Proof of $f(A)=f(A-B)+f(B)$ when $f$ is a injective map











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I want to prove the following proposition:



Let $A$ be a set, $B$ be a subset of $A$, and $f: Ato B$ be a injective map, then $f(A) = f(A-B) + f(B).$



Could you check my proof below?



Assume $f(A-B)cap f(B) neq emptyset$. For $xin f(A-B)cap f(B)$ there exist $ain A$ which satisfies $f(a)=x$ and $bin A-B$ which satisfies $f(b)=x$. However, this contradicts the original assumption that $f$ is injective: $forall a, b in A, f(a)=f(b)Rightarrow a=b$, thus the above is impossible. Thus, $f(A-B)cap f(B)=emptyset$ and hence $f(A)=f(A-B)+f(B)$.










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  • 3




    Terribly confusing. In the question statement you have a plus sign where a $cup$ is expected. Then right at the start of your proof you consider an intersection of sets, for some reason. This is followed by a dubious argument and you eventually finalize with incoherent conclusions with no sight of an expected double inclusion proof.
    – Git Gud
    Nov 17 at 9:38










  • @GitGud I am so sorry, it was a typo.
    – orematasaburou
    Nov 17 at 9:43






  • 1




    There is another typo: You want to say that there is an $b in B$ and an $a in A setminus B$ such that $f(a) = f(b)$.
    – Stefan Mesken
    Nov 17 at 9:59















up vote
3
down vote

favorite












I want to prove the following proposition:



Let $A$ be a set, $B$ be a subset of $A$, and $f: Ato B$ be a injective map, then $f(A) = f(A-B) + f(B).$



Could you check my proof below?



Assume $f(A-B)cap f(B) neq emptyset$. For $xin f(A-B)cap f(B)$ there exist $ain A$ which satisfies $f(a)=x$ and $bin A-B$ which satisfies $f(b)=x$. However, this contradicts the original assumption that $f$ is injective: $forall a, b in A, f(a)=f(b)Rightarrow a=b$, thus the above is impossible. Thus, $f(A-B)cap f(B)=emptyset$ and hence $f(A)=f(A-B)+f(B)$.










share|cite|improve this question




















  • 3




    Terribly confusing. In the question statement you have a plus sign where a $cup$ is expected. Then right at the start of your proof you consider an intersection of sets, for some reason. This is followed by a dubious argument and you eventually finalize with incoherent conclusions with no sight of an expected double inclusion proof.
    – Git Gud
    Nov 17 at 9:38










  • @GitGud I am so sorry, it was a typo.
    – orematasaburou
    Nov 17 at 9:43






  • 1




    There is another typo: You want to say that there is an $b in B$ and an $a in A setminus B$ such that $f(a) = f(b)$.
    – Stefan Mesken
    Nov 17 at 9:59













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I want to prove the following proposition:



Let $A$ be a set, $B$ be a subset of $A$, and $f: Ato B$ be a injective map, then $f(A) = f(A-B) + f(B).$



Could you check my proof below?



Assume $f(A-B)cap f(B) neq emptyset$. For $xin f(A-B)cap f(B)$ there exist $ain A$ which satisfies $f(a)=x$ and $bin A-B$ which satisfies $f(b)=x$. However, this contradicts the original assumption that $f$ is injective: $forall a, b in A, f(a)=f(b)Rightarrow a=b$, thus the above is impossible. Thus, $f(A-B)cap f(B)=emptyset$ and hence $f(A)=f(A-B)+f(B)$.










share|cite|improve this question















I want to prove the following proposition:



Let $A$ be a set, $B$ be a subset of $A$, and $f: Ato B$ be a injective map, then $f(A) = f(A-B) + f(B).$



Could you check my proof below?



Assume $f(A-B)cap f(B) neq emptyset$. For $xin f(A-B)cap f(B)$ there exist $ain A$ which satisfies $f(a)=x$ and $bin A-B$ which satisfies $f(b)=x$. However, this contradicts the original assumption that $f$ is injective: $forall a, b in A, f(a)=f(b)Rightarrow a=b$, thus the above is impossible. Thus, $f(A-B)cap f(B)=emptyset$ and hence $f(A)=f(A-B)+f(B)$.







proof-verification elementary-set-theory






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edited Nov 17 at 9:41

























asked Nov 17 at 9:29









orematasaburou

345




345








  • 3




    Terribly confusing. In the question statement you have a plus sign where a $cup$ is expected. Then right at the start of your proof you consider an intersection of sets, for some reason. This is followed by a dubious argument and you eventually finalize with incoherent conclusions with no sight of an expected double inclusion proof.
    – Git Gud
    Nov 17 at 9:38










