How many of these points can exist on a graph of $f:[0,1]\to [0,1]$? Say that $p$ is a _bumpy point_ if there is a collection of "bumps" limiting to it: . **Question:** If $f:[0,1]\to [0,1]$ is continuous, then must number of bumpy points on the graph of $f$ be countable? **Easier Question:** Does there exist a horizontal line with no bumpy points (other than the top line)? Note: I do not require that the "bumps" be semi-circles. They just have to be arcs beginning and ending at the same horizontal line as their limit point, and lie completely above that horizontal line.

No, there can be uncountably many bumpy points.

Consider the function $f$ defined as the uniform limit of the functions $f_n\colon[0,1]\mapsto [0,1]$ defined by $$ \begin{align} f_0(x)&=x\\\ f_{n+1}(x)&=\begin{cases} 4x & (0\leq x \leq \tfrac 1 4)\\\ 1-4(x-\tfrac 1 4) & (\tfrac 1 4\leq x \leq \tfrac 2 4)\\\ \tfrac 1 2 f_n(4(x-\tfrac 2 4)) & (\tfrac 2 4\leq t \leq \tfrac 3 4)\\\ \tfrac 1 2 + \tfrac 1 2 f_n(4(x-\tfrac 3 4)) & (\tfrac 3 4\leq t \leq 1).\\\ \end{cases} \end{align}$$

![bumpy function](

For each $k\geq 0$ and each choice of binary digits $b_1,\cdots,b_k\in\\{0,1\\}$, this function has separate bumps going from y-coordinate $\sum_{i=1}^k b_i 2^{-i}$ to $2^{-k} + \sum_{i=1}^k b_i 2^{-i}$ and back. So any y-coordinate in $[0,1)$ intersects infinitely many bumps. This also answers the "easier question" in the negative.