Number of continuous points of a graph (Problem 34 from 97-99 Math GRE practice questions booklet)

Question

Problem Statement: Let $f$ be a function with domain $[-1, 1]$ such that the coordinates of each point $(x, y)$ of its graph satisfy $x^2 + y^2 = 1$. What is the total number of points at which $f$ is necessarily continuous?

My confusion: I've been treating the first sentence as a roundabout way of saying that we have a circle of radius 1 centered at the origin. Then it seems "obvious" to me that $f$ is necessarily continuous at an infinite number of points. However, the correct answer is 2. What am I missing?

Update: I now get that it'd be a half-circle not a full circle (since it's a function), but even more than that, I see that I was imagining $f$ as a continuous function...now I'm seeing it as more of a connect-the-dots semi-circle function, but then in this case, now I'd like to think that $f$ has no continuous points. Any hints on how to get to the middle ground of 2 continuous points?

Answer 1

We furnish an example for which there are exactly two points of continuity, and no more:

$$f(x) = \begin{cases} \sqrt{1 - x^2}, & x \in [-1,1] \cap x \in \mathbb Q \\ -\sqrt{1 - x^2}, & x \in [-1,1] \cap x \not \in \mathbb Q \end{cases}$$

It is easy to see that $f$ meets the criterion $x^2 + f(x)^2 = 1$ for all $x \in [-1,1]$. For any $|x_0| < 1$, $f$ is discontinuous at $x_0$. For suppose $$\lim_{x \to x_0} f(x) = L = f(x_0).$$ Then for any $\epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - f(x_0)| < \epsilon$ whenever $0 < |x - x_0| < \delta$. But any such $\delta$-neighborhood of $x_0$ will contain a point, say $x'$, that is irrational if $x_0$ is rational, or vice versa, because the rationals are dense in the reals, and likewise for irrationals in the reals. So whenever $\epsilon < 2|f(x_0)| = 2\sqrt{1-x_0^2}$, it is impossible to find such a $\delta$-neighborhood; hence the supposed limit does not exist, and $f$ is discontinuous.

A slight modification of the above argument shows that $f$ is continuous at $x \in \{-1,1\}$, since in this case, the maximum difference between the "upper" and "lower" semicircles within a $\delta$-neighborhood of these points is $2\sqrt{1-(1-\delta)^2}$, hence the desired relationship between $\epsilon$ and $\delta$ is $\delta \le 1 - \sqrt{1 - (\epsilon/2)^2}$.

We cannot have less than $2$ points of continuity since the above argument shows that if $f$ is continuous at $x = 1$, it must also be continuous at $x = -1$. So $2$ is the minimum.

Answer 2

As JonathanZ said, we are dealing with a function, not a graph. Think about the vertical line test. There are only two points, $(-1,0)$ and $(1,0)$ that pass the vertical line test, therefore $f(x)$ is continuous at these points. That's how you get the two points.

In more detail: For a function to be continuous at x, it must be defined at x, its limit must exist at x, and the value of the function at that point must equal the value of the limit at x. If we have two possible points for our "function" $(x,y_1)$ and $(x,y_2)$, we can deduce that one of $y_1,y_2$ is positive and the other is negative. Then we can construct a function that satisfies the conditions for $f(x)$ that alternates between points below and above the $x-$ axis. Then none of our $x$ have a defined limit and therefore can't be continuous at $x$ unless there is only one possible $y$; this only occurs at $(−1,0)$ and $(1,0)$.

So many of the math GRE problems are just based on attention to sneaky details rather than real mathematical capability.