Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.












8












$begingroup$



Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.




I think this is false because of the counterexample



$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$



but I am not sure if I am right.






discrete-mathematics elementary-set-theory







share|cite|improve this question











$endgroup$








  • 9




    $begingroup$
    You are right.${}$
    $endgroup$
    – David Mitra
    Jan 11 at 19:54






  • 4




    $begingroup$
    Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 19:58
















8












$begingroup$



Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.




I think this is false because of the counterexample



$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$



but I am not sure if I am right.






discrete-mathematics elementary-set-theory







share|cite|improve this question











$endgroup$








  • 9




    $begingroup$
    You are right.${}$
    $endgroup$
    – David Mitra
    Jan 11 at 19:54






  • 4




    $begingroup$
    Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 19:58














8












8








8





$begingroup$



Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.




I think this is false because of the counterexample



$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$



but I am not sure if I am right.






discrete-mathematics elementary-set-theory







share|cite|improve this question











$endgroup$





Prove or give a counterexample: If $A$ is a subset of $B$ and $B$ belongs to $C$, then $A$ belongs to $C$.




I think this is false because of the counterexample



$$
begin{align}
A &= {1,2}\
B &= {1,2,3}\
C &= {{1,2,3}}
end{align}
$$



but I am not sure if I am right.






discrete-mathematics elementary-set-theory




discrete-mathematics 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 11 at 20:16









6005

36.2k751125




36.2k751125










asked Jan 11 at 19:53









hkdhkd

411




411








  • 9




    $begingroup$
    You are right.${}$
    $endgroup$
    – David Mitra
    Jan 11 at 19:54






  • 4




    $begingroup$
    Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 19:58














  • 9




    $begingroup$
    You are right.${}$
    $endgroup$
    – David Mitra
    Jan 11 at 19:54






  • 4




    $begingroup$
    Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
    $endgroup$
    – Hagen von Eitzen
    Jan 11 at 19:58








9




9




$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54




$begingroup$
You are right.${}$
$endgroup$
– David Mitra
Jan 11 at 19:54




4




4




$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58




$begingroup$
Smallest counter-example: $A={}$, $B={{}}$, $C={{{}}}$ :)
$endgroup$
– Hagen von Eitzen
Jan 11 at 19:58










1 Answer
1






active

oldest

votes


















4












$begingroup$

Nice counterexample, you are correct. Let us check each condition:




$A$ is a subset of $B$




That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.




$B$ belongs to $C$




That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.




then $A$ belongs to $C$




This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.



So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.






share|cite|improve this answer









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    4












    $begingroup$

    Nice counterexample, you are correct. Let us check each condition:




    $A$ is a subset of $B$




    That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.




    $B$ belongs to $C$




    That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.




    then $A$ belongs to $C$




    This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.



    So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Nice counterexample, you are correct. Let us check each condition:




      $A$ is a subset of $B$




      That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.




      $B$ belongs to $C$




      That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.




      then $A$ belongs to $C$




      This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.



      So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Nice counterexample, you are correct. Let us check each condition:




        $A$ is a subset of $B$




        That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.




        $B$ belongs to $C$




        That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.




        then $A$ belongs to $C$




        This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.



        So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.






        share|cite|improve this answer









        $endgroup$



        Nice counterexample, you are correct. Let us check each condition:




        $A$ is a subset of $B$




        That's true: you have $A = {1,2}$ and $B = {1,2,3}$. The elements of $A$ are $1$ and $2$, and both of them are also elements of $B$.




        $B$ belongs to $C$




        That's true also. You took $C = {{1,2,3}}$. $C$ has one element, and that's $B$. $C = {B}$.




        then $A$ belongs to $C$




        This is false -- $C$ only has one element, $B$. But $B$ isn't $A$, becuase $B$ has $3$ elements and $A$ only has $2$ elements. Specifically, $B$ has $3$ and $A$ doesn't.



        So your counterexample is correct: $A$ is a subset of $B$, and $B$ belongs to $C$, but that doesn't necessarily imply that $A$ belongs to $C$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 20:15









        60056005

        36.2k751125




        36.2k751125






























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