4
$\begingroup$

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.

$\endgroup$
5
  • 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

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.