My proof that the empty set is unique












0












$begingroup$


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




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










  • $begingroup$
    @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.
    $endgroup$
    – The Pointer
    Jan 6 at 17:25


















0












$begingroup$


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











$endgroup$








  • 4




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










  • $begingroup$
    @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.
    $endgroup$
    – The Pointer
    Jan 6 at 17:25
















0












0








0





$begingroup$


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











$endgroup$




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

2,60421436




2,60421436








  • 4




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










  • $begingroup$
    @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.
    $endgroup$
    – The Pointer
    Jan 6 at 17:25
















  • 4




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










  • $begingroup$
    @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.
    $endgroup$
    – The Pointer
    Jan 6 at 17:25










4




4




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




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












$begingroup$
@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.
$endgroup$
– The Pointer
Jan 6 at 17:25






$begingroup$
@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.
$endgroup$
– The Pointer
Jan 6 at 17:25












2 Answers
2






active

oldest

votes


















1












$begingroup$

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











$endgroup$





















    1












    $begingroup$

    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









    $endgroup$









    • 1




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










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











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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











    $endgroup$


















      1












      $begingroup$

      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











      $endgroup$
















        1












        1








        1





        $begingroup$

        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











        $endgroup$



        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

        482210




        482210























            1












            $begingroup$

            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









            $endgroup$









            • 1




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










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
















            1












            $begingroup$

            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









            $endgroup$









            • 1




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










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














            1












            1








            1





            $begingroup$

            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









            $endgroup$



            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

            4,1551411




            4,1551411








            • 1




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










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














            • 1




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










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








            1




            1




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




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












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




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


















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