Proving set identities: empty set case.












5














I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$



I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.



We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.



But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.



In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?



If so, some suggestions as to how to incorporate them into proofs would be helpful :-).






elementary-set-theory proof-writing







share|cite|improve this question


















  • 2




    Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
    – DonAntonio
    1 hour ago












  • Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
    – Mauro ALLEGRANZA
    44 mins ago










  • In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
    – Mauro ALLEGRANZA
    41 mins ago


















5














I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$



I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.



We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.



But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.



In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?



If so, some suggestions as to how to incorporate them into proofs would be helpful :-).






elementary-set-theory proof-writing







share|cite|improve this question


















  • 2




    Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
    – DonAntonio
    1 hour ago












  • Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
    – Mauro ALLEGRANZA
    44 mins ago










  • In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
    – Mauro ALLEGRANZA
    41 mins ago
















5












5








5


2





I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$



I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.



We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.



But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.



In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?



If so, some suggestions as to how to incorporate them into proofs would be helpful :-).






elementary-set-theory proof-writing







share|cite|improve this question













I am currently refreshing my knowledge in naive set theory, and would like to prove that for all sets $A,B,C$ we have $$Acap(B cup C) = (A cap B) cup (A cap C).$$



I understand that this can be done by proving both $$Acap(B cup C) subset (A cap B) cup (A cap C) text{and} (A cap B) cup (A cap C) subset Acap(B cup C) $$ hold true.



We can do this by letting $x$ be an arbitrary element of $Acap(B cup C)$ and showing that it is an element of $(A cap B) cup (A cap C)$, and vice versa.



But what about when $A cap (B cup C)$ is the empty set? Then I would think we can't let $x$ be an arbitrary element of $A cap (B cup C)$ since there are none. However, I am aware that $A cap (B cup C) subset (A cap B) cup (A cap C)$ is trivially true in this case.



In the discrete mathematics course I took at my university, I did not see such cases be brought to attention. Should they be mentioned in proofs of such identities? Why / why not?



If so, some suggestions as to how to incorporate them into proofs would be helpful :-).






elementary-set-theory proof-writing




elementary-set-theory proof-writing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









E-mu

405312




405312








  • 2




    Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
    – DonAntonio
    1 hour ago












  • Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
    – Mauro ALLEGRANZA
    44 mins ago










  • In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
    – Mauro ALLEGRANZA
    41 mins ago
















  • 2




    Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
    – DonAntonio
    1 hour ago












  • Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
    – Mauro ALLEGRANZA
    44 mins ago










  • In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
    – Mauro ALLEGRANZA
    41 mins ago










2




2




Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago






Because some of the two sides being the empty set is no problem. Your claim is: for any element here, I prove it also is an element there. If there is no element at all then that is not your problem (mathematicalwise, of course): the proof still holds! Anyway, if you still feel something else must be done (though it really mustn't), then you can make a separate case: if this side is empty then so and so. and also the other side is, and the other way around, too. But you really don't need that,
– DonAntonio
1 hour ago














Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago




Let $A cap (B cup C)$ empty. Thus, $x notin ldots$ that means : either (i) $x notin A$ or (ii) $x notin (B cup C)$. In the first case, $x notin (A cap B)$ and $x notin (A cap C)$ and thus $x notin$ their union.
– Mauro ALLEGRANZA
44 mins ago












In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago






In the secon case : from (ii) $x notin (B cup C)$ we have that $x notin B$ and $x notin C$ and again $x notin [(A cap B) cup (A cap C)]$. Conclusion : in both cases $x notin ldots$. But $x$ is whatever, and thus the result holds for every $x$, i.e. $forall x (x notin [(A cap B) cup (A cap C)])$, that means that the set is empty, and the result holds.
– Mauro ALLEGRANZA
41 mins ago












3 Answers
3






active

oldest

votes


















4














In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that




for every $x$, if $xin X$, then $xin Y$.




Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when




either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true




If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.



The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.






share|cite|improve this answer





















  • Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
    – E-mu
    13 mins ago



















4














If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.



It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.



So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.






share|cite|improve this answer































    2














    Since the empty set is a subset of any set, there is no need of including that in a formal proof.
    However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that




      for every $x$, if $xin X$, then $xin Y$.




      Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when




      either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true




      If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.



