Can the empty set be the image of a function on $mathbb{N}$? [duplicate]
$begingroup$
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?
functions elementary-set-theory
edited Nov 26 '18 at 15:02
user21820
39.4k543155
asked Nov 25 '18 at 18:32
avan1235avan1235
3297
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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.
add a comment |
$begingroup$
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?
functions elementary-set-theory
edited Nov 26 '18 at 15:02
user21820
39.4k543155
asked Nov 25 '18 at 18:32
avan1235avan1235
3297
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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
add a comment |
$begingroup$
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?
functions elementary-set-theory
edited Nov 26 '18 at 15:02
user21820
39.4k543155
asked Nov 25 '18 at 18:32
avan1235avan1235
3297
$endgroup$
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
functions elementary-set-theory
edited Nov 26 '18 at 15:02
user21820
39.4k543155
asked Nov 25 '18 at 18:32
avan1235avan1235
3297
edited Nov 26 '18 at 15:02
user21820
39.4k543155
asked Nov 25 '18 at 18:32
avan1235avan1235
3297
edited Nov 26 '18 at 15:02
user21820
39.4k543155
edited Nov 26 '18 at 15:02
user21820
39.4k543155
edited Nov 26 '18 at 15:02
user21820
39.4k543155
39.4k543155
asked Nov 25 '18 at 18:32
avan1235avan1235
3297
asked Nov 25 '18 at 18:32
avan1235avan1235
3297
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
add a comment |
$begingroup$
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
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
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.
answered Nov 25 '18 at 18:42
fleabloodfleablood
72.7k22788
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $f(1)$.
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 25 '18 at 18:39
furfurfurfur
869
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
$endgroup$
add a comment |
$begingroup$
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.
edited Nov 30 '18 at 3:20
Brahadeesh
6,50942364
answered Nov 25 '18 at 18:56
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Nov 25 '18 at 18:42
fleabloodfleablood
72.7k22788
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Nov 25 '18 at 18:42
fleabloodfleablood
72.7k22788
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
answered Nov 25 '18 at 18:42
fleabloodfleablood
72.7k22788
$endgroup$
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.
answered Nov 25 '18 at 18:42
fleabloodfleablood
72.7k22788
edited Nov 26 '18 at 22:19
answered Nov 25 '18 at 18:42
fleabloodfleablood
72.7k22788
answered Nov 25 '18 at 18:42
fleabloodfleablood
72.7k22788
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $f(1)$.
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
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 $f(1)$.
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
$endgroup$
add a comment |
$begingroup$
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 $f(1)$.
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
$endgroup$
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 $f(1)$.
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
answered Nov 25 '18 at 18:36
user3482749user3482749
4,3111019
4,3111019
add a comment |
add a comment |
$begingroup$
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.
answered Nov 25 '18 at 18:39
furfurfurfur
869
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 25 '18 at 18:39
furfurfurfur
869
$endgroup$
add a comment |
$begingroup$
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.
answered Nov 25 '18 at 18:39
furfurfurfur
869
$endgroup$
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.
answered Nov 25 '18 at 18:39
furfurfurfur
869
answered Nov 25 '18 at 18:39
furfurfurfur
869
answered Nov 25 '18 at 18:39
furfurfurfur
869
answered Nov 25 '18 at 18:39
furfurfurfur
869
869
add a comment |
add a comment |
$begingroup$
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.
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
$endgroup$
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$begingroup$
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.
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
$endgroup$
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$begingroup$
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.
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
$endgroup$
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.
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
answered Nov 25 '18 at 18:34
Maria MazurMaria Mazur
47.4k1260120
47.4k1260120
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$begingroup$
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.
edited Nov 30 '18 at 3:20
Brahadeesh
6,50942364
answered Nov 25 '18 at 18:56
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
add a comment |
$begingroup$
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.
edited Nov 30 '18 at 3:20
Brahadeesh
6,50942364
answered Nov 25 '18 at 18:56
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
add a comment |
$begingroup$
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.
edited Nov 30 '18 at 3:20
Brahadeesh
6,50942364
answered Nov 25 '18 at 18:56
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
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.
edited Nov 30 '18 at 3:20
Brahadeesh
6,50942364
answered Nov 25 '18 at 18:56
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
edited Nov 30 '18 at 3:20
Brahadeesh
6,50942364
edited Nov 30 '18 at 3:20
Brahadeesh
6,50942364
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
answered Nov 25 '18 at 18:56
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
answered Nov 25 '18 at 18:56
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
548
add a comment |
add a comment |
$begingroup$
If $D_f=emptyset$
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– hamam_Abdallah
Nov 25 '18 at 18:33
2
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By definition, the set $f(mathbb{N})$ contains each $f(n)$ for $nin mathbb{N}$.
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– anomaly
Nov 26 '18 at 0:46
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You say "an empty set" but note that there is only one empty set: "the empty set."
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– David Richerby
Nov 26 '18 at 13:18
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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.
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– fleablood
Nov 26 '18 at 16:45