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i have a small question but I can't make up my mind.

If we are given a function $f : X \to Y$

Because a function is in fact a set of ordered pairs, if we take $X$ to be the empty-set, we then say that $f$ does not exist or we say that it is equal to empty-set ? And more general, can you explain me the difference if any between "empty set" and "non-existence" ?

Thank you for your help!

Asaf Karagila
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InTheSearchForKnowledge
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    Well, the empty set is a set, whereas non-existence is a concept. – José Carlos Santos Aug 12 '23 at 16:49
  • An example might help: if $X$ is non-empty, and $Y$ is empty, then there does not exist a function $f:X\to Y$. On the other hand, if $X$ is empty, then there does exist a function $f:X\to Y$. It's just that the only such function is $\varnothing$. – Joe Aug 12 '23 at 16:53
  • @Joe: Hi, in your first example, why there does not exist a function, I think the function can be also the ∅ set, right ? – InTheSearchForKnowledge Aug 12 '23 at 17:10
  • @InTheSearchForKnowledge: Nope. The domain of $\varnothing$ is $\varnothing$, and in my first example, $X$ is non-empty. – Joe Aug 12 '23 at 18:00
  • @InTheSearchForKnowledge: See this answer for further details. – Joe Aug 12 '23 at 18:07
  • @Joe: hmm, as for me, a function is a set of well-ordered pairs that satisfy the uniqueness requirement. In this case, if $X$ is non-empty and $Y$ is empty, all the pairs in the empty-set satisfy the uniqueness requirement vaccously. – InTheSearchForKnowledge Aug 12 '23 at 18:08
  • @InTheSearchForKnowledge: You are missing the "existence of images" requirement (the "uniqueness of images" requirement does indeed hold vacuously). If $X$ is non-empty, then $\varnothing$ is not a function from $X$ to $\varnothing$ because it is not true that for all $x\in X$, there is a $y\in\varnothing$ such that $(x,y)\in\varnothing$. In other words, there is an $x\in X$ such that for all $y\in\varnothing$, we have $(x,y)\not\in\varnothing$. Proof: since $X$ is non-empty, it contains an element, say $x_0$. Note that for all $y\in\varnothing$, we have $(x_0,y)\not\in\varnothing$. – Joe Aug 12 '23 at 18:14
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    That final statement was vacuous, of course. More generally, if $X$ is non-empty, then there is no function $X\to\varnothing$. You can reason as follows: a function $f$ from $X$ to $\varnothing$ would be a subset of $X\times\varnothing=\varnothing$, and so we must have $f=\varnothing$. But, as we just showed, $\varnothing$ is not a function from $X$ to $\varnothing$. (However, $\varnothing$ is a function, and for every set $Y$, it is a function $\varnothing\to Y$). Does that answer your question? – Joe Aug 12 '23 at 18:20
  • @Joe: Yes, many thanks!! – InTheSearchForKnowledge Aug 12 '23 at 19:45
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    @InTheSearchForKnowledge: No problem. I'm glad I could help. I think what might have been confusing you to begin with is this: if there does not exist an object with a property $P$, then the set of objects with the property $P$ equals $\varnothing$. This is a completely different assertion to saying that $\varnothing$ satisfies the property $P$. In the latter case, there is an object with the property $P$, namely $\varnothing$, and so the set of objects satisfying $P$ is non-empty. – Joe Aug 12 '23 at 19:55

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