Typically, a function $f: D \mapsto R$ is described as $$\forall x \in D, \exists ! y \in R, \left(x,y\right) \in f \wedge P\left(x,y\right),$$ where $P$ is a predicate that specifies the relation between an input and the output, for instance, $y = 3x$ if $D \subseteq \mathbb{R}$ and $R \subseteq \mathbb{R}$. However, this definition does not specify what happens outside $D$. Should we explicitly specify that no element outside $D$ corresponds to an output under $f$? That is, should we use the following proposition $$\left(\forall x \in D, \exists ! y \in R, \left(x,y\right) \in f \wedge P\left(x,y\right) \right) \wedge \left(\forall x \not\in {D}, \lnot \exists y, \left(x,y\right)\in f\right)$$ to describe $f$?

My own understanding is as follows. In mathematical proofs, people are more concerned with the existence of such a function. That is, what happens outside $D$ does not matter. What matters is, such a function $f$ exists, so that the proof can move on. Specifically, when people talk about a function, they are saying: $$\exists f, \forall x \in D, \exists ! y \in R, \left(x,y\right) \in f \wedge P\left(x,y\right).$$ And they can proceed with a particular example of such a function. In such a circumstance, the information of existing $f$s outside $D$ is not of interest.

  • 1
    $\begingroup$ What is exactly $\overline D$? The complement of a set $D$ with respect to another set $A$ is defined as $A \setminus D = \{x \in A : x \notin D\}$. $\endgroup$
    – azif00
    Dec 22, 2020 at 2:33
  • $\begingroup$ Why do we have $P$ here? The relation is $f$ itself. $\endgroup$
    – Berci
    Dec 22, 2020 at 2:35
  • $\begingroup$ @azif00 Thanks for the remind. The previous statement is not rigorous. I have modifed my statement accordingly. $\endgroup$
    – Ziqi Fan
    Dec 22, 2020 at 2:37
  • $\begingroup$ "However, this definition does not specify what happens outside D. " It doesn't specify what happens to elephants on the Serangeti either. Only $D$ is relevant. "However, this definition does not specify what happens outside D" I'd say it does specify. The elements of $D$ are mapped, and therefore the elements not in $D$ are not mapped. And we don't care if $D$ swims inside the universe of $\mathbb R$ or if it fits into $\mathbb R \cup \{$ stars in in the milky way$\}\cup\{$ the elephants of the serengeti$\}$. The only thing that matters is the elements of $D$ are mapped. $\endgroup$
    – fleablood
    Dec 22, 2020 at 3:06
  • $\begingroup$ "outside" $D$ with respect to what. $\mathbb R$ is not the end all and be all and we have no way of knowing what "everything else" even means. If $0 \not \in D$ do we care what about $0$. What about $\frac {\sqrt 3}2 + \frac 12 \not \in D$ and $\frac {\sqrt 3}2 + \frac 12i\not \in \mathbb R$ do we care about that? What about the equilateral triangle $\triangle ABC$? That is something that isn't in $D$. Do we care about that? What about $Babar,\ the\ elephant\not \in D$? Do we care about that? $\endgroup$
    – fleablood
    Dec 22, 2020 at 3:11

1 Answer 1


Consider this relation $f$ from $ D= \{a, b,c\}$ to $T=\{0,1\}$ :

$f = \{(a,1), (b,1), (c,2)\}$

Is it true that

for all $x$, [ if $x\in D$ then, there is a unique $y$ in $T$ such that $(x,y)\in f ]$ ?

  • If $x = a$ or $x = b$ or $x=c$ , the antecedent is true and the consequent is true, so the whole condiitonal is true.

  • If $x$ is neither $a$ nor $b$ , nor $c$ ( that is if $x\notin D$) the antecedent is false, and ( by the truth table of the " if ... then" operator) the whole conditional is true.

So, the conditional is true for all $x$ ... and relation $f$ is a function.


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