How could empty set be unique if it could be vacuously falseHow does the Axiom of Extensionality prove the...
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How could empty set be unique if it could be vacuously false
How does the Axiom of Extensionality prove the uniqueness of Specified sets?For every set $A$, the empty set is a subset of $A$. The empty set is a set. Therefore, the empty set has a cardinality $geq 1ldots$Proving that the empty set is uniqueset definitions with empty setvacuous truth -> empty set is both included and not included in every set?If $A$ is an empty set, how should I understand $forall xin A$?What happens if the empty set is not a subset of every set?Empty set, subsets, and vacuous truthsIs empty set a subset of every set direct proof confusionAxiom of extensionality inconsistent with empty set?My proof that the empty set is unique
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
The argument that people use to prove that empty set is unique is that: Let $A$ and $B$ be two empty sets then $forall z : z in A implies z in B$ since there is no such $xin A$ hence this statement is vacuously true. Also the converse is true therefore $A=B$. My objection is we could have equally stated $forall z : z in A implies z notin B$ and thus conclude that $Aneq B$.
elementary-set-theory
edited 12 hours ago
Asaf Karagila♦
312k33446781
asked 12 hours ago
AnonymousAnonymous
565
$endgroup$
add a comment |
$begingroup$
The argument that people use to prove that empty set is unique is that: Let $A$ and $B$ be two empty sets then $forall z : z in A implies z in B$ since there is no such $xin A$ hence this statement is vacuously true. Also the converse is true therefore $A=B$. My objection is we could have equally stated $forall z : z in A implies z notin B$ and thus conclude that $Aneq B$.
elementary-set-theory
edited 12 hours ago
Asaf Karagila♦
312k33446781
asked 12 hours ago
AnonymousAnonymous
565
$endgroup$
1
$begingroup$
There is no way however that the antecedent $zin A$ is true. A false $rightarrow$ false conditional is true.
$endgroup$
– Robert Wolfe
12 hours ago
1
$begingroup$
See How does the Axiom of Extensionality prove the uniqueness of Specified sets?
$endgroup$
– Mauro ALLEGRANZA
11 hours ago
$begingroup$
Or, $forall z : z in A implies z notin B$ doesn't imply that $Aneq B$.
$endgroup$
– immibis
20 mins ago
add a comment |
$begingroup$
The argument that people use to prove that empty set is unique is that: Let $A$ and $B$ be two empty sets then $forall z : z in A implies z in B$ since there is no such $xin A$ hence this statement is vacuously true. Also the converse is true therefore $A=B$. My objection is we could have equally stated $forall z : z in A implies z notin B$ and thus conclude that $Aneq B$.
elementary-set-theory
edited 12 hours ago
Asaf Karagila♦
312k33446781
asked 12 hours ago
AnonymousAnonymous
565
$endgroup$
The argument that people use to prove that empty set is unique is that: Let $A$ and $B$ be two empty sets then $forall z : z in A implies z in B$ since there is no such $xin A$ hence this statement is vacuously true. Also the converse is true therefore $A=B$. My objection is we could have equally stated $forall z : z in A implies z notin B$ and thus conclude that $Aneq B$.
elementary-set-theory
elementary-set-theory
edited 12 hours ago
Asaf Karagila♦
312k33446781
asked 12 hours ago
AnonymousAnonymous
565
edited 12 hours ago
Asaf Karagila♦
312k33446781
asked 12 hours ago
AnonymousAnonymous
565
edited 12 hours ago
Asaf Karagila♦
312k33446781
edited 12 hours ago
Asaf Karagila♦
312k33446781
edited 12 hours ago
Asaf Karagila♦
312k33446781
312k33446781
asked 12 hours ago
AnonymousAnonymous
565
asked 12 hours ago
AnonymousAnonymous
565
asked 12 hours ago
AnonymousAnonymous
565
565
1
$begingroup$
There is no way however that the antecedent $zin A$ is true. A false $rightarrow$ false conditional is true.
$endgroup$
– Robert Wolfe
12 hours ago
1
$begingroup$
See How does the Axiom of Extensionality prove the uniqueness of Specified sets?
$endgroup$
– Mauro ALLEGRANZA
11 hours ago
$begingroup$
Or, $forall z : z in A implies z notin B$ doesn't imply that $Aneq B$.
$endgroup$
– immibis
20 mins ago
add a comment |
1
$begingroup$
There is no way however that the antecedent $zin A$ is true. A false $rightarrow$ false conditional is true.
$endgroup$
– Robert Wolfe
12 hours ago
1
$begingroup$
See How does the Axiom of Extensionality prove the uniqueness of Specified sets?
$endgroup$
– Mauro ALLEGRANZA
11 hours ago
$begingroup$
Or, $forall z : z in A implies z notin B$ doesn't imply that $Aneq B$.
$endgroup$
– immibis
20 mins ago
1
1
$begingroup$
There is no way however that the antecedent $zin A$ is true. A false $rightarrow$ false conditional is true.
$endgroup$
– Robert Wolfe
12 hours ago
$begingroup$
There is no way however that the antecedent $zin A$ is true. A false $rightarrow$ false conditional is true.
