Set Theory $Msetminusemptyset$ possible? [closed]












-1












$begingroup$


M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?






elementary-set-theory







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$endgroup$



closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    It is simply $$M setminus emptyset =M$$
    $endgroup$
    – Crostul
    Jan 10 at 12:01












  • $begingroup$
    thanks for the answer but why is it exactly the same ? i know it is a bit tautological
    $endgroup$
    – user627643
    Jan 10 at 12:02












  • $begingroup$
    Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
    $endgroup$
    – Git Gud
    Jan 10 at 12:03










  • $begingroup$
    Did you mean $Mbackslash{emptyset}$?
    $endgroup$
    – J.G.
    Jan 10 at 12:20
















-1












$begingroup$


M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?






elementary-set-theory







share|cite|improve this question











$endgroup$



closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    It is simply $$M setminus emptyset =M$$
    $endgroup$
    – Crostul
    Jan 10 at 12:01












  • $begingroup$
    thanks for the answer but why is it exactly the same ? i know it is a bit tautological
    $endgroup$
    – user627643
    Jan 10 at 12:02












  • $begingroup$
    Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
    $endgroup$
    – Git Gud
    Jan 10 at 12:03










  • $begingroup$
    Did you mean $Mbackslash{emptyset}$?
    $endgroup$
    – J.G.
    Jan 10 at 12:20














-1












-1








-1





$begingroup$


M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?






elementary-set-theory







share|cite|improve this question











$endgroup$




M is a non-empty set so is : $Msetminusemptyset$ possible to create a set like that?






elementary-set-theory




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 10 at 13:49









Aniruddh Agarwal

1218




1218










asked Jan 10 at 11:59







user627643











closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin Jan 10 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Adrian Keister, Holo, amWhy, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    It is simply $$M setminus emptyset =M$$
    $endgroup$
    – Crostul
    Jan 10 at 12:01












  • $begingroup$
    thanks for the answer but why is it exactly the same ? i know it is a bit tautological
    $endgroup$
    – user627643
    Jan 10 at 12:02












  • $begingroup$
    Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
    $endgroup$
    – Git Gud
    Jan 10 at 12:03










  • $begingroup$
    Did you mean $Mbackslash{emptyset}$?
    $endgroup$
    – J.G.
    Jan 10 at 12:20


















  • $begingroup$
    It is simply $$M setminus emptyset =M$$
    $endgroup$
    – Crostul
    Jan 10 at 12:01












  • $begingroup$
    thanks for the answer but why is it exactly the same ? i know it is a bit tautological
    $endgroup$
    – user627643
    Jan 10 at 12:02












  • $begingroup$
    Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
    $endgroup$
    – Git Gud
    Jan 10 at 12:03










  • $begingroup$
    Did you mean $Mbackslash{emptyset}$?
    $endgroup$
    – J.G.
    Jan 10 at 12:20
















$begingroup$
It is simply $$M setminus emptyset =M$$
$endgroup$
– Crostul
Jan 10 at 12:01






$begingroup$
It is simply $$M setminus emptyset =M$$
$endgroup$
– Crostul
Jan 10 at 12:01














$begingroup$
thanks for the answer but why is it exactly the same ? i know it is a bit tautological
$endgroup$
– user627643
Jan 10 at 12:02






$begingroup$
thanks for the answer but why is it exactly the same ? i know it is a bit tautological
$endgroup$
– user627643
Jan 10 at 12:02














$begingroup$
Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
$endgroup$
– Git Gud
Jan 10 at 12:03




$begingroup$
Given any two sets, $X$ and $Y$, $Xsetminus Y$ is also a set.
$endgroup$
– Git Gud
Jan 10 at 12:03












$begingroup$
Did you mean $Mbackslash{emptyset}$?
$endgroup$
– J.G.
Jan 10 at 12:20




$begingroup$
Did you mean $Mbackslash{emptyset}$?
$endgroup$
– J.G.
Jan 10 at 12:20










3 Answers
3






active

oldest

votes


















2












$begingroup$

We have $M setminus emptyset =M$.



Proof:




  1. If $x in M setminus emptyset$, then $x in M$


  2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$







share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    We know that,
    $$AB=Acap B^c$$
    $$Longrightarrow Msetminusemptyset=Mcap U=M$$
    where $U$ is the universal set.



    Hope it is helpful






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.






      share|cite|improve this answer









      $endgroup$



















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        We have $M setminus emptyset =M$.



        Proof:




        1. If $x in M setminus emptyset$, then $x in M$


        2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$







        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          We have $M setminus emptyset =M$.



          Proof:




          1. If $x in M setminus emptyset$, then $x in M$


          2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$







          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            We have $M setminus emptyset =M$.



            Proof:




            1. If $x in M setminus emptyset$, then $x in M$


            2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$







            share|cite|improve this answer









            $endgroup$



            We have $M setminus emptyset =M$.



            Proof:




            1. If $x in M setminus emptyset$, then $x in M$


            2. If $x in M$, then $x in M$ and $x notin emptyset$, hence $x in M setminus emptyset.$








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 10 at 12:13









            FredFred

            45.7k1848




            45.7k1848























                1












                $begingroup$

                We know that,
                $$AB=Acap B^c$$
                $$Longrightarrow Msetminusemptyset=Mcap U=M$$
                where $U$ is the universal set.



                Hope it is helpful






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  We know that,
                  $$AB=Acap B^c$$
                  $$Longrightarrow Msetminusemptyset=Mcap U=M$$
                  where $U$ is the universal set.



                  Hope it is helpful






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    We know that,
                    $$AB=Acap B^c$$
                    $$Longrightarrow Msetminusemptyset=Mcap U=M$$
                    where $U$ is the universal set.



                    Hope it is helpful






                    share|cite|improve this answer









                    $endgroup$



                    We know that,
                    $$AB=Acap B^c$$
                    $$Longrightarrow Msetminusemptyset=Mcap U=M$$
                    where $U$ is the universal set.



                    Hope it is helpful







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 10 at 12:20









                    MartundMartund

                    1,623213




                    1,623213























                        0












                        $begingroup$

                        $X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.






                            share|cite|improve this answer









                            $endgroup$



                            $X setminus Y$ is the set of all elements in $X$ and not in $Y$ (i.e. $X setminus Y = {x in X : x notin Y}$). So if $X cap Y = emptyset$, then $X setminus Y = X$. In your case, $M cap emptyset$ is really easy to observe. So you have $M setminus emptyset = M$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 10 at 12:16









                            BermudesBermudes

                            17713




                            17713















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