Uniqueness proof : $a = a'$ so $a$ is unique. Is the proof absolutely rigorous? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Definition of “unique” as it relates to sets?Cardinality of Sets ProofDirect proof of empty set being subset of every setProof about SetsHow do I write a rigorous proof of the following problem: $(A triangle B) cup C neq (A cup C) triangle (B cup C)$Rigorous proof that surjectivity implies injectivity for finite setsRigorous proof that countable union of countable sets is countableWhat does it mean for the empty set to be connected and totally disconnected?Let $f$ be a family of sets. Prove that there is a unique set $A$ such that $f subseteq mathscr P (A)$.We have $kge2$ sets, and none of them are equal. Show that at least one of the sets contains none of the other sets.
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Uniqueness proof : $a = a'$ so $a$ is unique. Is the proof absolutely rigorous?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Definition of “unique” as it relates to sets?Cardinality of Sets ProofDirect proof of empty set being subset of every setProof about SetsHow do I write a rigorous proof of the following problem: $(A triangle B) cup C neq (A cup C) triangle (B cup C)$Rigorous proof that surjectivity implies injectivity for finite setsRigorous proof that countable union of countable sets is countableWhat does it mean for the empty set to be connected and totally disconnected?Let $f$ be a family of sets. Prove that there is a unique set $A$ such that $f subseteq mathscr P (A)$.We have $kge2$ sets, and none of them are equal. Show that at least one of the sets contains none of the other sets.
$begingroup$
My question deals with uniqueness proofs. For example the proof of the uniqueness of the empty set, or the proof of the uniqueness of the identity element in a group.
These proofs are convincing of course, but are they absolutely rigorous?
Is it possible to consider the following objection: when I prove that empty-set- 1 and ( hypothetical ) empty-set-2 are in fact equal, I prove that the total number of empty sets is not equal to 2, but is this the same as showing that the total number of empty sets is equal to 1?
How to make totally explicit the logic that is behind uniqueness proofs?
Is this logic questionable?
abstract-algebra elementary-set-theory
$endgroup$
add a comment |
$begingroup$
My question deals with uniqueness proofs. For example the proof of the uniqueness of the empty set, or the proof of the uniqueness of the identity element in a group.
These proofs are convincing of course, but are they absolutely rigorous?
Is it possible to consider the following objection: when I prove that empty-set- 1 and ( hypothetical ) empty-set-2 are in fact equal, I prove that the total number of empty sets is not equal to 2, but is this the same as showing that the total number of empty sets is equal to 1?
How to make totally explicit the logic that is behind uniqueness proofs?
Is this logic questionable?
abstract-algebra elementary-set-theory
$endgroup$
$begingroup$
The uniqueness-proof shows that the neutral element in a group is unique. I am not sure whether this fits with the given situation (two null-sets).
$endgroup$
– Peter
21 hours ago
$begingroup$
What I meant was that the proof method is the same. Something like : "Suppose there is a second null set, a second identity element...In that case ... so the " second" is in fact identical with the first one... so the first is unique"
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
Did you actually mean the empty set (terminology used in the answer) ? In this case, null-set is an unlucky terminology because it could be mixed with the kernel or the set of roots of an equation, or other things.
$endgroup$
– Peter
21 hours ago
$begingroup$
@Peter - Yes, I meant " empty set", maybe "null set" is old fashioned. I'll edit as soon as possible.
$endgroup$
– Eleonore Saint James
21 hours ago
add a comment |
$begingroup$
My question deals with uniqueness proofs. For example the proof of the uniqueness of the empty set, or the proof of the uniqueness of the identity element in a group.
These proofs are convincing of course, but are they absolutely rigorous?
Is it possible to consider the following objection: when I prove that empty-set- 1 and ( hypothetical ) empty-set-2 are in fact equal, I prove that the total number of empty sets is not equal to 2, but is this the same as showing that the total number of empty sets is equal to 1?
How to make totally explicit the logic that is behind uniqueness proofs?
Is this logic questionable?
abstract-algebra elementary-set-theory
$endgroup$
My question deals with uniqueness proofs. For example the proof of the uniqueness of the empty set, or the proof of the uniqueness of the identity element in a group.
