Can the empty set be the image of a function on $mathbb{N}$? [duplicate]












5












$begingroup$



This question already has an answer here:




  • Why is there no function with a nonempty domain and an empty range?

    5 answers




I cannot find any example of function $f:mathbb{N}rightarrowmathbb{N}$ of which we can say that
$$
f(mathbb{N})=emptyset
$$

Does there exists any?










share|cite|improve this question











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marked as duplicate by Asaf Karagila elementary-set-theory
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Nov 26 '18 at 15:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    If $D_f=emptyset$
    $endgroup$
    – hamam_Abdallah
    Nov 25 '18 at 18:33






  • 2




    $begingroup$
    By definition, the set $f(mathbb{N})$ contains each $f(n)$ for $nin mathbb{N}$.
    $endgroup$
    – anomaly
    Nov 26 '18 at 0:46










  • $begingroup$
    You say "an empty set" but note that there is only one empty set: "the empty set."
    $endgroup$
    – David Richerby
    Nov 26 '18 at 13:18










  • $begingroup$
    Not sure that this counts as a duplicate. The other question was a technical question of why the axiom of specification fails to claim "$Fsubset Atimes emptyset$ so that $F$ is a function" exists and required a far more technical and obtuse than this rather practical question. That's my two cents. But I'm certainly not going to vote to reopen.
    $endgroup$
    – fleablood
    Nov 26 '18 at 16:45
















5












$begingroup$



This question already has an answer here:




  • Why is there no function with a nonempty domain and an empty range?

    5 answers




I cannot find any example of function $f:mathbb{N}rightarrowmathbb{N}$ of which we can say that
$$
f(mathbb{N})=emptyset
$$

Does there exists any?










share|cite|improve this question











$endgroup$



marked as duplicate by Asaf Karagila elementary-set-theory
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Nov 26 '18 at 15:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    If $D_f=emptyset$
    $endgroup$
    – hamam_Abdallah
    Nov 25 '18 at 18:33






  • 2




    $begingroup$
    By definition, the set $f(mathbb{N})$ contains each $f(n)$ for $nin mathbb{N}$.
    $endgroup$
    – anomaly
    Nov 26 '18 at 0:46










  • $begingroup$
    You say "an empty set" but note that there is only one empty set: "the empty set."
    $endgroup$
    – David Richerby
    Nov 26 '18 at 13:18










  • $begingroup$
    Not sure that this counts as a duplicate. The other question was a technical question of why the axiom of specification fails to claim "$Fsubset Atimes emptyset$ so that $F$ is a function" exists and required a far more technical and obtuse than this rather practical question. That's my two cents. But I'm certainly not going to vote to reopen.
    $endgroup$
    – fleablood
    Nov 26 '18 at 16:45














5












5








5





$begingroup$



This question already has an answer here:




  • Why is there no function with a nonempty domain and an empty range?

    5 answers




I cannot find any example of function $f:mathbb{N}rightarrowmathbb{N}$ of which we can say that
$$
f(mathbb{N})=emptyset
$$

Does there exists any?










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Why is there no function with a nonempty domain and an empty range?

    5 answers




I cannot find any example of function $f:mathbb{N}rightarrowmathbb{N}$ of which we can say that
$$
f(mathbb{N})=emptyset
$$

Does there exists any?





This question already has an answer here:




  • Why is there no function with a nonempty domain and an empty range?

    5 answers








functions elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 15:02









user21820

39.4k543155




39.4k543155










asked Nov 25 '18 at 18:32









avan1235avan1235

3297




3297




marked as duplicate by Asaf Karagila elementary-set-theory
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Nov 26 '18 at 15:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Asaf Karagila elementary-set-theory
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Nov 26 '18 at 15:22


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    If $D_f=emptyset$
    $endgroup$
    – hamam_Abdallah
    Nov 25 '18 at 18:33






  • 2




    $begingroup$
    By definition, the set $f(mathbb{N})$ contains each $f(n)$ for $nin mathbb{N}$.
    $endgroup$
    – anomaly
    Nov 26 '18 at 0:46










  • $begingroup$
    You say "an empty set" but note that there is only one empty set: "the empty set."
    $endgroup$
    – David Richerby
    Nov 26 '18 at 13:18