  • @GitGud I am so sorry, it was a typo.
    – orematasaburou
    Nov 17 at 9:43






  • 1




    There is another typo: You want to say that there is an $b in B$ and an $a in A setminus B$ such that $f(a) = f(b)$.
    – Stefan Mesken
    Nov 17 at 9:59














  • 3




    Terribly confusing. In the question statement you have a plus sign where a $cup$ is expected. Then right at the start of your proof you consider an intersection of sets, for some reason. This is followed by a dubious argument and you eventually finalize with incoherent conclusions with no sight of an expected double inclusion proof.
    – Git Gud
    Nov 17 at 9:38










  • @GitGud I am so sorry, it was a typo.
    – orematasaburou
    Nov 17 at 9:43






  • 1




    There is another typo: You want to say that there is an $b in B$ and an $a in A setminus B$ such that $f(a) = f(b)$.
    – Stefan Mesken
    Nov 17 at 9:59








3




3




Terribly confusing. In the question statement you have a plus sign where a $cup$ is expected. Then right at the start of your proof you consider an intersection of sets, for some reason. This is followed by a dubious argument and you eventually finalize with incoherent conclusions with no sight of an expected double inclusion proof.
– Git Gud
Nov 17 at 9:38




Terribly confusing. In the question statement you have a plus sign where a $cup$ is expected. Then right at the start of your proof you consider an intersection of sets, for some reason. This is followed by a dubious argument and you eventually finalize with incoherent conclusions with no sight of an expected double inclusion proof.
– Git Gud
Nov 17 at 9:38












@GitGud I am so sorry, it was a typo.
– orematasaburou
Nov 17 at 9:43




@GitGud I am so sorry, it was a typo.
– orematasaburou
Nov 17 at 9:43




1




1




There is another typo: You want to say that there is an $b in B$ and an $a in A setminus B$ such that $f(a) = f(b)$.
– Stefan Mesken
Nov 17 at 9:59




There is another typo: You want to say that there is an $b in B$ and an $a in A setminus B$ such that $f(a) = f(b)$.
– Stefan Mesken
Nov 17 at 9:59










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Other than the typo I've pointed out in a comment, your proof is fine.



Note, however, that you don't need that $B subseteq A$. The following is true:



Lemma. Let $f colon A to B$ be injective, $C subseteq D subseteq A$ Then
$$f[D] = f[D setminus C] mathbin{dot{cup}} f[C].$$



(The proof is virtually the same as the one you've given.)



Also note that injectivity is necessary:



Lemma. If $f colon A to B$ is not injective, then there is some $C subseteq A$ such that
$$f[A setminus C] cap f[C] neq emptyset.$$



I'll leave the easy proof to you as an exercise. (Hint: You can choose $C$ to be a singleton.)






share|cite|improve this answer



















  • 1




    Thank you for the answer and exercise! That actually wasn't typo for me (I meant it). Would you explain why I have to write $bin B, ain A-B$ instead of just $a, bin A$?
    – orematasaburou
    Nov 17 at 10:12










  • @oremata You want to show that $f[A setminus B] cap f[B] = emptyset$. Toward a contradiction you must assume there is some $a in A setminus B$ and some $b in B$ such that $f(a) = f(b)$. If $a in A cap B$, it could be that $a = b$ and then there is no contradiction.
    – Stefan Mesken
    Nov 17 at 10:18










  • I understand the point. But, before writing about the injection, I mentioned in the proof that there exist $ain A$ and $bin A-B$, so I think I don't have to write it again... Is it wrong? Sorry, I'm new to proof stuff.
    – orematasaburou
    Nov 17 at 10:27










  • @oremata The point is that you could have $a = b$ and then you don't get your contradiction. If you fix $a in B$ and $b in A setminus B$, you know that $a neq b$ (because $B$ and $A setminus B$ don't have a common element). If you then have $f(a) = f(b)$, this is a contradiction to injectivity.
    – Stefan Mesken
    Nov 17 at 10:30










  • If you want to use a binary operation symbol with an accent on it, it should be coded as mathbin{dot{cup}} so the spacing is right.
    – egreg
    Nov 17 at 10:41


















up vote
0
down vote













I assume that $+$ denotes disjoint union. Your idea is good, but you lose yourself in explanations.



The proof that $f(A)=f(A-B)cup f(B)$ is easy and doesn't require injectivity.



Now the proof the sets are disjoint. More generally,




if $X$ and $Y$ are disjoint sets and $f(X)cap f(Y)neemptyset$, then $f$ is not injective.




