Why is the domain of a circle with a defined radius the set of all real numbers?

Question

I'm working through a textbook and a problem asked about the domain and co-domain of $x^2 + y^2 = 1$. My understanding is that the domain of a function is the set of possible numbers $x$ can take on to make the original expression valid.

Because the left-hand side is the sum of two squared terms, the lowest value $x$ can take, while remaining a true expression, is $-1$ and the highest is $1$.

Given this, I think the correct answer for the domain is $\{-1 < x < 1\}$, but the textbook says the domain is the set of all real numbers; why is this the case?

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Here is a screenshot of the textbook:

Answer 1

The author is using an uncommon meaning for "domain" and "codomain".

In this problem $C$ is the set of all pairs $(x,y)$ for which $$ x^2 + y^2 = 1 . $$

If he means by "the domain of $C$" (or of any relation) "the set of values that can appear as $x$" then his answer to (b) is wrong. You correctly note that only $x \in [-1,1]$ will work.

He might mean "the set of values for which the expression makes algebraic sense, even if the equality fails", in which case the domain is indeed the whole real line.

I do not like this usage. It's commonly reserved for relations that are functions, and the codomain then has a slightly different meaning.

Answer 2

Okay, so you're talking about the domain of a set-theoretic relation, not a function, so that clears up a lot of initial confusion.

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So, the domain of a relation $R$ on a set $A$ (i.e. $R \subseteq A \times A$) is the collection of all lefthand components of its members (which are ordered pairs).

Here, then, $\text{domain}(C)= [-1,1]$. This is easy to see for two reasons:

• Graphed a relation in $\mathbb{R}^2$, the circle itself only encompasses the $x$ values in $[-1,1]$
• Algebraically, if $|x|>1$, then $x^2 > 1$. Since $y^2 \ge 0$, there is no $y$ for which such an $x$ satisfies $(x,y) \in C$.

This essentially matches up with your ideas.

I say the closed interval $[-1,1]$ as opposed to the open $(-1,1)$, however, since $(\pm 1,0) \in C$ though.

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Addendum:

I'm adding this after reading Ethan Bolker's answer and the text in question (Discrete Mathematics with Applications by Susanna Epp).

It seems that the definition of domain used in this case is not the usual one I see.

The definition I tend to see is this: let $R$ be a relation from $A$ to $B$, so that $R \subseteq A \times B$. Then

$$\text{domain}(R) := \{ a \in A \mid \text{there is a $b \in B$ such that $(a,b) \in R$\}}$$

Epp's text uses the simpler $\text{domain}(R) := A$, which can be seen in this definition (bottom of page $14$):

It somewhat makes me think of the distinction between a function and a partial function.

Answer 3

A relation is just a subset of the Cartesian product of two sets, $X\times Y$, with the first set $X$ being called the domain and the second set $Y$ being called the codomain. You're told that the circle $C$ is the set of pairs $(x,y)\in \Bbb R\times \Bbb R$ such that $x^2 + y^2 = 1$. Thus, $C\subset \Bbb R \times \Bbb R$ and so the domain and range of $C$ are both $\Bbb R$.

A function is a special kind of relation, one where there is exactly one pair $(x,y)$ for each $x\in X$. Because of this, the domain of a function can't include any values which aren't represented in the relation at all, which results in your intuition about the domain containing just the $x$-values which can satisfy the equation.

In general, the domain and range of any relation is what they tell you it is, and then whether or not that relation is a function is determined by the above criterion.