  • $begingroup$
    Not sure that this counts as a duplicate. The other question was a technical question of why the axiom of specification fails to claim "$Fsubset Atimes emptyset$ so that $F$ is a function" exists and required a far more technical and obtuse than this rather practical question. That's my two cents. But I'm certainly not going to vote to reopen.
    $endgroup$
    – fleablood
    Nov 26 '18 at 16:45


















  • $begingroup$
    If $D_f=emptyset$
    $endgroup$
    – hamam_Abdallah
    Nov 25 '18 at 18:33






  • 2




    $begingroup$
    By definition, the set $f(mathbb{N})$ contains each $f(n)$ for $nin mathbb{N}$.
    $endgroup$
    – anomaly
    Nov 26 '18 at 0:46










  • $begingroup$
    You say "an empty set" but note that there is only one empty set: "the empty set."
    $endgroup$
    – David Richerby
    Nov 26 '18 at 13:18










  • $begingroup$
    Not sure that this counts as a duplicate. The other question was a technical question of why the axiom of specification fails to claim "$Fsubset Atimes emptyset$ so that $F$ is a function" exists and required a far more technical and obtuse than this rather practical question. That's my two cents. But I'm certainly not going to vote to reopen.
    $endgroup$
    – fleablood
    Nov 26 '18 at 16:45
















$begingroup$
If $D_f=emptyset$
$endgroup$
– hamam_Abdallah
Nov 25 '18 at 18:33




$begingroup$
If $D_f=emptyset$
$endgroup$
– hamam_Abdallah
Nov 25 '18 at 18:33




2




2




$begingroup$
By definition, the set $f(mathbb{N})$ contains each $f(n)$ for $nin mathbb{N}$.
$endgroup$
– anomaly
Nov 26 '18 at 0:46




$begingroup$
By definition, the set $f(mathbb{N})$ contains each $f(n)$ for $nin mathbb{N}$.
$endgroup$
– anomaly
Nov 26 '18 at 0:46












$begingroup$
You say "an empty set" but note that there is only one empty set: "the empty set."
$endgroup$
– David Richerby
Nov 26 '18 at 13:18




$begingroup$
You say "an empty set" but note that there is only one empty set: "the empty set."
$endgroup$
– David Richerby
Nov 26 '18 at 13:18












$begingroup$
Not sure that this counts as a duplicate. The other question was a technical question of why the axiom of specification fails to claim "$Fsubset Atimes emptyset$ so that $F$ is a function" exists and required a far more technical and obtuse than this rather practical question. That's my two cents. But I'm certainly not going to vote to reopen.
$endgroup$
– fleablood
Nov 26 '18 at 16:45




$begingroup$
Not sure that this counts as a duplicate. The other question was a technical question of why the axiom of specification fails to claim "$Fsubset Atimes emptyset$ so that $F$ is a function" exists and required a far more technical and obtuse than this rather practical question. That's my two cents. But I'm certainly not going to vote to reopen.
$endgroup$
– fleablood
Nov 26 '18 at 16:45










5 Answers
5






active

oldest

votes


















15












$begingroup$

No. By definition of $f:A to B$, then for every $ain A$ then $f(a)$ must exist and $f(a) in B$. So if $A$ is not empty then $f(A)$ is not empty (although it can have a few as only one element.)



However it is possible that $A$ is empty in which case $f(A)$ is (obviously) also empty.



$f: emptyset to B$ is the empty function in this case.



=====



Or to put it really simple $f(1)$ has to be in the image so the image can't be empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes".
    $endgroup$
    – Ister
    Nov 26 '18 at 9:19






  • 1




    $begingroup$
    The OP states that the domain is $mathbb{N}$...
    $endgroup$
    – Vincent
    Nov 26 '18 at 9:53










  • $begingroup$
    @Ister "You should say 'No if the domain is non-empty.'" Do you really think it needs to be proven that $Bbb N$ is non-empty?
    $endgroup$
    – fleablood
    Nov 26 '18 at 16:21












  • $begingroup$
    @Vincent was your comment addressed at me or Ister. If at me, well, ... so? I've shown it is false for any non-empty domain. $Bbb N$ is a non-empty domain. So the answer to the OP's question is "no".
    $endgroup$
    – fleablood
    Nov 26 '18 at 16:23










  • $begingroup$
    @fleablood the comment was directed at Ister.
    $endgroup$
    – Vincent
    Nov 26 '18 at 16:25





















9












$begingroup$

There is no such function: in particular, the only function whose image is the empty set is the empty function, whose domain is also the empty set. In particular, $f(mathbb{N})$ cannot be the empty set, because contains $f(1)$.






