# Can you use level sets to suggest discontinuity?

Consider the function

$$f(x, y) = \frac{x^2 + y^2}{y}$$

for which I already showed that the level set of height $c$ is given by a circle of the form

$$C_c: (x - 0)^2 + (y - \frac{c}{2})^2 = \frac{c^2}{4}$$

Next, observe a plot of the level sets $C_1$ (red), $C_2$ (blue) and $C_3$ (green).

Note that the point $(0, 0)$ is contained in all three sets; in other words, $f$ seems to have more than one value at $(0, 0)$ (this, I haven't quite grasped yet).

Is it valid to make the argument that $f$ is discontinuous in $(0, 0)$, since it can be shown that some level sets intersect?

More generally, is any intersection between two level sets a point of discontinuity?

• Doesn't that imply that f isn't invertable not that f is discontinuous? Commented May 22, 2016 at 15:52
• Oops. I'm an idiot. The level set doesn't really imply that because the level set isn't continuous at (0,0). Consider what conditions would have to be true for the level set to be continuous at (0,0). You are changing the domain of the function by creating the level set which causes the overlaps. I am inclined to say that the overlapping of level sets comes from the type of discontinuity and does not work as a general rule. Basically a level set overlapping implies that f is not continuous but the reverse is not true. Commented May 22, 2016 at 15:56

Let $C_1$ denote the level set where $f(x,y) = c_1$, and $C_2$ the level set where $f(x,y) = c_2$ ($c_1 \neq c_2$). Suppose that $(0,0)$ is a limit point of $C_1$ and $C_2$ (i.e. the level sets 'intersect'). Then I can take the limits $$\lim_{(x,y) \to (0,0), (x,y) \in C_1}f(x,y) = c_1 \neq c_2 = \lim_{(x,y) \to (0,0), (x,y) \in C_2}f(x,y).$$
So $f$ cannot be continuous at $(0,0)$.