My proof that the empty set is unique












0














I'm trying to prove that the empty set is unique.



Proof:



Let $U = { a }$ be the universal set.



Assume $a notin emptyset '$ and $a notin emptyset$.



Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement



$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$



is vacuously true.



Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$



I would appreciate it if people could please provide feedback as to the correctness of my proof.



EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.






proof-verification elementary-set-theory







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  • 4




    You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
    – Hagen von Eitzen
    Jan 6 at 17:22










  • @HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
    – The Pointer
    Jan 6 at 17:25


















0














I'm trying to prove that the empty set is unique.



Proof:



Let $U = { a }$ be the universal set.



Assume $a notin emptyset '$ and $a notin emptyset$.



Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement



$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$



is vacuously true.



Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$



I would appreciate it if people could please provide feedback as to the correctness of my proof.



EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.






proof-verification elementary-set-theory







share|cite|improve this question




















  • 4




    You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
    – Hagen von Eitzen
    Jan 6 at 17:22










  • @HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
    – The Pointer
    Jan 6 at 17:25
















0












0








0







I'm trying to prove that the empty set is unique.



Proof:



Let $U = { a }$ be the universal set.



Assume $a notin emptyset '$ and $a notin emptyset$.



Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement



$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$



is vacuously true.



Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$



I would appreciate it if people could please provide feedback as to the correctness of my proof.



EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.






proof-verification elementary-set-theory







share|cite|improve this question















I'm trying to prove that the empty set is unique.



Proof:



Let $U = { a }$ be the universal set.



Assume $a notin emptyset '$ and $a notin emptyset$.



Without loss of generality, since $a notin emptyset'$, $emptyset '$ does not contain any elements. Since $emptyset '$ does not contain any elements, it must by default be a subset of $emptyset$, since the conditional statement



$$a in emptyset ' Rightarrow emptyset ' subseteq emptyset$$



is vacuously true.



Therefore, since $emptyset' subseteq emptyset$ and $emptyset subseteq emptyset '$, we have that $emptyset ' = emptyset$. $tag*{$blacksquare$}$



I would appreciate it if people could please provide feedback as to the correctness of my proof.



EDIT: Please be specific about what is incorrect and why. That way, I can learn what I did wrong and improve much more effectively.






proof-verification elementary-set-theory




proof-verification elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 17:37







The Pointer

















asked Jan 6 at 17:11









The PointerThe Pointer

2,60421335




2,60421335








  • 4




    You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
    – Hagen von Eitzen
    Jan 6 at 17:22










  • @HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
    – The Pointer
    Jan 6 at 17:25
















  • 4




    You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
    – Hagen von Eitzen
    Jan 6 at 17:22










  • @HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
    – The Pointer
    Jan 6 at 17:25










4




4




You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
– Hagen von Eitzen
Jan 6 at 17:22




You should not need a universal set. Also, please be more careful with writing down the definition of what being empty means
– Hagen von Eitzen
Jan 6 at 17:22












@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
– The Pointer
Jan 6 at 17:25






@HagenvonEitzen Are you saying that my proof is wrong, or that there are better ways to prove the theorem? Please be more specific about which part is incorrect.
– The Pointer
Jan 6 at 17:25












2 Answers
2






active

oldest

votes


















1














The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.



EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.






share|cite|improve this answer































    1














    Let $A$ and $B$ be two empty sets.
    Then the assertions $xin A$ and $xin B$ are logically equivalent.
    By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.






    share|cite|improve this answer

















    • 1




      This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
      – Holo
      Jan 6 at 17:25










    • @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
      – Holo
      Jan 6 at 17:45











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    1














    The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.



    EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.






    share|cite|improve this answer




























      1














      The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.



      EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.






      share|cite|improve this answer


























        1












        1








        1






        The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.



        EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.






        share|cite|improve this answer














        The empty set is a subset of any set. Let $A$ and $B$ be two empty sets. Since $A$ is empty, then $A subseteq B$. Similarly, $B subseteq A$. Hence $A=B$.



        EDIT: Your assumptions are a bit suspicious and the use of the universal set is really unnecessary. Basically the part: assume $anotin emptyset'$ and $emptyset'$ does not contain any elements is a bit wordy and I am not sure if it is a valid logic flow. The rest of your solution is pretty much the idea that I uncover above. All you need to claim is that two sets are empty and then use the fact that they are subsets of each-other.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 7 at 16:37

























        answered Jan 6 at 17:32









        EuxhenHEuxhenH

        472210




        472210























            1














            Let $A$ and $B$ be two empty sets.
            Then the assertions $xin A$ and $xin B$ are logically equivalent.
            By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.






            share|cite|improve this answer

















            • 1




              This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
              – Holo
              Jan 6 at 17:25










            • @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
              – Holo
              Jan 6 at 17:45
















            1














            Let $A$ and $B$ be two empty sets.
            Then the assertions $xin A$ and $xin B$ are logically equivalent.
            By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.






            share|cite|improve this answer

















            • 1




              This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
              – Holo
              Jan 6 at 17:25










            • @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
              – Holo
              Jan 6 at 17:45














            1












            1








            1






            Let $A$ and $B$ be two empty sets.
            Then the assertions $xin A$ and $xin B$ are logically equivalent.
            By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.






            share|cite|improve this answer












            Let $A$ and $B$ be two empty sets.
            Then the assertions $xin A$ and $xin B$ are logically equivalent.
            By the definition of equality of sets, $A=B$ iff $forall x(xin ALongleftrightarrow xin B)$, it follows that $A=B$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 17:23









            WuestenfuxWuestenfux

            3,7861411




            3,7861411








            • 1




              This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
              – Holo
              Jan 6 at 17:25










            • @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
              – Holo
              Jan 6 at 17:45














            • 1




              This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
              – Holo
              Jan 6 at 17:25










            • @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
              – Holo
              Jan 6 at 17:45








            1




            1




            This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
            – Holo
            Jan 6 at 17:25




            This is not definition, this is an axiom. The definition will be closer to $forall x(xin ALongleftrightarrow xin B)land forall w(Ain wiff Bin w)$
            – Holo
            Jan 6 at 17:25












            @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
            – Holo
            Jan 6 at 17:45




            @SvanN because axiom and definition are different things, there are models(not of ZF) where extensionality is not an axiom
            – Holo
            Jan 6 at 17:45


















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