These proofs are convincing of course, but are they absolutely rigorous?
Is it possible to consider the following objection: when I prove that empty-set- 1 and ( hypothetical ) empty-set-2 are in fact equal, I prove that the total number of empty sets is not equal to 2, but is this the same as showing that the total number of empty sets is equal to 1?
How to make totally explicit the logic that is behind uniqueness proofs?
Is this logic questionable?
abstract-algebra elementary-set-theory
abstract-algebra elementary-set-theory
edited 7 hours ago
Eleonore Saint James
asked 21 hours ago
Eleonore Saint JamesEleonore Saint James
18910
18910
$begingroup$
The uniqueness-proof shows that the neutral element in a group is unique. I am not sure whether this fits with the given situation (two null-sets).
$endgroup$
– Peter
21 hours ago
$begingroup$
What I meant was that the proof method is the same. Something like : "Suppose there is a second null set, a second identity element...In that case ... so the " second" is in fact identical with the first one... so the first is unique"
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
Did you actually mean the empty set (terminology used in the answer) ? In this case, null-set is an unlucky terminology because it could be mixed with the kernel or the set of roots of an equation, or other things.
$endgroup$
– Peter
21 hours ago
$begingroup$
@Peter - Yes, I meant " empty set", maybe "null set" is old fashioned. I'll edit as soon as possible.
$endgroup$
– Eleonore Saint James
21 hours ago
add a comment |
$begingroup$
The uniqueness-proof shows that the neutral element in a group is unique. I am not sure whether this fits with the given situation (two null-sets).
$endgroup$
– Peter
21 hours ago
$begingroup$
What I meant was that the proof method is the same. Something like : "Suppose there is a second null set, a second identity element...In that case ... so the " second" is in fact identical with the first one... so the first is unique"
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
Did you actually mean the empty set (terminology used in the answer) ? In this case, null-set is an unlucky terminology because it could be mixed with the kernel or the set of roots of an equation, or other things.
$endgroup$
– Peter
21 hours ago
$begingroup$
@Peter - Yes, I meant " empty set", maybe "null set" is old fashioned. I'll edit as soon as possible.
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
The uniqueness-proof shows that the neutral element in a group is unique. I am not sure whether this fits with the given situation (two null-sets).
$endgroup$
– Peter
21 hours ago
$begingroup$
The uniqueness-proof shows that the neutral element in a group is unique. I am not sure whether this fits with the given situation (two null-sets).
$endgroup$
– Peter
21 hours ago
$begingroup$
What I meant was that the proof method is the same. Something like : "Suppose there is a second null set, a second identity element...In that case ... so the " second" is in fact identical with the first one... so the first is unique"
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
What I meant was that the proof method is the same. Something like : "Suppose there is a second null set, a second identity element...In that case ... so the " second" is in fact identical with the first one... so the first is unique"
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
Did you actually mean the empty set (terminology used in the answer) ? In this case, null-set is an unlucky terminology because it could be mixed with the kernel or the set of roots of an equation, or other things.
$endgroup$
– Peter
21 hours ago
$begingroup$
Did you actually mean the empty set (terminology used in the answer) ? In this case, null-set is an unlucky terminology because it could be mixed with the kernel or the set of roots of an equation, or other things.
$endgroup$
– Peter
21 hours ago
$begingroup$
@Peter - Yes, I meant " empty set", maybe "null set" is old fashioned. I'll edit as soon as possible.
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
@Peter - Yes, I meant " empty set", maybe "null set" is old fashioned. I'll edit as soon as possible.
$endgroup$
– Eleonore Saint James
21 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Sorry, but the only correct way to fully understand uniqueness is to look at the logical definition. We say that there is a unique object satisfying a property $Q$ iff we can prove the following:
$∃x ( Q(x) ∧ ∀y ( Q(y) ⇒ x=y ) )$.
(There exists an object $x$ such that $x$ satisfies $Q$ and every object that satisfies $Q$ is equal to $x$.)
It turns out that you can prove in pure first-order logic that this is equivalent to:
$∃x ( Q(x) ) ∧ ∀x,y ( Q(x) ∧ Q(y) ⇒ x=y )$.
