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I've just started my Bachelor studies in Math, and on my first week in the first quiz there was a question where my answer did not agree with the author's.

After a long conversation with my Teaching Assistant, I do not doubt anymore that my answer is wrong, but I'm still having a hard time understanding why. Could you please help me better understand where am I wrong?

The question in the quiz was the following:

Can the set $\{(x, y) | x, y \in \mathbb{R}, y = f(x), x^2 + y^2 = 1\}$ be the graph of a function $f: \mathbb{R} \mapsto \mathbb{R}$?

My answer was yes.

For example, if I define f to be $f(x) := \sqrt{1 - x^2}$, then the set becomes $\{(x, y) | x, y \in \mathbb{R}, y = \sqrt{1-x^2}, x^2 + y^2 = 1\}$, and because the second constraint is more restrictive than the third, the set becomes exactly the graph of $f(x)$.

But apparently, I don't quite understand something about the set-builder notation, because the right answer to this quiz question is: no.

My TA explained to me, that the right side of the | sign in the set-builder notation are not "constraints" but rather "assumptions" or "definitions", therefore the set is a circle, and - well, at least this part is obvious to me as well - a circle cannot be the graph of a function.

I also mentioned to my TA that maybe they wanted to ask something like this:

Can the set $\{(x, y) | x, y \in \mathbb{R}, x^2 + y^2 = 1\}$ be the graph of a function $f: \mathbb{R} \mapsto \mathbb{R}$?

Without the $y = f(x)$ predicate in the set-builder notation, the set is not ambiguous anymore. It's indeed a circle, therefore indeed it cannot be the graph of an f function. My TA's response to this was, that there are multiple possible equivalent definitions of the same thing, and my definition is equivalent with the one in the quiz.

Could you please shed some light on this for me?

Daniel Pek
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  • One way to make sense of this question is to interpret it as "is there a function $f: \Bbb R \to \Bbb R$ st ${ \dotsb, y = f(x), \dotsb }$ is the graph of $f$?" This is how I would interpret the question as you've written it, and the answer is still no. You should be careful to show that functions you introduce are actually well-defined - are you sure your $f(x)$ is defined on all of $\Bbb R$? – Izaak van Dongen Sep 03 '22 at 16:42
  • I'm sure that the question in the quiz was exactly the one I've written down, if that's what you're asking. However, can't a function $f: \mathbb{R} \mapsto \mathbb{R}$ be a partial function, in which case the pre-image is $[-1, 1]$, but the domain is still $\mathbb{R}$? – Daniel Pek Sep 03 '22 at 17:28
  • In analysis we don't use partial functions. They do appear in computer science, which is arguably a branch of math, so I hesitate to say that we don't use them in math. But we definitely don't use them in analysis. – Ian Sep 03 '22 at 17:39
  • What about $\sqrt{x}$ and $1/x$, then? – Daniel Pek Sep 03 '22 at 18:04
  • I'm afraid you seem to have a bit of a misunderstanding about functions and domains. In maths, if you say $f$ is a function with no further qualifiers then that means $f$ must be a total function. Neither of $\sqrt x$ or $1/x$ define a function from $\Bbb R$ to $\Bbb R$ - you would have to specify what values the function takes for negative $x$ and $x = 0$ respectively. None of the definitions or theorems in analysis would make any sense if we worked with partial functions! Have you been given a definition of function yet? If not, it was definitely weird for you to be asked this question. – Izaak van Dongen Sep 04 '22 at 10:50
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    Yes, we were given the definition of a function, and now that I went back, it indeed specifies that every element of the domain must be mapped to something by the function. Coming from software engineering, I guess it'll take for me some getting used to the nuances where math definitions differ from my existing intuitions. :) Thanks for explaining! – Daniel Pek Sep 04 '22 at 11:04

7 Answers7

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The question is horribly phrased. If you ask if a set can be the graph of a function $f$, then $f$ should not appear in the definition of the set.