$endgroup$
– Robert Wolfe
12 hours ago
1
1
$begingroup$
See How does the Axiom of Extensionality prove the uniqueness of Specified sets?
$endgroup$
– Mauro ALLEGRANZA
11 hours ago
$begingroup$
See How does the Axiom of Extensionality prove the uniqueness of Specified sets?
$endgroup$
– Mauro ALLEGRANZA
11 hours ago
$begingroup$
Or, $forall z : z in A implies z notin B$ doesn't imply that $Aneq B$.
$endgroup$
– immibis
20 mins ago
$begingroup$
Or, $forall z : z in A implies z notin B$ doesn't imply that $Aneq B$.
$endgroup$
– immibis
20 mins ago
add a comment |
2 Answers
2
active
oldest
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$begingroup$
But that's not how you negate $A = B$. For two sets to be non-equal, you have to actually find an element which is in one of the two sets, and not the other. The negation of a statement beginning with $forall$ is a statement beginning with $exists$. And if you carefully state "$Aneq B$" with the correct quantifiers and implications, and assume $A$ and $B$ are both empty sets, then you will find that $Aneq B$ is actually not true.
answered 12 hours ago
ArthurArthur
129k7124216
$endgroup$
add a comment |
$begingroup$
Indeed, $forall z:zin Aimplies znotin B$. And, by the same argument, $forall z:zin Bimplies znotin A$. But all that you deduce from this is that $Acap B=emptyset$. There is no contradiction here.
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
But that's not how you negate $A = B$. For two sets to be non-equal, you have to actually find an element which is in one of the two sets, and not the other. The negation of a statement beginning with $forall$ is a statement beginning with $exists$. And if you carefully state "$Aneq B$" with the correct quantifiers and implications, and assume $A$ and $B$ are both empty sets, then you will find that $Aneq B$ is actually not true.
answered 12 hours ago
ArthurArthur
129k7124216
$endgroup$
add a comment |
$begingroup$
But that's not how you negate $A = B$. For two sets to be non-equal, you have to actually find an element which is in one of the two sets, and not the other. The negation of a statement beginning with $forall$ is a statement beginning with $exists$. And if you carefully state "$Aneq B$" with the correct quantifiers and implications, and assume $A$ and $B$ are both empty sets, then you will find that $Aneq B$ is actually not true.
answered 12 hours ago
ArthurArthur
129k7124216
$endgroup$
add a comment |
$begingroup$
But that's not how you negate $A = B$. For two sets to be non-equal, you have to actually find an element which is in one of the two sets, and not the other. The negation of a statement beginning with $forall$ is a statement beginning with $exists$. And if you carefully state "$Aneq B$" with the correct quantifiers and implications, and assume $A$ and $B$ are both empty sets, then you will find that $Aneq B$ is actually not true.
answered 12 hours ago
ArthurArthur
129k7124216
$endgroup$
But that's not how you negate $A = B$. For two sets to be non-equal, you have to actually find an element which is in one of the two sets, and not the other. The negation of a statement beginning with $forall$ is a statement beginning with $exists$. And if you carefully state "$Aneq B$" with the correct quantifiers and implications, and assume $A$ and $B$ are both empty sets, then you will find that $Aneq B$ is actually not true.
answered 12 hours ago
ArthurArthur
129k7124216
edited 11 hours ago
answered 12 hours ago
ArthurArthur
129k7124216
answered 12 hours ago
ArthurArthur
129k7124216
answered 12 hours ago
ArthurArthur
129k7124216
129k7124216
add a comment |
add a comment |
$begingroup$
Indeed, $forall z:zin Aimplies znotin B$. And, by the same argument, $forall z:zin Bimplies znotin A$. But all that you deduce from this is that $Acap B=emptyset$. There is no contradiction here.
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
add a comment |
$begingroup$
Indeed, $forall z:zin Aimplies znotin B$. And, by the same argument, $forall z:zin Bimplies znotin A$. But all that you deduce from this is that $Acap B=emptyset$. There is no contradiction here.
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
add a comment |
$begingroup$
Indeed, $forall z:zin Aimplies znotin B$. And, by the same argument, $forall z:zin Bimplies znotin A$. But all that you deduce from this is that $Acap B=emptyset$. There is no contradiction here.
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
$endgroup$
Indeed, $forall z:zin Aimplies znotin B$. And, by the same argument, $forall z:zin Bimplies znotin A$. But all that you deduce from this is that $Acap B=emptyset$. There is no contradiction here.
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
answered 12 hours ago
José Carlos SantosJosé Carlos Santos
193k24148266
193k24148266
add a comment |
add a comment |
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1
$begingroup$
There is no way however that the antecedent $zin A$ is true. A false $rightarrow$ false conditional is true.
$endgroup$
– Robert Wolfe
12 hours ago
1
$begingroup$
See How does the Axiom of Extensionality prove the uniqueness of Specified sets?
$endgroup$
– Mauro ALLEGRANZA
11 hours ago
$begingroup$
Or, $forall z : z in A implies z notin B$ doesn't imply that $Aneq B$.
$endgroup$
– immibis
20 mins ago