Prove $f[A] subseteq f[B] Rightarrow A ⊆ B$, given f is injective.












1












$begingroup$


I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:



(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)



Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.






discrete-mathematics proof-verification elementary-set-theory







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$endgroup$












  • $begingroup$
    You should write the question properly. What's $f,A,B$?
    $endgroup$
    – Yanko
    Jan 4 at 16:31






  • 1




    $begingroup$
    How do you conclude that $ain B$? That's where you should use injectivity
    $endgroup$
    – Exodd
    Jan 4 at 16:33






  • 1




    $begingroup$
    The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
    $endgroup$
    – Michael Burr
    Jan 4 at 16:40
















1












$begingroup$


I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:



(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)



Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.






discrete-mathematics proof-verification elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    You should write the question properly. What's $f,A,B$?
    $endgroup$
    – Yanko
    Jan 4 at 16:31






  • 1




    $begingroup$
    How do you conclude that $ain B$? That's where you should use injectivity
    $endgroup$
    – Exodd
    Jan 4 at 16:33






  • 1




    $begingroup$
    The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
    $endgroup$
    – Michael Burr
    Jan 4 at 16:40














1












1








1





$begingroup$


I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:



(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)



Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.






discrete-mathematics proof-verification elementary-set-theory







share|cite|improve this question











$endgroup$




I have constructed a proof for the above question, but do not know where I must use the hypothesis of injectivity:



(Note: Let $X,Y$ be sets with $A,B ⊆ X$ and $f:X to Y $.)



Let $y∈f[A]$; then, there exists $a∈A$ s.t. $f(a) = y$. Since $f[A]⊆ f[B]$, then, we also have $y∈f[B]$ and so, $f(a)∈ f[B]$. We conclude $a∈B$. Thus, $a∈A Rightarrow a∈B$, so $A⊆B$.






discrete-mathematics proof-verification elementary-set-theory




discrete-mathematics proof-verification elementary-set-theory






share|cite|improve this question















share|cite|improve this question













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edited Jan 4 at 16:40









Foobaz John

22.9k41552




22.9k41552










asked Jan 4 at 16:29









Julia KimJulia Kim

174




174












  • $begingroup$
    You should write the question properly. What's $f,A,B$?
    $endgroup$
    – Yanko
    Jan 4 at 16:31






  • 1




    $begingroup$
    How do you conclude that $ain B$? That's where you should use injectivity
    $endgroup$
    – Exodd
    Jan 4 at 16:33






  • 1




    $begingroup$
    The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
    $endgroup$
    – Michael Burr
    Jan 4 at 16:40


















  • $begingroup$
    You should write the question properly. What's $f,A,B$?
    $endgroup$
    – Yanko
    Jan 4 at 16:31






  • 1




    $begingroup$
    How do you conclude that $ain B$? That's where you should use injectivity
    $endgroup$
    – Exodd
    Jan 4 at 16:33






  • 1




    $begingroup$
    The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
    $endgroup$
    – Michael Burr
    Jan 4 at 16:40
















$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31




$begingroup$
You should write the question properly. What's $f,A,B$?
$endgroup$
– Yanko
Jan 4 at 16:31




1




1




$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33




$begingroup$
How do you conclude that $ain B$? That's where you should use injectivity
$endgroup$
– Exodd
Jan 4 at 16:33




1




1




$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40




$begingroup$
The proof must be wrong as written since you never use the assumption of injectivity. That being said, you only need one sentence to fix this up.
$endgroup$
– Michael Burr
Jan 4 at 16:40










2 Answers
2






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$begingroup$

Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Attempt:



    Assume $B subset A$ ($B$ proper subset of $A$).
    ($A$ $B not = emptyset$)



    There exists an $a in A$ $B.$



    $f(a) in f(A$ $B) subset f(A)$, i.e.



    $f(a) in f(A)$;



    since $f$ is injective: $f(a) not in f(B)$,



    contradicts the assumption $f(A)subset f(B).$






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      3












      $begingroup$

      Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.






          share|cite|improve this answer









          $endgroup$



          Suppose that $xin A$. Then $y=f(x)in f(A)$ and because $f(A)subseteq f(B)$ we have that $yin f(B)$. So there exists $w$ in $B$ such that $f(w)=y=f(x)$. But $f$ is injective, whence $w=x$ and $xin B$ as desired.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 16:39









          Foobaz JohnFoobaz John

          22.9k41552




          22.9k41552























              0












              $begingroup$

              Attempt:



              Assume $B subset A$ ($B$ proper subset of $A$).
              ($A$ $B not = emptyset$)



              There exists an $a in A$ $B.$



              $f(a) in f(A$ $B) subset f(A)$, i.e.



              $f(a) in f(A)$;



              since $f$ is injective: $f(a) not in f(B)$,



              contradicts the assumption $f(A)subset f(B).$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Attempt:



                Assume $B subset A$ ($B$ proper subset of $A$).
                ($A$ $B not = emptyset$)



                There exists an $a in A$ $B.$



                $f(a) in f(A$ $B) subset f(A)$, i.e.



                $f(a) in f(A)$;



                since $f$ is injective: $f(a) not in f(B)$,



                contradicts the assumption $f(A)subset f(B).$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Attempt:



                  Assume $B subset A$ ($B$ proper subset of $A$).
                  ($A$ $B not = emptyset$)



                  There exists an $a in A$ $B.$



                  $f(a) in f(A$ $B) subset f(A)$, i.e.



                  $f(a) in f(A)$;



                  since $f$ is injective: $f(a) not in f(B)$,



                  contradicts the assumption $f(A)subset f(B).$






                  share|cite|improve this answer









                  $endgroup$



                  Attempt:



                  Assume $B subset A$ ($B$ proper subset of $A$).
                  ($A$ $B not = emptyset$)



                  There exists an $a in A$ $B.$



                  $f(a) in f(A$ $B) subset f(A)$, i.e.



                  $f(a) in f(A)$;



                  since $f$ is injective: $f(a) not in f(B)$,



                  contradicts the assumption $f(A)subset f(B).$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 18:51









                  Peter SzilasPeter Szilas

                  11.7k2822




                  11.7k2822






























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