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This question already has an answer here:

I cannot find any example of function $f:\mathbb{N}\rightarrow\mathbb{N}$ of which we can say that $$ f(\mathbb{N})=\emptyset $$ Does there exists any?

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marked as duplicate by Asaf Karagila elementary-set-theory 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
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    $\begingroup$ By definition, the set $f(\mathbb{N})$ contains each $f(n)$ for $n\in \mathbb{N}$. $\endgroup$ – 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 "$F\subset A\times \emptyset$ 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
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No. By definition of $f:A \to B$, then for every $a\in A$ 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.

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $f(1)$.

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

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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.

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