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    Well, the definition state that for any x natural number there exists a uniquely determined natural number y such as f(x)=y. Since f(N) has no elements, it means that there doesn’t exist any y such as f(x)=y. Contradiction! So such functions do not exist :) The image of a function has at least one element as Df has at least 1 element.






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Since for every $ain mathbb{N}$ the $f(a)$ is also in $mathbb{N}$ we see that $f(mathbb{N})ne emptyset $ so there is no such function.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        This case is not possible since, by definition of range of a function, $f(a)$ should belong to the codomain for each $a$ in the domain.






        share|cite|improve this answer











        $endgroup$




















          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          15












          $begingroup$

          No. By definition of $f:A to B$, then for every $ain A$ then $f(a)$ must exist and $f(a) in B$. So if $A$ is not empty then $f(A)$ is not empty (although it can have a few as only one element.)



          However it is possible that $A$ is empty in which case $f(A)$ is (obviously) also empty.



          $f: emptyset to B$ is the empty function in this case.



          =====



          Or to put it really simple $f(1)$ has to be in the image so the image can't be empty.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes".
            $endgroup$
            – Ister
            Nov 26 '18 at 9:19






          • 1




            $begingroup$
            The OP states that the domain is $mathbb{N}$...
            $endgroup$
            – Vincent
            Nov 26 '18 at 9:53










          • $begingroup$
            @Ister "You should say 'No if the domain is non-empty.'" Do you really think it needs to be proven that $Bbb N$ is non-empty?
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:21












          • $begingroup$
            @Vincent was your comment addressed at me or Ister. If at me, well, ... so? I've shown it is false for any non-empty domain. $Bbb N$ is a non-empty domain. So the answer to the OP's question is "no".
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:23










          • $begingroup$
            @fleablood the comment was directed at Ister.
            $endgroup$
            – Vincent
            Nov 26 '18 at 16:25


















          15












          $begingroup$

          No. By definition of $f:A to B$, then for every $ain A$ then $f(a)$ must exist and $f(a) in B$. So if $A$ is not empty then $f(A)$ is not empty (although it can have a few as only one element.)



          However it is possible that $A$ is empty in which case $f(A)$ is (obviously) also empty.



          $f: emptyset to B$ is the empty function in this case.



          =====



          Or to put it really simple $f(1)$ has to be in the image so the image can't be empty.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes".
            $endgroup$
            – Ister
            Nov 26 '18 at 9:19






          • 1




            $begingroup$
            The OP states that the domain is $mathbb{N}$...
            $endgroup$
            – Vincent
            Nov 26 '18 at 9:53










          • $begingroup$
            @Ister "You should say 'No if the domain is non-empty.'" Do you really think it needs to be proven that $Bbb N$ is non-empty?
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:21












          • $begingroup$
            @Vincent was your comment addressed at me or Ister. If at me, well, ... so? I've shown it is false for any non-empty domain. $Bbb N$ is a non-empty domain. So the answer to the OP's question is "no".
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:23










          • $begingroup$
            @fleablood the comment was directed at Ister.
            $endgroup$
            – Vincent
            Nov 26 '18 at 16:25
















          15












          15








          15





          $begingroup$

          No. By definition of $f:A to B$, then for every $ain A$ then $f(a)$ must exist and $f(a) in B$. So if $A$ is not empty then $f(A)$ is not empty (although it can have a few as only one element.)



          However it is possible that $A$ is empty in which case $f(A)$ is (obviously) also empty.



          $f: emptyset to B$ is the empty function in this case.



          =====



          Or to put it really simple $f(1)$ has to be in the image so the image can't be empty.






          share|cite|improve this answer











          $endgroup$



          No. By definition of $f:A to B$, then for every $ain A$ then $f(a)$ must exist and $f(a) in B$. So if $A$ is not empty then $f(A)$ is not empty (although it can have a few as only one element.)



          However it is possible that $A$ is empty in which case $f(A)$ is (obviously) also empty.