(Some object satisfies $Q$ and every objects $x,y$ that satisfy $Q$ are equal.)
Hence you may see an author/teacher prove either version to prove:
$∃!x ( Q(x) )$.
(This is the symbolic representation of "There exists a unique $x$ that satisfies $Q$.".)
Your question is a bit too vague for me to be able to tell which version your teacher/textbook is using. But the point is that uniqueness is defined as per above, and not in terms of 'number of' things that satisfy $Q$, because in general that is not well-defined.
$endgroup$
2
$begingroup$
The only answer that adresses the fact that it is the definition of uniqueness. +1
$endgroup$
– Max
17 hours ago
add a comment |
$begingroup$
When you show that $a = a'$ in the proof of uniqueness, $=$ is an equivalence relation. Hence if you have a collection of $a$s which are "unique" candidates that actually exist, no matter how large (can be more than two), you partition them into equivalence class(es), and there is only one since all of them are equal.
$endgroup$
1
$begingroup$
Ooooo I love this answer
$endgroup$
– Eevee Trainer
21 hours ago
add a comment |
$begingroup$
Take the collection of all the different $a$'s satisfying the property of interest. If the size of this collection is at-least two (can be infinite) we can take two distinct $a$'s out of this collection. Let's call them $a_x$ and $a_y$.
By your proof $a_x=a_y$ , which is a contradiction since they are distinct!
Hence, your proof doesn't merely dismiss the possibility of there being two (for instance) null sets, but actually dismisses the possiblity of there being two or more of them.
The immediate conclusion that follows is that there is either one object satisfying the property or none at all. Showing that it is the former and not the latter is where the existence proof comes in.
$endgroup$
add a comment |
$begingroup$
Yes, because you can sort of proceed in a pseudo-inductive manner ("pseudo-" because it has that flavor, at least to me, but it's not quite induction).
Since you assume nothing special about, say, $emptyset_1, emptyset_2$ and show that $emptyset = emptyset_1 = emptyset_2$, there's no real difference by considering arbitrary indices. In general you could say $emptyset_i = emptyset_j = emptyset$ for all $i,j$ in some indexing set $I$.
Thus, you conclude that there can be no more than one empty set.
This means there's either one or no empty sets. By showing the existence of the empty set, you get that there is exactly one empty set. Thus, you get uniqueness.
$endgroup$
1
$begingroup$
@Eavee Trainer. Thanks. Nice answer.
$endgroup$
– Eleonore Saint James
21 hours ago
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sorry, but the only correct way to fully understand uniqueness is to look at the logical definition. We say that there is a unique object satisfying a property $Q$ iff we can prove the following:
$∃x ( Q(x) ∧ ∀y ( Q(y) ⇒ x=y ) )$.
(There exists an object $x$ such that $x$ satisfies $Q$ and every object that satisfies $Q$ is equal to $x$.)
It turns out that you can prove in pure first-order logic that this is equivalent to:
$∃x ( Q(x) ) ∧ ∀x,y ( Q(x) ∧ Q(y) ⇒ x=y )$.
(Some object satisfies $Q$ and every objects $x,y$ that satisfy $Q$ are equal.)
Hence you may see an author/teacher prove either version to prove:
$∃!x ( Q(x) )$.
(This is the symbolic representation of "There exists a unique $x$ that satisfies $Q$.".)
Your question is a bit too vague for me to be able to tell which version your teacher/textbook is using. But the point is that uniqueness is defined as per above, and not in terms of 'number of' things that satisfy $Q$, because in general that is not well-defined.
$endgroup$
2
$begingroup$
The only answer that adresses the fact that it is the definition of uniqueness. +1
$endgroup$
– Max
17 hours ago
add a comment |
$begingroup$
Sorry, but the only correct way to fully understand uniqueness is to look at the logical definition. We say that there is a unique object satisfying a property $Q$ iff we can prove the following:
$∃x ( Q(x) ∧ ∀y ( Q(y) ⇒ x=y ) )$.
(There exists an object $x$ such that $x$ satisfies $Q$ and every object that satisfies $Q$ is equal to $x$.)
