How to prove that the set-theoretic difference operation $setminus$ cannot be defined through $cap$ and $cup$
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How does one go about proving that the set-theoretic difference operation $setminus$ cannot be defined through the operations $cap$ and $cup$?
My thoughts: I first assumed $A$ and $B$ are two non-disjoint non-empty sets since if they are disjoint and non-empty, then we have that $Asetminus B= A =Acap A=(Acap B) cup A$. Therefore we have defined, in this case, $setminus$ in terms of $cap$ and $cup$.
Next, I drew three Venn diagrams for $Acap B $, $Acup B $ and $Asetminus B $ and made the observation that $A setminus B$ involves an exclusion of a part of $A$. When I looked at the definitions, I could see this clearer:
$$Acap B = {xmid (xin A) land (xin B)}$$
$$Acup B = {xmid (xin A) lor (xin B)}$$
$$Asetminus B = {xmid (xin A) land mathbf{(xnotin B)}}$$
From this, I decided to conclude that since the intersection and union operations have the condition that $(xin B)$ whereas the set difference operation requires $(xnotin B)$, we cannot define set difference through the operations union and intersection only.
This is as far as I could go. How can I prove this formally?
discrete-mathematics elementary-set-theory
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
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add a comment |
$begingroup$
How does one go about proving that the set-theoretic difference operation $setminus$ cannot be defined through the operations $cap$ and $cup$?
My thoughts: I first assumed $A$ and $B$ are two non-disjoint non-empty sets since if they are disjoint and non-empty, then we have that $Asetminus B= A =Acap A=(Acap B) cup A$. Therefore we have defined, in this case, $setminus$ in terms of $cap$ and $cup$.
Next, I drew three Venn diagrams for $Acap B $, $Acup B $ and $Asetminus B $ and made the observation that $A setminus B$ involves an exclusion of a part of $A$. When I looked at the definitions, I could see this clearer:
$$Acap B = {xmid (xin A) land (xin B)}$$
$$Acup B = {xmid (xin A) lor (xin B)}$$
$$Asetminus B = {xmid (xin A) land mathbf{(xnotin B)}}$$
From this, I decided to conclude that since the intersection and union operations have the condition that $(xin B)$ whereas the set difference operation requires $(xnotin B)$, we cannot define set difference through the operations union and intersection only.
This is as far as I could go. How can I prove this formally?
discrete-mathematics elementary-set-theory
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
$endgroup$
add a comment |
$begingroup$
How does one go about proving that the set-theoretic difference operation $setminus$ cannot be defined through the operations $cap$ and $cup$?
My thoughts: I first assumed $A$ and $B$ are two non-disjoint non-empty sets since if they are disjoint and non-empty, then we have that $Asetminus B= A =Acap A=(Acap B) cup A$. Therefore we have defined, in this case, $setminus$ in terms of $cap$ and $cup$.
Next, I drew three Venn diagrams for $Acap B $, $Acup B $ and $Asetminus B $ and made the observation that $A setminus B$ involves an exclusion of a part of $A$. When I looked at the definitions, I could see this clearer:
$$Acap B = {xmid (xin A) land (xin B)}$$
$$Acup B = {xmid (xin A) lor (xin B)}$$
$$Asetminus B = {xmid (xin A) land mathbf{(xnotin B)}}$$
From this, I decided to conclude that since the intersection and union operations have the condition that $(xin B)$ whereas the set difference operation requires $(xnotin B)$, we cannot define set difference through the operations union and intersection only.
This is as far as I could go. How can I prove this formally?
discrete-mathematics elementary-set-theory
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
$endgroup$
How does one go about proving that the set-theoretic difference operation $setminus$ cannot be defined through the operations $cap$ and $cup$?
My thoughts: I first assumed $A$ and $B$ are two non-disjoint non-empty sets since if they are disjoint and non-empty, then we have that $Asetminus B= A =Acap A=(Acap B) cup A$. Therefore we have defined, in this case, $setminus$ in terms of $cap$ and $cup$.
Next, I drew three Venn diagrams for $Acap B $, $Acup B $ and $Asetminus B $ and made the observation that $A setminus B$ involves an exclusion of a part of $A$. When I looked at the definitions, I could see this clearer:
$$Acap B = {xmid (xin A) land (xin B)}$$
$$Acup B = {xmid (xin A) lor (xin B)}$$
$$Asetminus B = {xmid (xin A) land mathbf{(xnotin B)}}$$
From this, I decided to conclude that since the intersection and union operations have the condition that $(xin B)$ whereas the set difference operation requires $(xnotin B)$, we cannot define set difference through the operations union and intersection only.
This is as far as I could go. How can I prove this formally?
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
asked Dec 8 '18 at 13:04
E.NoleE.Nole
141114
141114
add a comment |
add a comment |
2 Answers
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All you need is a counterexample.
Let $A = B = { 0 }$. Then all of the sets
$$A,~ B,~ A cup A,~ A cap A,~ A cup B,~ A cap B,~ B cup A,~ B cap A,~ B cup B,~ B cup B$$
are equal to ${ 0 }$, and so any set built out of $A$, $B$ and the operations $cup$ and $cap$ is equal to ${ 0 }$.