          $f: emptyset to B$ is the empty function in this case.



          =====



          Or to put it really simple $f(1)$ has to be in the image so the image can't be empty.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 '18 at 22:19

























          answered Nov 25 '18 at 18:42









          fleabloodfleablood

          72.7k22788




          72.7k22788












          • $begingroup$
            You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes".
            $endgroup$
            – Ister
            Nov 26 '18 at 9:19






          • 1




            $begingroup$
            The OP states that the domain is $mathbb{N}$...
            $endgroup$
            – Vincent
            Nov 26 '18 at 9:53










          • $begingroup$
            @Ister "You should say 'No if the domain is non-empty.'" Do you really think it needs to be proven that $Bbb N$ is non-empty?
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:21












          • $begingroup$
            @Vincent was your comment addressed at me or Ister. If at me, well, ... so? I've shown it is false for any non-empty domain. $Bbb N$ is a non-empty domain. So the answer to the OP's question is "no".
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:23










          • $begingroup$
            @fleablood the comment was directed at Ister.
            $endgroup$
            – Vincent
            Nov 26 '18 at 16:25




















          • $begingroup$
            You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes".
            $endgroup$
            – Ister
            Nov 26 '18 at 9:19






          • 1




            $begingroup$
            The OP states that the domain is $mathbb{N}$...
            $endgroup$
            – Vincent
            Nov 26 '18 at 9:53










          • $begingroup$
            @Ister "You should say 'No if the domain is non-empty.'" Do you really think it needs to be proven that $Bbb N$ is non-empty?
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:21












          • $begingroup$
            @Vincent was your comment addressed at me or Ister. If at me, well, ... so? I've shown it is false for any non-empty domain. $Bbb N$ is a non-empty domain. So the answer to the OP's question is "no".
            $endgroup$
            – fleablood
            Nov 26 '18 at 16:23










          • $begingroup$
            @fleablood the comment was directed at Ister.
            $endgroup$
            – Vincent
            Nov 26 '18 at 16:25


















          $begingroup$
          You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes".
          $endgroup$
          – Ister
          Nov 26 '18 at 9:19




          $begingroup$
          You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes".
          $endgroup$
          – Ister
          Nov 26 '18 at 9:19




          1




          1




          $begingroup$
          The OP states that the domain is $mathbb{N}$...
          $endgroup$
          – Vincent
          Nov 26 '18 at 9:53




          $begingroup$
          The OP states that the domain is $mathbb{N}$...
          $endgroup$
          – Vincent
          Nov 26 '18 at 9:53












          $begingroup$
          @Ister "You should say 'No if the domain is non-empty.'" Do you really think it needs to be proven that $Bbb N$ is non-empty?
          $endgroup$
          – fleablood
          Nov 26 '18 at 16:21






          $begingroup$
          @Ister "You should say 'No if the domain is non-empty.'" Do you really think it needs to be proven that $Bbb N$ is non-empty?
          $endgroup$
          – fleablood
          Nov 26 '18 at 16:21














          $begingroup$
          @Vincent was your comment addressed at me or Ister. If at me, well, ... so? I've shown it is false for any non-empty domain. $Bbb N$ is a non-empty domain. So the answer to the OP's question is "no".
          $endgroup$
          – fleablood
          Nov 26 '18 at 16:23




          $begingroup$
          @Vincent was your comment addressed at me or Ister. If at me, well, ... so? I've shown it is false for any non-empty domain. $Bbb N$ is a non-empty domain. So the answer to the OP's question is "no".
          $endgroup$
          – fleablood
          Nov 26 '18 at 16:23












          $begingroup$
          @fleablood the comment was directed at Ister.
          $endgroup$
          – Vincent
          Nov 26 '18 at 16:25






          $begingroup$
          @fleablood the comment was directed at Ister.
          $endgroup$
          – Vincent
          Nov 26 '18 at 16:25













          9












          $begingroup$

          There is no such function: in particular, the only function whose image is the empty set is the empty function, whose domain is also the empty set. In particular, $f(mathbb{N})$ cannot be the empty set, because contains $f(1)$.






          share|cite|improve this answer









          $endgroup$


















            9












            $begingroup$

            There is no such function: in particular, the only function whose image is the empty set is the empty function, whose domain is also the empty set. In particular, $f(mathbb{N})$ cannot be the empty set, because contains $f(1)$.