Also, note that the question was whether the set can be the graph of a function $f : \Bbb{R}\to\Bbb{R}$. This cannot be the case (ignoring the issue mentioned in the beginning) since if $(x,y)$ in your set, then $x \in [-1,1]$, so that if your set was the graph of a function $f$, then the domain of $f$ would necessarily be a subset of $[-1,1]$.

PhoemueX
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  • In my opinion the domain of $f$ still can be R, it's just that $f$ happens to be a partial function, and its pre-image would necessarily be a subset of [-1, 1]. – Daniel Pek Sep 03 '22 at 17:25
  • To answer my own question here as well, functions must be total, and preimage is the same as the domain in this case, therefore if the domain is R, then the set cannot be a graph of f. – Daniel Pek Sep 04 '22 at 12:48
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These kinds of issues don't have 100% definite answers but I think your re-writing of the question is indeed what should have been asked. The original version of the question is misleading or incorrect in my opinion.

SBK
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there's a certain amount of confusion regarding that set-builder notation. to figure out who's confused, ask the TA about the set $$S=\{(x,y)\in\Bbb R^2:x^2+y^2=1,y\ge0\}.$$Q: is that a circle or a semicircle? A: It's a semicircle.

David C. Ullrich
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    I've already asked my TA about this, and the answer was the same as yours. However I'm not sure I understand your point. – Daniel Pek Sep 03 '22 at 17:32
  • My point is you said he said that ${(x, y) | x, y \in \mathbb{R}, y = \sqrt{1-x^2}, x^2 + y^2 = 1}$ was a circle; if so he doesn't understand the notation. (But if he says that set is a circle I would have expected him to also say that set I defined was a circle...) – David C. Ullrich Sep 03 '22 at 20:29
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Your TA is wrong. The expressions in set-builder notation are constraints, and the notation denotes the set of points that satisfy the conjunction of all of them.

Your answer would be correct if the question allowed any domain for $f$, but since it requires $f$ to have domain $\Bbb R$, the answer is no. The graph of a function $\Bbb R\to\Bbb R$ has a point $(x,f(x))$ for every $x\in\Bbb R$, whereas the given set can't have any points with $|x|>1$.

Karl
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  • In my opinion the domain of $f$ still can be $\mathbb{R}$, it's just that $f$ happens to be a partial function, and its pre-image would necessarily be a subset of $[-1, 1]$. – Daniel Pek Sep 03 '22 at 17:35
  • With standard terminology, a partial function that is not defined on all of $\Bbb R$ is not a function $\Bbb R\to\Bbb R$. – Karl Sep 03 '22 at 17:39
  • Does that mean, that the domains of $\sqrt{x}$ and $1/x$ are not $\mathbb{R}$? – Daniel Pek Sep 03 '22 at 18:03
  • @DanielPek No, the domain of $\sqrt{x}$ is $\left[0,\infty\right)$ and the domain of $1/x$ is all real numbers except 0. The domain of $f$ can still be all real numbers depending on what $f$ is. – Accelerator Sep 03 '22 at 22:21
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It's a lousy question, your TA is afraid to back down, and you're going to get too many answers here that will reach for some horribly convoluted interpretation rather than go with the incredibly likely alternative that it's poorly written.

Sorry I can't give you a better answer. Your interpretation is very good, and the only question left to you is whether you want to escalate the issue. And that's down to how many points this was worth, how open the professor seems, and his much time and social capital you want to spend - all interpersonal questions, not math overs. Good luck.

JonathanZ
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The question itself is unnecessarily confusing. The function $f(x)$ shouldn't be in that set if the question is asking if the set represents the graph of $f$.

(Answer) But if we disregard that, then the answer is no. The function $f$ shouldn't be mapped from $\mathbb{R} \to \mathbb{R}$ if the graph of $x^2+y^2 = 1$ only has that $x \in \left[-1,1\right]$.

Even if you were trying to prove the question is true (ignoring the confusion), just giving one example, like letting $f(x) = \sqrt{1-x^2}$, wouldn't suffice.