      The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.






      share|cite|improve this answer





















      • Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
        – E-mu
        13 mins ago
















      4














      In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that




      for every $x$, if $xin X$, then $xin Y$.




      Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when




      either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true




      If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.



      The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.






      share|cite|improve this answer





















      • Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
        – E-mu
        13 mins ago














      4












      4








      4






      In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that




      for every $x$, if $xin X$, then $xin Y$.




      Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when




      either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true




      If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.



      The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.






      share|cite|improve this answer












      In order to show that for two sets $X$ and $Y$ it holds that $Xsubseteq Y$, you have to prove that




      for every $x$, if $xin X$, then $xin Y$.




      Note that a statement of the form “if $mathscr{A}$ then $mathscr{B}$” is true when




      either $mathscr{A}$ is false or both $mathscr{A}$ and $mathscr{B}$ are true




      If $X$ is the empty set, then “$xin X$” is false for every $x$; hence “if $xin X$ then $xin Y$” is true.



      The phrase “take an arbitrary element $xin X$” is possibly misleading, but its intended meaning is “suppose $xin X$”.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 1 hour ago









      egreg

      177k1484198




      177k1484198












      • Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
        – E-mu
        13 mins ago


















      • Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
        – E-mu
        13 mins ago
















      Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
      – E-mu
      13 mins ago




      Ah, so if we write "suppose $x in X$" and successfully deduce that "$x in Y$" , we have shown that "if $x in X$, then $x in Y$" is true. But, if "$x in X$" is always false, we have still shown "if $x in X$ then $x in Y$" is true, since as you wrote, the implication would be true for all $x$ in that case. I.e. we are allowed to suppose things are true, even when they may never be. Is my understanding correct?
      – E-mu
      13 mins ago











      4














      If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.



      It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.



      So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.






      share|cite|improve this answer




























        4














        If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.



        It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.



        So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.






        share|cite|improve this answer


























          4












          4








          4






          If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.



          It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.



          So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.






          share|cite|improve this answer














          If $D$ is the empty set then the following statement is always true:$$text{If }xin Dtext{ then }cdots$$no matter what the dots are standing for.



          It is false that $xin D$ and "ex falso sequitur quodlibet". A false statement implies whatever you want.



          So the assumption will not bring you into troubles, but - on the contrary - will give you freedom to accept whatever you want.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          drhab

          96.7k544128




          96.7k544128























              2














              Since the empty set is a subset of any set, there is no need of including that in a formal proof.
              However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .






              share|cite|improve this answer


























                2














                Since the empty set is a subset of any set, there is no need of including that in a formal proof.
                However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .






                share|cite|improve this answer
























                  2












                  2








                  2






                  Since the empty set is a subset of any set, there is no need of including that in a formal proof.
                  However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .






                  share|cite|improve this answer












                  Since the empty set is a subset of any set, there is no need of including that in a formal proof.
                  However it is a good idea to be aware of the fact that the equality holds even in the case of empty sets .







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Mohammad Riazi-Kermani

                  40.5k42058




                  40.5k42058






























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                      A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

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                      0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{...
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                      Calculate evaluation metrics using cross_val_predict sklearn

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                      1 In the sklearn.model_selection.cross_val_predict page it is stated: Generate cross-validated estimates for each input data point. It is not appropriate to pass these predictions into an evaluation metric. Can someone explain what does it mean? If this gives estimate of Y (y prediction) for every Y (true Y), why can't I calculate metrics such as RMSE or coefficient of determination using these results? python scikit-learn cross-validation share | improve this question edited Nov 28 '18 at 17:52 desertnaut 20.3k 7 43 79 ...
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                      Insert data from modal to MySQL (multiple modal on website)

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                      0 I have problem with inserting data to MySQL from modal. My modal: <a href="#" class="badge badge-pill badge-success">6 komentarzy</a> <a href="#" class="badge badge-pill badge-danger">brak komentarzy</a> <a data-toggle="modal" href="#add_desk_comm_{$desk_['desk_id']}" data-target="#add_desk_comm_{$desk_['desk_id']}" class="ediiit">(dodaj)</a> <div class="modal fade" id="add_desk_comm_{$desk_['desk_id']}" tabindex="-1" role="dialog" aria-labelledby="edit_printer" aria-hidden="true"> <div class="modal-dialog" role="document"> ...
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