            share|cite|improve this answer









            $endgroup$
















              9












              9








              9





              $begingroup$

              There is no such function: in particular, the only function whose image is the empty set is the empty function, whose domain is also the empty set. In particular, $f(mathbb{N})$ cannot be the empty set, because contains $f(1)$.






              share|cite|improve this answer









              $endgroup$



              There is no such function: in particular, the only function whose image is the empty set is the empty function, whose domain is also the empty set. In particular, $f(mathbb{N})$ cannot be the empty set, because contains $f(1)$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 25 '18 at 18:36









              user3482749user3482749

              4,3111019




              4,3111019























                  7












                  $begingroup$

                  Well, the definition state that for any x natural number there exists a uniquely determined natural number y such as f(x)=y. Since f(N) has no elements, it means that there doesn’t exist any y such as f(x)=y. Contradiction! So such functions do not exist :) The image of a function has at least one element as Df has at least 1 element.






                  share|cite|improve this answer









                  $endgroup$


















                    7












                    $begingroup$

                    Well, the definition state that for any x natural number there exists a uniquely determined natural number y such as f(x)=y. Since f(N) has no elements, it means that there doesn’t exist any y such as f(x)=y. Contradiction! So such functions do not exist :) The image of a function has at least one element as Df has at least 1 element.






                    share|cite|improve this answer









                    $endgroup$
















                      7












                      7








                      7





                      $begingroup$

                      Well, the definition state that for any x natural number there exists a uniquely determined natural number y such as f(x)=y. Since f(N) has no elements, it means that there doesn’t exist any y such as f(x)=y. Contradiction! So such functions do not exist :) The image of a function has at least one element as Df has at least 1 element.






                      share|cite|improve this answer









                      $endgroup$



                      Well, the definition state that for any x natural number there exists a uniquely determined natural number y such as f(x)=y. Since f(N) has no elements, it means that there doesn’t exist any y such as f(x)=y. Contradiction! So such functions do not exist :) The image of a function has at least one element as Df has at least 1 element.







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                      share|cite|improve this answer










                      answered Nov 25 '18 at 18:39









                      furfurfurfur

                      869




                      869























                          3












                          $begingroup$

                          Since for every $ain mathbb{N}$ the $f(a)$ is also in $mathbb{N}$ we see that $f(mathbb{N})ne emptyset $ so there is no such function.






                          share|cite|improve this answer









                          $endgroup$


















                            3












                            $begingroup$

                            Since for every $ain mathbb{N}$ the $f(a)$ is also in $mathbb{N}$ we see that $f(mathbb{N})ne emptyset $ so there is no such function.






                            share|cite|improve this answer









                            $endgroup$
















                              3












                              3








                              3





                              $begingroup$

                              Since for every $ain mathbb{N}$ the $f(a)$ is also in $mathbb{N}$ we see that $f(mathbb{N})ne emptyset $ so there is no such function.






                              share|cite|improve this answer









                              $endgroup$



                              Since for every $ain mathbb{N}$ the $f(a)$ is also in $mathbb{N}$ we see that $f(mathbb{N})ne emptyset $ so there is no such function.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 25 '18 at 18:34









                              Maria MazurMaria Mazur

                              47.4k1260120




                              47.4k1260120























                                  1












                                  $begingroup$

                                  This case is not possible since, by definition of range of a function, $f(a)$ should belong to the codomain for each $a$ in the domain.






                                  share|cite|improve this answer











                                  $endgroup$


















                                    1












                                    $begingroup$

                                    This case is not possible since, by definition of range of a function, $f(a)$ should belong to the codomain for each $a$ in the domain.






                                    share|cite|improve this answer











                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      This case is not possible since, by definition of range of a function, $f(a)$ should belong to the codomain for each $a$ in the domain.






                                      share|cite|improve this answer











                                      $endgroup$



                                      This case is not possible since, by definition of range of a function, $f(a)$ should belong to the codomain for each $a$ in the domain.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited Nov 30 '18 at 3:20









                                      Brahadeesh

                                      6,50942364




                                      6,50942364










                                      answered Nov 25 '18 at 18:56









                                      MANI SHANKAR PANDEYMANI SHANKAR PANDEY

                                      548




                                      548















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