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I think that the following definitions of the domain of a function are inconsistent:

Definition A

A function $f$ from a set $X$ to set $Y$ is defined by a set $G$ of ordered pairs $(x,y)$ such that $\forall x\in X$ and $\forall y\in Y$, and every element of $X$ is the first component of exactly one ordered pair in $G$. The set $X$ is called the domain of the function $f$.

Definition B

Let $f: X\mapsto Y$. Then $X$ is the set of all inputs for which $f$ provides an image in $Y$.

Let $f(x)=x^2$ be a function over the reals.

Using definition A, the domain of this function can be any subset of $\mathbb R$, because every element in any subset of $\mathbb R$ satisfies the condition that it is the first component of exactly one ordered pair.

But using definition B, the domain of this function is precisely the set $\mathbb{R}$. Therefore, these definitions are inconsistent. Which one should be used?

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    $\begingroup$ Definition A is the one that should be used in a rigorous mathematical setting. Definition B is the one that is most commonly used in US calculus textbooks $\endgroup$ Commented Jul 15, 2020 at 16:59
  • $\begingroup$ When you specify 'over the reals' you are imposing that $X = \mathbb{R}$. $\endgroup$ Commented Jul 15, 2020 at 17:00
  • $\begingroup$ @Omnomnomnom And do these definitions contradict each other? $\endgroup$ Commented Jul 15, 2020 at 17:35
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    $\begingroup$ Technically, yes. Definition A implies that a function is completely specified once a formula is given, definition B implies that a formula (such as $f(x) = \log(x)$) is insufficient $\endgroup$ Commented Jul 15, 2020 at 18:30
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    $\begingroup$ A bit more problematic is the fact that the definitions of "discontinuous" depend on context. In calculus textbooks, it is often said that the function $f(x) = 1/x$ over the real numbers is discontinuous because of its "discontinuity" at $x = 0$. Where definition $B$ is used, we would say that the function $f:\Bbb R \setminus \{0\} \to \Bbb R$ defined by $f(x) = 1/x$ is continuous because it is continuous at all points in its domain. $\endgroup$ Commented Jul 15, 2020 at 18:33

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Technically, for every subset $X$ of $\Bbb R$ (including $X=\Bbb R$ itself, there exists a function $f\colon X\to\Bbb R$ defined by $f(x)=x^2$; and these are all distinct and perfectly valid functions (they're related to one another of course).

The two definitions are not inconsistent, actually. For example, define $X=\{1,2,3\}$, and define $f\colon X\to\Bbb R$ by $f(x)=x^2$ for all $x\in X$. For this function, the set of all inputs for which $f$ provides an image in $\Bbb R$ is exactly $\{1,2,3\}=X$. The fact that we can think of a function that extends this $f$ to a larger domain is conceptually nice but mathematically irrelevant to that individual fact.

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