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

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?

## marked as duplicate by Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 26 '18 at 15:22

• If $D_f=\emptyset$ – hamam_Abdallah Nov 25 '18 at 18:33
• By definition, the set $f(\mathbb{N})$ contains each $f(n)$ for $n\in \mathbb{N}$. – anomaly Nov 26 '18 at 0:46
• You say "an empty set" but note that there is only one empty set: "the empty set." – David Richerby Nov 26 '18 at 13:18
• 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. – fleablood Nov 26 '18 at 16:45

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.

• You should say "No if the domain is non-empty.". As you've clearly proven in the last two lines the answer is "Yes". – Ister Nov 26 '18 at 9:19
• The OP states that the domain is $\mathbb{N}$... – Vincent Nov 26 '18 at 9:53
• @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? – fleablood Nov 26 '18 at 16:21
• @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". – fleablood Nov 26 '18 at 16:23
• @fleablood the comment was directed at Ister. – Vincent Nov 26 '18 at 16:25

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 $$f(1)$$.

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.

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.

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.