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.










1












$begingroup$


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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    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















1












$begingroup$


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?










share|cite|improve this question











$endgroup$











  • $begingroup$
    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













1












1








1


0



$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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
















  • $begingroup$
    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










4 Answers
4






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    The only answer that adresses the fact that it is the definition of uniqueness. +1
    $endgroup$
    – Max
    17 hours ago


















5












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Ooooo I love this answer
    $endgroup$
    – Eevee Trainer
    21 hours ago


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      @Eavee Trainer. Thanks. Nice answer.
      $endgroup$
      – Eleonore Saint James
      21 hours ago











    Your Answer








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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$








    • 2




      $begingroup$
      The only answer that adresses the fact that it is the definition of uniqueness. +1
      $endgroup$
      – Max
      17 hours ago















    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$








    • 2




      $begingroup$
      The only answer that adresses the fact that it is the definition of uniqueness. +1
      $endgroup$
      – Max
      17 hours ago













    1












    1








    1





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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












    • 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











    5












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      Ooooo I love this answer
      $endgroup$
      – Eevee Trainer
      21 hours ago















    5












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$








    • 1




      $begingroup$
      Ooooo I love this answer
      $endgroup$
      – Eevee Trainer
      21 hours ago













    5












    5








    5





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 21 hours ago









    tonychow0929tonychow0929

    42137




    42137







    • 1




      $begingroup$
      Ooooo I love this answer
      $endgroup$
      – Eevee Trainer
      21 hours ago












    • 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











    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 17 hours ago









        Bar AlonBar Alon

        529217




        529217





















            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              @Eavee Trainer. Thanks. Nice answer.
              $endgroup$
              – Eleonore Saint James
              21 hours ago















            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              @Eavee Trainer. Thanks. Nice answer.
              $endgroup$
              – Eleonore Saint James
              21 hours ago













            0












            0








            0





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 21 hours ago









            Eevee TrainerEevee Trainer

            10.5k31842




            10.5k31842







            • 1




              $begingroup$
              @Eavee Trainer. Thanks. Nice answer.
              $endgroup$
              – Eleonore Saint James
              21 hours ago












            • 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

















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