Talk:Empty function

What does the empty function evaluate to, for a particular set A?--ASL 00:51, 8 July 2007 (UTC)

Since the empty function has an empty domain, the empty function doesn't "evaluate to" anything. Paul August ☎ 02:10, 8 July 2007 (UTC)

I just wasn't used to defining a function in terms of a graph (which is a set, and therefore may be empty)--ASL 21:03, 8 July 2007 (UTC)

The idea is not that the empty function is empty as a set, nor does it depend upon defining a function in terms of a graph. The idea is that the domain is empty, and it applies equally well using the informal definition of a function as a rule which associates each element of its domain with a unique element of its codomain. If you define a function as a graph then an empty function will, incidentally, also be empty as a set. However it is common to formally define a function as an ordered triple (A, B, G) where A (the domain) and B (the codomain) are sets and G (the graph) is a subset of A X B (the set of ordered pairs with first element in A and second element in B) satisfying the property that if (r, t) and (s, t) are in G then r = s. In this case, for any set B, (∅, B, ∅) is an empty function, but of course not the empty set. Notice this means that although there is only one empty set, there are an infinite number of empty functions, one for each set B. Paul August ☎ 16:29, 9 July 2007 (UTC)

Interesting. I have as yet never encountered such a definition of function, it was always just a subset of the Cartesian product. What are the upsides? This definition clearly distinguishes between

$f:\mathbb {R} \rightarrow \mathbb {R}$

$g:\mathbb {R} \rightarrow \mathbb {N}$

$h:\mathbb {R} \rightarrow \{1\}$

such as f(x)=g(x)=h(x)=1 for every real x. Is there any situation where such a distinction is necessary or convenient? Because I can imagine situation where it would be convenient to use functions which differ only in codomain interchangeably. For example, there is a proof of Cantor-Bernstein theorem which utilizes a bijection created from injection by restricting its codomain to its range. Then, because by this definition those function would not be equal, the fact the created function was a bijection would not be trivial and had to be proved, which would approximately double the length of the proof. 109.73.215.142 (talk) 11:52, 20 December 2014 (UTC)

Uniquely determined

For each set A, there is exactly one empty function

$f_{A}:\varnothing \rightarrow A,$

as

$g:\varnothing \rightarrow A$,

then

$\forall x\in \varnothing :f_{A}(x)=g(x)$

Simpler:

$g:\varnothing \rightarrow A$,

then

$g=\varnothing =f_{A}$,

On the other hand, let

$g:\varnothing \rightarrow A$,

then

$\forall x\in \varnothing :f_{A}(x)\neq g(x)$