a set.
But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?
What did I do wrong?
elementary-set-theory logic
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
asked Dec 6 '18 at 7:15
bbwbbw
50039
$endgroup$
add a comment |
$begingroup$
Consider the following problem:
(*) If $A$ is a non-empty set, then $cap A$ is a set.
But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?
What did I do wrong?
elementary-set-theory logic
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
asked Dec 6 '18 at 7:15
bbwbbw
50039
$endgroup$
Consider the following problem:
(*) If $A$ is a non-empty set, then $cap A$ is a set.
But I don't think we need the hypothesis "$A$ is non-empty". I can simply write $${x in A mid forall y (y in A ) rightarrow (x in y)}$$ This is a set by the separation axiom and it is the intersection set $cap A$, isn't it?
What did I do wrong?
elementary-set-theory logic
elementary-set-theory logic
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
asked Dec 6 '18 at 7:15
bbwbbw
50039
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
asked Dec 6 '18 at 7:15
bbwbbw
50039
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
edited Dec 17 '18 at 12:29
Andrés E. Caicedo
65.6k8159250
65.6k8159250
asked Dec 6 '18 at 7:15
bbwbbw
50039
asked Dec 6 '18 at 7:15
bbwbbw
50039
asked Dec 6 '18 at 7:15
bbwbbw
50039
50039
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
definition gives $emptyset$.
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
$endgroup$
– Arthur
Dec 6 '18 at 7:24
$begingroup$
If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
$endgroup$
– bbw
Dec 6 '18 at 7:38
1
$begingroup$
@bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
$endgroup$
– Arthur
Dec 6 '18 at 7:40
add a comment |
$begingroup$
If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
$endgroup$
$begingroup$
thank! but if $A$ is non-empty, is there a problem with my proposed definition?
$endgroup$
– bbw
Dec 6 '18 at 7:36
$begingroup$
Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
$endgroup$
– Carlos Jiménez
Dec 6 '18 at 7:43
add a comment |
$begingroup$
Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
definition gives $emptyset$.
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
$endgroup$
– Arthur
Dec 6 '18 at 7:24
$begingroup$
If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
$endgroup$
– bbw
Dec 6 '18 at 7:38
1
$begingroup$
@bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
$endgroup$
– Arthur
Dec 6 '18 at 7:40
add a comment |
$begingroup$
Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
definition gives $emptyset$.
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
$endgroup$
– Arthur
Dec 6 '18 at 7:24
$begingroup$
If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
$endgroup$
– bbw
Dec 6 '18 at 7:38
1
$begingroup$
@bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
$endgroup$
– Arthur
Dec 6 '18 at 7:40
add a comment |
$begingroup$
Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
definition gives $emptyset$.
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
Let $A={{a}}$, say. Then $bigcap A={a}$ but $anotin A$, so your proposed
definition gives $emptyset$.
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 6 '18 at 7:18
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
$endgroup$
– Arthur
Dec 6 '18 at 7:24
$begingroup$
If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
$endgroup$
– bbw
Dec 6 '18 at 7:38
1
$begingroup$
@bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
$endgroup$
– Arthur
Dec 6 '18 at 7:40
add a comment |
$begingroup$
Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
$endgroup$
– Arthur
Dec 6 '18 at 7:24
$begingroup$
If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
$endgroup$
– bbw
Dec 6 '18 at 7:38
1
$begingroup$
@bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
$endgroup$
– Arthur
Dec 6 '18 at 7:40
$begingroup$
Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
$endgroup$
– Arthur
Dec 6 '18 at 7:24
$begingroup$
Could you add a more correct definition of $bigcap A$? That would make this answer much better in my opinion.
$endgroup$
– Arthur
Dec 6 '18 at 7:24
$begingroup$
If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
$endgroup$
– bbw
Dec 6 '18 at 7:38
$begingroup$
If $A={{a}}$ then according to my proposed definition, we have $cap A$ is the set of $x$ such that $x={a}$ and $x in {a}$ which gives $a in {a}$. Is this really a contradiction?
