Prove that if $A preceq B$ and $B approx C$ then $A preceq C$












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Could anyone check my working please?




Prove that if $A preceq B$ and $B approx C$ then $A preceq C$




Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.



Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.



We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.










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    $begingroup$


    Could anyone check my working please?




    Prove that if $A preceq B$ and $B approx C$ then $A preceq C$




    Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.



    Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.



    We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.










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      0





      $begingroup$


      Could anyone check my working please?




      Prove that if $A preceq B$ and $B approx C$ then $A preceq C$




      Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.



      Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.



      We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.










      share|cite|improve this question











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      Could anyone check my working please?




      Prove that if $A preceq B$ and $B approx C$ then $A preceq C$




      Assume $A preceq B$ (ie. there is a $f:Ato B$ such that it is one-one) and $B approx C$ (ie. there is a $g:B to C$ that is one-one and onto), we need to show that there is a $h:Ato C$ that is one-one.



      Let this $h$ be $gcirc f$, and show that it is one-one. ie. If $g(f(a))=g(f(a'))$ then $f(a)=f(a')$.



      We know that $g(f(a))=g(f(a'))$, but since we know that $g$ is one-one, ie. if $g(b)=g(b')$ then $b=b'$, therefore $f(a)=f(a')$.







      proof-verification elementary-set-theory






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      edited Dec 30 '18 at 20:29









      Andrés E. Caicedo

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      66.1k8160252










      asked Dec 30 '18 at 16:14









      Daniel MakDaniel Mak

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      520517






















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          $begingroup$

          If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.




          Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.







          share|cite|improve this answer









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            $begingroup$

            If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.




            Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.







            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.




              Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.







              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.




                Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.







                share|cite|improve this answer









                $endgroup$



                If you want to show $g circ f$ is one-to-one, you want to show that $g(f(a)) = g(f(a'))$ implies $a = a'$. Notice you haven't used anything about $f$ in your argument.




                Your argument correctly shows that $f(a) = f(a')$. Now use the fact that $f$ is one-to-one to conclude that $a=a'$.








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                share|cite|improve this answer










                answered Dec 30 '18 at 17:01









                angryavianangryavian

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