But $B setminus A = varnothing ne { 0 }$, and so the set difference operator $setminus$ can't be expressed in terms of $cup$ and $cap$ alone.
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
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add a comment |
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The union and intersection operators are monotone. If $Asubseteq A'$
and $Bsubseteq B'$ then $Acup Bsubseteq A'cup B'$ and $Acap Bsubseteq A'cap B'$. If $f(A,B)$ is a formal combination of $A$s and $B$s with unions and intersections, then it is monotone: $f(A,B)subseteq f(A',B')$ under the above
assumptions. But $Asetminus B$ is not a monotone operation.
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
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2 Answers
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2 Answers
2
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active
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$begingroup$
All you need is a counterexample.
Let $A = B = { 0 }$. Then all of the sets
$$A,~ B,~ A cup A,~ A cap A,~ A cup B,~ A cap B,~ B cup A,~ B cap A,~ B cup B,~ B cup B$$
are equal to ${ 0 }$, and so any set built out of $A$, $B$ and the operations $cup$ and $cap$ is equal to ${ 0 }$.
But $B setminus A = varnothing ne { 0 }$, and so the set difference operator $setminus$ can't be expressed in terms of $cup$ and $cap$ alone.
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
$endgroup$
add a comment |
$begingroup$
All you need is a counterexample.
Let $A = B = { 0 }$. Then all of the sets
$$A,~ B,~ A cup A,~ A cap A,~ A cup B,~ A cap B,~ B cup A,~ B cap A,~ B cup B,~ B cup B$$
are equal to ${ 0 }$, and so any set built out of $A$, $B$ and the operations $cup$ and $cap$ is equal to ${ 0 }$.
But $B setminus A = varnothing ne { 0 }$, and so the set difference operator $setminus$ can't be expressed in terms of $cup$ and $cap$ alone.
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
$endgroup$
add a comment |
$begingroup$
All you need is a counterexample.
Let $A = B = { 0 }$. Then all of the sets
$$A,~ B,~ A cup A,~ A cap A,~ A cup B,~ A cap B,~ B cup A,~ B cap A,~ B cup B,~ B cup B$$
are equal to ${ 0 }$, and so any set built out of $A$, $B$ and the operations $cup$ and $cap$ is equal to ${ 0 }$.
But $B setminus A = varnothing ne { 0 }$, and so the set difference operator $setminus$ can't be expressed in terms of $cup$ and $cap$ alone.
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
$endgroup$
All you need is a counterexample.
Let $A = B = { 0 }$. Then all of the sets
$$A,~ B,~ A cup A,~ A cap A,~ A cup B,~ A cap B,~ B cup A,~ B cap A,~ B cup B,~ B cup B$$
are equal to ${ 0 }$, and so any set built out of $A$, $B$ and the operations $cup$ and $cap$ is equal to ${ 0 }$.
But $B setminus A = varnothing ne { 0 }$, and so the set difference operator $setminus$ can't be expressed in terms of $cup$ and $cap$ alone.
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
answered Dec 8 '18 at 14:48
Clive NewsteadClive Newstead
51.2k474135
51.2k474135
add a comment |
add a comment |
$begingroup$
The union and intersection operators are monotone. If $Asubseteq A'$
and $Bsubseteq B'$ then $Acup Bsubseteq A'cup B'$ and $Acap Bsubseteq A'cap B'$. If $f(A,B)$ is a formal combination of $A$s and $B$s with unions and intersections, then it is monotone: $f(A,B)subseteq f(A',B')$ under the above
assumptions. But $Asetminus B$ is not a monotone operation.
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
The union and intersection operators are monotone. If $Asubseteq A'$
and $Bsubseteq B'$ then $Acup Bsubseteq A'cup B'$ and $Acap Bsubseteq A'cap B'$. If $f(A,B)$ is a formal combination of $A$s and $B$s with unions and intersections, then it is monotone: $f(A,B)subseteq f(A',B')$ under the above
assumptions. But $Asetminus B$ is not a monotone operation.
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
The union and intersection operators are monotone. If $Asubseteq A'$
and $Bsubseteq B'$ then $Acup Bsubseteq A'cup B'$ and $Acap Bsubseteq A'cap B'$. If $f(A,B)$ is a formal combination of $A$s and $B$s with unions and intersections, then it is monotone: $f(A,B)subseteq f(A',B')$ under the above
assumptions. But $Asetminus B$ is not a monotone operation.
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
The union and intersection operators are monotone. If $Asubseteq A'$
and $Bsubseteq B'$ then $Acup Bsubseteq A'cup B'$ and $Acap Bsubseteq A'cap B'$. If $f(A,B)$ is a formal combination of $A$s and $B$s with unions and intersections, then it is monotone: $f(A,B)subseteq f(A',B')$ under the above
assumptions. But $Asetminus B$ is not a monotone operation.
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Dec 8 '18 at 13:14
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
add a comment |
add a comment |
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