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$(a)$ How many functions are there from a nonempty set $S$ into $\emptyset$?

$(b)$ How many functions are there from $\emptyset$ into an arbitrary set $S$?

This question seems very simplistic but I don't know the answer. I think for $(a)$ that there isn't a function that maps a set $S$ into a empty set? For $(b)$ I assume it to be all function that map the empty set to an arbitrary set since all sets contain the empty set?

Tom
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2 Answers2

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A function from a set $A$ to a set $B$ is a subset of $A\times B$ satisfying certain conditions, one of which is that its domain is $A$. If either $A$ or $B$ is empty, $A\times B=\varnothing$, and $\varnothing$ is therefore the only subset of $A\times B$. If $A\ne\varnothing$, $\varnothing$ is not a function with domain $A$, so you’re quite right about $(a)$: there are no such functions. If $A=\varnothing$, though, it’s a different story. The domain of the function $\varnothing$ is $\{a:\langle a,b\rangle\in\varnothing\}$, which is ... ?

Brian M. Scott
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  • Can you elaborate more on why a set $A$ to a set $B$ is a subset of $A\times B$ and why if either $A$ or $B$ is empty then $A\times B =\emptyset $? – Tom Aug 25 '13 at 07:58
  • @Tom: The first is just a matter of definition: a function from $A$ to $B$ is a particular kind of relation from $A$ to $B$, and by definition the relations from $A$ to $B$ are precisely the subsets of $A\times B$. The second is really also just a matter of definition: $A\times B$ is by definition the set of all ordered pairs $\langle a,b\rangle$ such that $a\in A$ and $b\in B$. If either $A$ or $B$ is empty, then there clearly are no such pairs, so $A\times B=\varnothing$ as well. – Brian M. Scott Aug 25 '13 at 08:00
  • Is there an explicit, obvious difference between $A$ to $B$ and $A\times B$? Because I can't see none. They both seem to be the same to me. – Tom Aug 25 '13 at 08:03
  • @Tom: $A\times B$ is a particular set; I don’t know what you mean by ‘$A$ to $B$’, because there is no mathematical object with that name. – Brian M. Scott Aug 25 '13 at 08:04
  • Sorry, I meant when you said, "a function from a set $A$ to a set $B$". So pretty much $f: A\to B$. – Tom Aug 25 '13 at 08:06
  • @Tom: They are very different things. $A\times B$ is the collection of all ordered pairs with first element from $A$ and second from $B$; a function $f:A\to B$ is a subset of $A\times B$ that satisfies certain conditions, and those conditions guarantee that apart from a few trivial cases, $A\times B$ is never a function from $A$ to $B$. For instance, if $A={0,1}$ and $B={a,b}$, $$A\times B={\langle 0,a\rangle,\langle 0,b\rangle,\langle 1,a\rangle,\langle 1,b\rangle};,$$ but that’s not a function from $A$ to $B$, because (for instance) it sends $0$ both to $a$ and to $b$. – Brian M. Scott Aug 25 '13 at 08:09
  • @Tom: In that example the functions from $A$ to $B$ are the following subsets of $A\times B$: $${\langle 0,a\rangle,\langle 1,a\rangle};;$$ $${\langle 0,a\rangle,\langle 1,b\rangle};;$$ $${\langle 0,b\rangle,\langle 1,a\rangle};;$$ $${\langle 0,b\rangle,\langle 1,b\rangle};.$$ – Brian M. Scott Aug 25 '13 at 08:10
  • Oh, I got it now. Very well put. Last couple of questions. Why did you and Asaf mention $A\times B$; why would that be necessary to answer the question I gave? Would I be able to answer it without mentioning $A\times B$ and just by mentioning a set A to a set B? – Tom Aug 25 '13 at 08:16
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    @Tom: In general you should expect that in order to prove something about a mathematical object, you’ll have to use its definition in some way. Since a function from $A$ to $B$ is by definition a certain kind of subset of $A\times B$, you should expect to have to consider what $\varnothing\times S$ and $S\times\varnothing$ are in order to answer questions about functions from $\varnothing$ to $S$ or from $S$ to $\varnothing$. I cannot emphasize this strongly enough: when you’re just beginning and therefore haven’t much of a feel for what to look for, your automatic reaction should be ... – Brian M. Scott Aug 25 '13 at 08:20
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    ... to start with the definitions of the objects involved. Here that immediately tells you that $f$ can only be the empty set, and the questions then become: Is the empty set a function from a non-empty $S$ to $\varnothing$? (No.) Is the empty set a function from $\varnothing$ to a non-empty set $S$? – Brian M. Scott Aug 25 '13 at 08:22
  • Thanks for your very well detailed explanation! You've cleared many misconceptions that I had and to answer your question "Is the empty set a function from ∅ to a non-empty set S "? Is yes, there is one function which is the empty set? Correct? – Tom Aug 25 '13 at 08:29
  • @Tom: Yep, you’ve got it. You’re welcome! – Brian M. Scott Aug 25 '13 at 08:31
  • Awesome. I cant express enough how thankful I am. Thanks again! – Tom Aug 25 '13 at 08:33
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HINT: Recall that a function from $A$ to $B$ is a subset of $A\times B$, whose domain equals $A$.

Asaf Karagila
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