It turns out that you can prove in pure first-order logic that this is equivalent to:
$∃x ( Q(x) ) ∧ ∀x,y ( Q(x) ∧ Q(y) ⇒ x=y )$.
(Some object satisfies $Q$ and every objects $x,y$ that satisfy $Q$ are equal.)
Hence you may see an author/teacher prove either version to prove:
$∃!x ( Q(x) )$.
(This is the symbolic representation of "There exists a unique $x$ that satisfies $Q$.".)
Your question is a bit too vague for me to be able to tell which version your teacher/textbook is using. But the point is that uniqueness is defined as per above, and not in terms of 'number of' things that satisfy $Q$, because in general that is not well-defined.
$endgroup$
2
$begingroup$
The only answer that adresses the fact that it is the definition of uniqueness. +1
$endgroup$
– Max
17 hours ago
add a comment |
$begingroup$
Sorry, but the only correct way to fully understand uniqueness is to look at the logical definition. We say that there is a unique object satisfying a property $Q$ iff we can prove the following:
$∃x ( Q(x) ∧ ∀y ( Q(y) ⇒ x=y ) )$.
(There exists an object $x$ such that $x$ satisfies $Q$ and every object that satisfies $Q$ is equal to $x$.)
It turns out that you can prove in pure first-order logic that this is equivalent to:
$∃x ( Q(x) ) ∧ ∀x,y ( Q(x) ∧ Q(y) ⇒ x=y )$.
(Some object satisfies $Q$ and every objects $x,y$ that satisfy $Q$ are equal.)
Hence you may see an author/teacher prove either version to prove:
$∃!x ( Q(x) )$.
(This is the symbolic representation of "There exists a unique $x$ that satisfies $Q$.".)
Your question is a bit too vague for me to be able to tell which version your teacher/textbook is using. But the point is that uniqueness is defined as per above, and not in terms of 'number of' things that satisfy $Q$, because in general that is not well-defined.
$endgroup$
Sorry, but the only correct way to fully understand uniqueness is to look at the logical definition. We say that there is a unique object satisfying a property $Q$ iff we can prove the following:
$∃x ( Q(x) ∧ ∀y ( Q(y) ⇒ x=y ) )$.
(There exists an object $x$ such that $x$ satisfies $Q$ and every object that satisfies $Q$ is equal to $x$.)
It turns out that you can prove in pure first-order logic that this is equivalent to:
$∃x ( Q(x) ) ∧ ∀x,y ( Q(x) ∧ Q(y) ⇒ x=y )$.
(Some object satisfies $Q$ and every objects $x,y$ that satisfy $Q$ are equal.)
Hence you may see an author/teacher prove either version to prove:
$∃!x ( Q(x) )$.
(This is the symbolic representation of "There exists a unique $x$ that satisfies $Q$.".)
Your question is a bit too vague for me to be able to tell which version your teacher/textbook is using. But the point is that uniqueness is defined as per above, and not in terms of 'number of' things that satisfy $Q$, because in general that is not well-defined.
answered 17 hours ago
user21820user21820
40.2k544163
40.2k544163
2
$begingroup$
The only answer that adresses the fact that it is the definition of uniqueness. +1
$endgroup$
– Max
17 hours ago
add a comment |
2
$begingroup$
The only answer that adresses the fact that it is the definition of uniqueness. +1
$endgroup$
– Max
17 hours ago
2
2
$begingroup$
The only answer that adresses the fact that it is the definition of uniqueness. +1
$endgroup$
– Max
17 hours ago
$begingroup$
The only answer that adresses the fact that it is the definition of uniqueness. +1
$endgroup$
– Max
17 hours ago
add a comment |
$begingroup$
When you show that $a = a'$ in the proof of uniqueness, $=$ is an equivalence relation. Hence if you have a collection of $a$s which are "unique" candidates that actually exist, no matter how large (can be more than two), you partition them into equivalence class(es), and there is only one since all of them are equal.