(Opinion) I'm not sure if your TA is correct about the right side of $|$ being "assumptions" and "definitions," either. I think a better term is constraints because the right side is more of a "rule." (In general, a set written in set-builder notation has the syntax "$\left\{\text{expression}\mid \text{rule}\right\}$" where its elements are all values of the expression specified by that rule.) But that depends on what your TA meant by "assumptions" and "definitions."

Accelerator
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This answer provides some additional information, which might also be helpful.

  • Here we have given a function \begin{align*} \color{blue}{f:\mathbb{R}\to \mathbb{R}} \end{align*} from which we know domain and codomain, but nothing else.

  • And we have given a set \begin{align*} \color{blue}{\{(x, y) | x, y \in \mathbb{R}, y=f(x), x^2 + y^2 = 1\}}\tag{1} \end{align*}

At first I will have a closer look at the set and the elements which are specified by this set-builder notation. OP mentioned them with - not "constraints" but rather "assumptions" or "definitions" -

A somewhat closer look to the set:

I'd like to cite P.R. Halmos who tells us in his Naive Set Theory in section 2 Specification:

... We are now ready fo formulate the major principle of set theory, ... Axiom of specification. To every set $A$ and to every condition $S(x)$ there corresponds a set $B$ whose elements are exactly those elements $x$ of $A$ for which $S(x)$ holds.

A condition here is just a sentence. The symbolism is intended to indicate that the letter $x$ is free in the sentence $S(x)$; that means that $x$ occurs in $S(x)$ at least once without being introduced by one of the phrases for some $x$ or for all $x$. It is an immediate consequence of the axiom of extension that the axiom of specification determines the set $B$ uniquely. To indicate the way $B$ is obtained from $A$ and from $S(x)$ it is customary to write \begin{align*} \color{blue}{B=\{x\in A:S(x)\}}\tag{2} \end{align*}

Here the term condition is used by Halmos to specify the sentence $S(x)$. In our example (1) we have two conditions \begin{align*} y=f(x)\qquad \mathrm{and}\qquad x^2+y^2=1 \end{align*} which determine pairs $(x,y)\in\mathbb{R}^2$ which are to select to uniquely obtain the elements of the set (1).

The graph of $f$:

According to the definition of a graph of a function $f:\mathbb{R}\to\mathbb{R}$, the graph is the set \begin{align*} \color{blue}{\{(x,y) | x,y\in \mathbb{R}, y=f(x)\}}\tag{3} \end{align*} which is, using the wording of Halmos, pretty much the first condition of the sentence $S(x,y)$ in (1). We can also write the set (1) in the form \begin{align*} \{(x, y) | x, y \in \mathbb{R}, y=f(x)\}\cap\{(x, y) | x, y \in \mathbb{R}, x^2 + y^2 = 1\} \end{align*} and see when interpreting them geometrically as points in the $(x,y)$-plane that the set (1) consists of all points $(x,y)$ which are formed by intersecting the graph of $f$ with the unit circle. Depending from the graph of $f$ the set may either be empty or it may contain some points from the circle, which are then also points in the graph of $f$.

The function $f$:

Here I'd like to cite P.R. Halmos again, who writes in Section 8, Functions:

If $X$ and $Y$ are sets, a function from (or on) $X$ to (or into) $Y$ is a relation $f$ such that $\mathrm{dom} f=X$ and such that for each $x$ in $X$ there is a unique element $y$ in $Y$ with $(x,y)\in f$.

...

The symbol \begin{align*} f:X\to Y \end{align*} is sometimes used as an abbreviation for $f$ is a function from $X$ to $Y$.

Conclusion:

  • Here we clearly see, that the domain $\mathrm{dom} f$ of a function is an essential part when specifying a function. We can't ignore it, or parts of it.

  • Since the graph of a function in (3) uses the whole domain of $f$ in its definition we conclude that the set (1) can't be a graph of a function $f$ which has domain $\mathbb{R}$. For instance $(-2,f(x))$ is not an element of the set (1) but it is an element of the graph of $f$.

Markus Scheuer
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