$Ysubseteq X$ iff $Xcup Y^c=Omega$, and $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$

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Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:



1) $Ysubseteq X$ iff $Xcup Y^c=Omega$



2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$



Here $^c$ denotes the complement



The statements do logically makes sense, however I'm having trouble proving it formally.










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    up vote
    2
    down vote

    favorite












    Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:



    1) $Ysubseteq X$ iff $Xcup Y^c=Omega$



    2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$



    Here $^c$ denotes the complement



    The statements do logically makes sense, however I'm having trouble proving it formally.










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:



      1) $Ysubseteq X$ iff $Xcup Y^c=Omega$



      2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$



      Here $^c$ denotes the complement



      The statements do logically makes sense, however I'm having trouble proving it formally.










      share|cite|improve this question















      Let $X$ and $Y$ be subsets of the universe $Omega$. Prove the following:



      1) $Ysubseteq X$ iff $Xcup Y^c=Omega$



      2) $Xcap Y=emptyset$ iff $X^ccup Y^c=Omega$



      Here $^c$ denotes the complement



      The statements do logically makes sense, however I'm having trouble proving it formally.







      discrete-mathematics






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      edited Sep 9 at 11:35

























      asked Sep 9 at 11:27









      Mictej

      304




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          Let's prove the first complication.



          If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.



          To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$



          The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$



          By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.



          The second statement is similar to prove, but if you want I'll make it.






          share|cite|improve this answer




















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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

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            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Let's prove the first complication.



            If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.



            To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$



            The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$



            By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.



            The second statement is similar to prove, but if you want I'll make it.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              Let's prove the first complication.



              If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.



              To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$



              The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$



              By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.



              The second statement is similar to prove, but if you want I'll make it.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Let's prove the first complication.



                If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.



                To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$



                The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$



                By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.



                The second statement is similar to prove, but if you want I'll make it.






                share|cite|improve this answer












                Let's prove the first complication.



                If $Ysubseteq X$ then we have to prove that $Xcup Y^c=Omega$.



                To prove it let's show both inclusions: it is clear that $Xcup Y^csubseteq Omega,$ so we only have to prove the other one. Take $xin Omega:$ if it belongs to $X$ we're done, otherwise it belongs to $X^c,$ that by hypothesis on $Y$ is a subset of $Y^c.$ So it certainly belongs to the union of $X$ and $Y^c.$



                The other arrow is similar: if $Xcup Y^c=Omega$ then we have to prove that $Ysubseteq X.$



                By absurd, suppose that $exists xin Ysetminus X.$ Then it wouldn't belong to $Xcup Y^c$ because it is nor in $X$ neither in $Y^c.$ So it is an element of $Omega$ which doesn't belong to $Xcup Y^c,$ which goes against the hypotheses.



                The second statement is similar to prove, but if you want I'll make it.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 9 at 11:40









                Riccardo Ceccon

                875320




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