$endgroup$
– bbw
Dec 6 '18 at 7:38
1
1
$begingroup$
@bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
$endgroup$
– Arthur
Dec 6 '18 at 7:40
$begingroup$
@bbw No, that's not what your definition says. Your definition says that $bigcap A$ will be the set of elements of $A$ (this is what "${xin Amid{}$" says), which is an element of all the elements of $A$. And ${a}notin {a}$, so by your definition, $bigcap A$ is empty.
$endgroup$
– Arthur
Dec 6 '18 at 7:40
add a comment |
$begingroup$
If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
$endgroup$
$begingroup$
thank! but if $A$ is non-empty, is there a problem with my proposed definition?
$endgroup$
– bbw
Dec 6 '18 at 7:36
$begingroup$
Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
$endgroup$
– Carlos Jiménez
Dec 6 '18 at 7:43
add a comment |
$begingroup$
If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
$endgroup$
$begingroup$
thank! but if $A$ is non-empty, is there a problem with my proposed definition?
$endgroup$
– bbw
Dec 6 '18 at 7:36
$begingroup$
Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
$endgroup$
– Carlos Jiménez
Dec 6 '18 at 7:43
add a comment |
$begingroup$
If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
$endgroup$
If you don't impose the condition that $A$ is non-empty then $displaystylebigcap A$ is not necessarily a set. Consider $A=emptyset$. Then, by definition $displaystylebigcap A=left{wmidforall x(xin Ato win x) right}$. If $V$ is the universe of sets, then $displaystylebigcap emptyset=left{wmidforall x(xin emptysetto win x) right}$. But the formula $forall x(xin emptysetto win x)$ is vacuously true. No matter who is $w$. Then, for all $win V$ we have that $win displaystylebigcap emptyset$. Therefore $displaystylebigcap emptyset=V$ and clearly $V$ is not a set.
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
answered Dec 6 '18 at 7:24
Carlos JiménezCarlos Jiménez
2,4051620
2,4051620
$begingroup$
thank! but if $A$ is non-empty, is there a problem with my proposed definition?
$endgroup$
– bbw
Dec 6 '18 at 7:36
$begingroup$
Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
$endgroup$
– Carlos Jiménez
Dec 6 '18 at 7:43
add a comment |
$begingroup$
thank! but if $A$ is non-empty, is there a problem with my proposed definition?
$endgroup$
– bbw
Dec 6 '18 at 7:36
$begingroup$
Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
$endgroup$
– Carlos Jiménez
Dec 6 '18 at 7:43
$begingroup$
thank! but if $A$ is non-empty, is there a problem with my proposed definition?
$endgroup$
– bbw
Dec 6 '18 at 7:36
$begingroup$
thank! but if $A$ is non-empty, is there a problem with my proposed definition?
$endgroup$
– bbw
Dec 6 '18 at 7:36
$begingroup$
Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
$endgroup$
– Carlos Jiménez
Dec 6 '18 at 7:43
$begingroup$
Yes. Your definition is wrong. See the answer of Lord Shar the Unknown. The definition of intersection is in my answer.
$endgroup$
– Carlos Jiménez
Dec 6 '18 at 7:43
add a comment |
$begingroup$
Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
add a comment |
$begingroup$
Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
$endgroup$
Axiom: $cup A={x:exists yin A,(xin y)}$ exists. So let $B={xin cup A: forall yin A,(xin y)},$ which exists by Comprehension (Separation). If $A=emptyset$ then $cup A=emptyset$ so $B=emptyset.$ If $Ane emptyset$ then $B={x:forall yin A,(xin y)}=cap A$.
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
answered Dec 17 '18 at 10:32
DanielWainfleetDanielWainfleet
35.3k31648
35.3k31648
add a comment |
add a comment |
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