Indeed, if $zin f(X)cap f(Y)$, then $z=f(x)=f(y)$, with $xin X$ and $yin Y$. Since $X$ and $Y$ are disjoint, $xne y$.






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    2 Answers
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    up vote
    2
    down vote



    accepted










    Other than the typo I've pointed out in a comment, your proof is fine.



    Note, however, that you don't need that $B subseteq A$. The following is true:



    Lemma. Let $f colon A to B$ be injective, $C subseteq D subseteq A$ Then
    $$f[D] = f[D setminus C] mathbin{dot{cup}} f[C].$$



    (The proof is virtually the same as the one you've given.)



    Also note that injectivity is necessary:



    Lemma. If $f colon A to B$ is not injective, then there is some $C subseteq A$ such that
    $$f[A setminus C] cap f[C] neq emptyset.$$



    I'll leave the easy proof to you as an exercise. (Hint: You can choose $C$ to be a singleton.)






    share|cite|improve this answer



















    • 1




      Thank you for the answer and exercise! That actually wasn't typo for me (I meant it). Would you explain why I have to write $bin B, ain A-B$ instead of just $a, bin A$?
      – orematasaburou
      Nov 17 at 10:12










    • @oremata You want to show that $f[A setminus B] cap f[B] = emptyset$. Toward a contradiction you must assume there is some $a in A setminus B$ and some $b in B$ such that $f(a) = f(b)$. If $a in A cap B$, it could be that $a = b$ and then there is no contradiction.
      – Stefan Mesken
      Nov 17 at 10:18










    • I understand the point. But, before writing about the injection, I mentioned in the proof that there exist $ain A$ and $bin A-B$, so I think I don't have to write it again... Is it wrong? Sorry, I'm new to proof stuff.
      – orematasaburou
      Nov 17 at 10:27










    • @oremata The point is that you could have $a = b$ and then you don't get your contradiction. If you fix $a in B$ and $b in A setminus B$, you know that $a neq b$ (because $B$ and $A setminus B$ don't have a common element). If you then have $f(a) = f(b)$, this is a contradiction to injectivity.
      – Stefan Mesken
      Nov 17 at 10:30










    • If you want to use a binary operation symbol with an accent on it, it should be coded as mathbin{dot{cup}} so the spacing is right.
      – egreg
      Nov 17 at 10:41















    up vote
    2
    down vote



    accepted










    Other than the typo I've pointed out in a comment, your proof is fine.



    Note, however, that you don't need that $B subseteq A$. The following is true:



    Lemma. Let $f colon A to B$ be injective, $C subseteq D subseteq A$ Then
    $$f[D] = f[D setminus C] mathbin{dot{cup}} f[C].$$



    (The proof is virtually the same as the one you've given.)



    Also note that injectivity is necessary:



    Lemma. If $f colon A to B$ is not injective, then there is some $C subseteq A$ such that
    $$f[A setminus C] cap f[C] neq emptyset.$$



    I'll leave the easy proof to you as an exercise. (Hint: You can choose $C$ to be a singleton.)






    share|cite|improve this answer



















    • 1




      Thank you for the answer and exercise! That actually wasn't typo for me (I meant it). Would you explain why I have to write $bin B, ain A-B$ instead of just $a, bin A$?
      – orematasaburou
      Nov 17 at 10:12










    • @oremata You want to show that $f[A setminus B] cap f[B] = emptyset$. Toward a contradiction you must assume there is some $a in A setminus B$ and some $b in B$ such that $f(a) = f(b)$. If $a in A cap B$, it could be that $a = b$ and then there is no contradiction.
      – Stefan Mesken
      Nov 17 at 10:18










    • I understand the point. But, before writing about the injection, I mentioned in the proof that there exist $ain A$ and $bin A-B$, so I think I don't have to write it again... Is it wrong? Sorry, I'm new to proof stuff.
      – orematasaburou
      Nov 17 at 10:27










    • @oremata The point is that you could have $a = b$ and then you don't get your contradiction. If you fix $a in B$ and $b in A setminus B$, you know that $a neq b$ (because $B$ and $A setminus B$ don't have a common element). If you then have $f(a) = f(b)$, this is a contradiction to injectivity.
      – Stefan Mesken
      Nov 17 at 10:30










    • If you want to use a binary operation symbol with an accent on it, it should be coded as mathbin{dot{cup}} so the spacing is right.
      – egreg
      Nov 17 at 10:41













    up vote
    2
    down vote



    accepted







    up vote
    2
    down vote



    accepted






    Other than the typo I've pointed out in a comment, your proof is fine.