$endgroup$
1
$begingroup$
Ooooo I love this answer
$endgroup$
– Eevee Trainer
21 hours ago
add a comment |
$begingroup$
When you show that $a = a'$ in the proof of uniqueness, $=$ is an equivalence relation. Hence if you have a collection of $a$s which are "unique" candidates that actually exist, no matter how large (can be more than two), you partition them into equivalence class(es), and there is only one since all of them are equal.
$endgroup$
1
$begingroup$
Ooooo I love this answer
$endgroup$
– Eevee Trainer
21 hours ago
add a comment |
$begingroup$
When you show that $a = a'$ in the proof of uniqueness, $=$ is an equivalence relation. Hence if you have a collection of $a$s which are "unique" candidates that actually exist, no matter how large (can be more than two), you partition them into equivalence class(es), and there is only one since all of them are equal.
$endgroup$
When you show that $a = a'$ in the proof of uniqueness, $=$ is an equivalence relation. Hence if you have a collection of $a$s which are "unique" candidates that actually exist, no matter how large (can be more than two), you partition them into equivalence class(es), and there is only one since all of them are equal.
answered 21 hours ago
tonychow0929tonychow0929
42137
42137
1
$begingroup$
Ooooo I love this answer
$endgroup$
– Eevee Trainer
21 hours ago
add a comment |
1
$begingroup$
Ooooo I love this answer
$endgroup$
– Eevee Trainer
21 hours ago
1
1
$begingroup$
Ooooo I love this answer
$endgroup$
– Eevee Trainer
21 hours ago
$begingroup$
Ooooo I love this answer
$endgroup$
– Eevee Trainer
21 hours ago
add a comment |
$begingroup$
Take the collection of all the different $a$'s satisfying the property of interest. If the size of this collection is at-least two (can be infinite) we can take two distinct $a$'s out of this collection. Let's call them $a_x$ and $a_y$.
By your proof $a_x=a_y$ , which is a contradiction since they are distinct!
Hence, your proof doesn't merely dismiss the possibility of there being two (for instance) null sets, but actually dismisses the possiblity of there being two or more of them.
The immediate conclusion that follows is that there is either one object satisfying the property or none at all. Showing that it is the former and not the latter is where the existence proof comes in.
$endgroup$
add a comment |
$begingroup$
Take the collection of all the different $a$'s satisfying the property of interest. If the size of this collection is at-least two (can be infinite) we can take two distinct $a$'s out of this collection. Let's call them $a_x$ and $a_y$.
By your proof $a_x=a_y$ , which is a contradiction since they are distinct!
Hence, your proof doesn't merely dismiss the possibility of there being two (for instance) null sets, but actually dismisses the possiblity of there being two or more of them.
The immediate conclusion that follows is that there is either one object satisfying the property or none at all. Showing that it is the former and not the latter is where the existence proof comes in.
$endgroup$
add a comment |
$begingroup$
Take the collection of all the different $a$'s satisfying the property of interest. If the size of this collection is at-least two (can be infinite) we can take two distinct $a$'s out of this collection. Let's call them $a_x$ and $a_y$.
By your proof $a_x=a_y$ , which is a contradiction since they are distinct!
Hence, your proof doesn't merely dismiss the possibility of there being two (for instance) null sets, but actually dismisses the possiblity of there being two or more of them.
The immediate conclusion that follows is that there is either one object satisfying the property or none at all. Showing that it is the former and not the latter is where the existence proof comes in.
$endgroup$
Take the collection of all the different $a$'s satisfying the property of interest. If the size of this collection is at-least two (can be infinite) we can take two distinct $a$'s out of this collection. Let's call them $a_x$ and $a_y$.
By your proof $a_x=a_y$ , which is a contradiction since they are distinct!
Hence, your proof doesn't merely dismiss the possibility of there being two (for instance) null sets, but actually dismisses the possiblity of there being two or more of them.
The immediate conclusion that follows is that there is either one object satisfying the property or none at all. Showing that it is the former and not the latter is where the existence proof comes in.
answered 17 hours ago
Bar AlonBar Alon
529217
529217
add a comment |
add a comment |
$begingroup$
Yes, because you can sort of proceed in a pseudo-inductive manner ("pseudo-" because it has that flavor, at least to me, but it's not quite induction).