    Note, however, that you don't need that $B subseteq A$. The following is true:



    Lemma. Let $f colon A to B$ be injective, $C subseteq D subseteq A$ Then
    $$f[D] = f[D setminus C] mathbin{dot{cup}} f[C].$$



    (The proof is virtually the same as the one you've given.)



    Also note that injectivity is necessary:



    Lemma. If $f colon A to B$ is not injective, then there is some $C subseteq A$ such that
    $$f[A setminus C] cap f[C] neq emptyset.$$



    I'll leave the easy proof to you as an exercise. (Hint: You can choose $C$ to be a singleton.)






    share|cite|improve this answer














    Other than the typo I've pointed out in a comment, your proof is fine.



    Note, however, that you don't need that $B subseteq A$. The following is true:



    Lemma. Let $f colon A to B$ be injective, $C subseteq D subseteq A$ Then
    $$f[D] = f[D setminus C] mathbin{dot{cup}} f[C].$$



    (The proof is virtually the same as the one you've given.)



    Also note that injectivity is necessary:



    Lemma. If $f colon A to B$ is not injective, then there is some $C subseteq A$ such that
    $$f[A setminus C] cap f[C] neq emptyset.$$



    I'll leave the easy proof to you as an exercise. (Hint: You can choose $C$ to be a singleton.)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 17 at 10:46

























    answered Nov 17 at 10:03









    Stefan Mesken

    14.4k32046




    14.4k32046








    • 1




      Thank you for the answer and exercise! That actually wasn't typo for me (I meant it). Would you explain why I have to write $bin B, ain A-B$ instead of just $a, bin A$?
      – orematasaburou
      Nov 17 at 10:12










    • @oremata You want to show that $f[A setminus B] cap f[B] = emptyset$. Toward a contradiction you must assume there is some $a in A setminus B$ and some $b in B$ such that $f(a) = f(b)$. If $a in A cap B$, it could be that $a = b$ and then there is no contradiction.
      – Stefan Mesken
      Nov 17 at 10:18










    • I understand the point. But, before writing about the injection, I mentioned in the proof that there exist $ain A$ and $bin A-B$, so I think I don't have to write it again... Is it wrong? Sorry, I'm new to proof stuff.
      – orematasaburou
      Nov 17 at 10:27










    • @oremata The point is that you could have $a = b$ and then you don't get your contradiction. If you fix $a in B$ and $b in A setminus B$, you know that $a neq b$ (because $B$ and $A setminus B$ don't have a common element). If you then have $f(a) = f(b)$, this is a contradiction to injectivity.
      – Stefan Mesken
      Nov 17 at 10:30










    • If you want to use a binary operation symbol with an accent on it, it should be coded as mathbin{dot{cup}} so the spacing is right.
      – egreg
      Nov 17 at 10:41














    • 1




      Thank you for the answer and exercise! That actually wasn't typo for me (I meant it). Would you explain why I have to write $bin B, ain A-B$ instead of just $a, bin A$?
      – orematasaburou
      Nov 17 at 10:12










    • @oremata You want to show that $f[A setminus B] cap f[B] = emptyset$. Toward a contradiction you must assume there is some $a in A setminus B$ and some $b in B$ such that $f(a) = f(b)$. If $a in A cap B$, it could be that $a = b$ and then there is no contradiction.
      – Stefan Mesken
      Nov 17 at 10:18










    • I understand the point. But, before writing about the injection, I mentioned in the proof that there exist $ain A$ and $bin A-B$, so I think I don't have to write it again... Is it wrong? Sorry, I'm new to proof stuff.
      – orematasaburou
      Nov 17 at 10:27










    • @oremata The point is that you could have $a = b$ and then you don't get your contradiction. If you fix $a in B$ and $b in A setminus B$, you know that $a neq b$ (because $B$ and $A setminus B$ don't have a common element). If you then have $f(a) = f(b)$, this is a contradiction to injectivity.
      – Stefan Mesken
      Nov 17 at 10:30










    • If you want to use a binary operation symbol with an accent on it, it should be coded as mathbin{dot{cup}} so the spacing is right.
      – egreg
      Nov 17 at 10:41








    1




    1




    Thank you for the answer and exercise! That actually wasn't typo for me (I meant it). Would you explain why I have to write $bin B, ain A-B$ instead of just $a, bin A$?
    – orematasaburou
    Nov 17 at 10:12




    Thank you for the answer and exercise! That actually wasn't typo for me (I meant it). Would you explain why I have to write $bin B, ain A-B$ instead of just $a, bin A$?
    – orematasaburou
    Nov 17 at 10:12












    @oremata You want to show that $f[A setminus B] cap f[B] = emptyset$. Toward a contradiction you must assume there is some $a in A setminus B$ and some $b in B$ such that $f(a) = f(b)$. If $a in A cap B$, it could be that $a = b$ and then there is no contradiction.
    – Stefan Mesken
    Nov 17 at 10:18




    @oremata You want to show that $f[A setminus B] cap f[B] = emptyset$. Toward a contradiction you must assume there is some $a in A setminus B$ and some $b in B$ such that $f(a) = f(b)$. If $a in A cap B$, it could be that $a = b$ and then there is no contradiction.
    – Stefan Mesken
    Nov 17 at 10:18












    I understand the point. But, before writing about the injection, I mentioned in the proof that there exist $ain A$ and $bin A-B$, so I think I don't have to write it again... Is it wrong? Sorry, I'm new to proof stuff.
    – orematasaburou
    Nov 17 at 10:27




    I understand the point. But, before writing about the injection, I mentioned in the proof that there exist $ain A$ and $bin A-B$, so I think I don't have to write it again... Is it wrong? Sorry, I'm new to proof stuff.
    – orematasaburou
    Nov 17 at 10:27












    @oremata The point is that you could have $a = b$ and then you don't get your contradiction. If you fix $a in B$ and $b in A setminus B$, you know that $a neq b$ (because $B$ and $A setminus B$ don't have a common element). If you then have $f(a) = f(b)$, this is a contradiction to injectivity.
    – Stefan Mesken
    Nov 17 at 10:30




    @oremata The point is that you could have $a = b$ and then you don't get your contradiction. If you fix $a in B$ and $b in A setminus B$, you know that $a neq b$ (because $B$ and $A setminus B$ don't have a common element). If you then have $f(a) = f(b)$, this is a contradiction to injectivity.
    – Stefan Mesken
    Nov 17 at 10:30












    If you want to use a binary operation symbol with an accent on it, it should be coded as mathbin{dot{cup}} so the spacing is right.
    – egreg
    Nov 17 at 10:41




    If you want to use a binary operation symbol with an accent on it, it should be coded as mathbin{dot{cup}} so the spacing is right.
    – egreg
    Nov 17 at 10:41










    up vote
    0
    down vote













    I assume that $+$ denotes disjoint union. Your idea is good, but you lose yourself in explanations.



    The proof that $f(A)=f(A-B)cup f(B)$ is easy and doesn't require injectivity.



    Now the proof the sets are disjoint. More generally,




    if $X$ and $Y$ are disjoint sets and $f(X)cap f(Y)neemptyset$, then $f$ is not injective.




    Indeed, if $zin f(X)cap f(Y)$, then $z=f(x)=f(y)$, with $xin X$ and $yin Y$. Since $X$ and $Y$ are disjoint, $xne y$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      I assume that $+$ denotes disjoint union. Your idea is good, but you lose yourself in explanations.



      The proof that $f(A)=f(A-B)cup f(B)$ is easy and doesn't require injectivity.



      Now the proof the sets are disjoint. More generally,




      if $X$ and $Y$ are disjoint sets and $f(X)cap f(Y)neemptyset$, then $f$ is not injective.




      Indeed, if $zin f(X)cap f(Y)$, then $z=f(x)=f(y)$, with $xin X$ and $yin Y$. Since $X$ and $Y$ are disjoint, $xne y$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        I assume that $+$ denotes disjoint union. Your idea is good, but you lose yourself in explanations.



        The proof that $f(A)=f(A-B)cup f(B)$ is easy and doesn't require injectivity.



        Now the proof the sets are disjoint. More generally,




        if $X$ and $Y$ are disjoint sets and $f(X)cap f(Y)neemptyset$, then $f$ is not injective.




        Indeed, if $zin f(X)cap f(Y)$, then $z=f(x)=f(y)$, with $xin X$ and $yin Y$. Since $X$ and $Y$ are disjoint, $xne y$.






        share|cite|improve this answer












        I assume that $+$ denotes disjoint union. Your idea is good, but you lose yourself in explanations.



        The proof that $f(A)=f(A-B)cup f(B)$ is easy and doesn't require injectivity.



        Now the proof the sets are disjoint. More generally,




        if $X$ and $Y$ are disjoint sets and $f(X)cap f(Y)neemptyset$, then $f$ is not injective.




        Indeed, if $zin f(X)cap f(Y)$, then $z=f(x)=f(y)$, with $xin X$ and $yin Y$. Since $X$ and $Y$ are disjoint, $xne y$.







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        answered Nov 17 at 10:51









        egreg

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