Since you assume nothing special about, say, $emptyset_1, emptyset_2$ and show that $emptyset = emptyset_1 = emptyset_2$, there's no real difference by considering arbitrary indices. In general you could say $emptyset_i = emptyset_j = emptyset$ for all $i,j$ in some indexing set $I$.
Thus, you conclude that there can be no more than one empty set.
This means there's either one or no empty sets. By showing the existence of the empty set, you get that there is exactly one empty set. Thus, you get uniqueness.
$endgroup$
1
$begingroup$
@Eavee Trainer. Thanks. Nice answer.
$endgroup$
– Eleonore Saint James
21 hours ago
add a comment |
$begingroup$
Yes, because you can sort of proceed in a pseudo-inductive manner ("pseudo-" because it has that flavor, at least to me, but it's not quite induction).
Since you assume nothing special about, say, $emptyset_1, emptyset_2$ and show that $emptyset = emptyset_1 = emptyset_2$, there's no real difference by considering arbitrary indices. In general you could say $emptyset_i = emptyset_j = emptyset$ for all $i,j$ in some indexing set $I$.
Thus, you conclude that there can be no more than one empty set.
This means there's either one or no empty sets. By showing the existence of the empty set, you get that there is exactly one empty set. Thus, you get uniqueness.
$endgroup$
1
$begingroup$
@Eavee Trainer. Thanks. Nice answer.
$endgroup$
– Eleonore Saint James
21 hours ago
add a comment |
$begingroup$
Yes, because you can sort of proceed in a pseudo-inductive manner ("pseudo-" because it has that flavor, at least to me, but it's not quite induction).
Since you assume nothing special about, say, $emptyset_1, emptyset_2$ and show that $emptyset = emptyset_1 = emptyset_2$, there's no real difference by considering arbitrary indices. In general you could say $emptyset_i = emptyset_j = emptyset$ for all $i,j$ in some indexing set $I$.
Thus, you conclude that there can be no more than one empty set.
This means there's either one or no empty sets. By showing the existence of the empty set, you get that there is exactly one empty set. Thus, you get uniqueness.
$endgroup$
Yes, because you can sort of proceed in a pseudo-inductive manner ("pseudo-" because it has that flavor, at least to me, but it's not quite induction).
Since you assume nothing special about, say, $emptyset_1, emptyset_2$ and show that $emptyset = emptyset_1 = emptyset_2$, there's no real difference by considering arbitrary indices. In general you could say $emptyset_i = emptyset_j = emptyset$ for all $i,j$ in some indexing set $I$.
Thus, you conclude that there can be no more than one empty set.
This means there's either one or no empty sets. By showing the existence of the empty set, you get that there is exactly one empty set. Thus, you get uniqueness.
answered 21 hours ago
Eevee TrainerEevee Trainer
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@Eavee Trainer. Thanks. Nice answer.
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– Eleonore Saint James
21 hours ago
add a comment |
1
$begingroup$
@Eavee Trainer. Thanks. Nice answer.
$endgroup$
– Eleonore Saint James
21 hours ago
1
1
$begingroup$
@Eavee Trainer. Thanks. Nice answer.
$endgroup$
– Eleonore Saint James
21 hours ago
$begingroup$
@Eavee Trainer. Thanks. Nice answer.
$endgroup$
– Eleonore Saint James
21 hours ago
add a comment |
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-abstract-algebra, elementary-set-theory
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The uniqueness-proof shows that the neutral element in a group is unique. I am not sure whether this fits with the given situation (two null-sets).
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– Peter
21 hours ago
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What I meant was that the proof method is the same. Something like : "Suppose there is a second null set, a second identity element...In that case ... so the " second" is in fact identical with the first one... so the first is unique"
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– Eleonore Saint James
21 hours ago
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Did you actually mean the empty set (terminology used in the answer) ? In this case, null-set is an unlucky terminology because it could be mixed with the kernel or the set of roots of an equation, or other things.
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– Peter
21 hours ago
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@Peter - Yes, I meant " empty set", maybe "null set" is old fashioned. I'll edit as soon as possible.
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– Eleonore Saint James
21 hours ago