Let $f$ be a continuous function defined on the closed interval $[0,1]$. Clearly $f$ is bounded and attains its bounds.
Then my question is how often can $f$ take a value in its range countably many times?
Formally let $Z(f,p)=\{x\in \text{Image}(f):|f^{-1}(x)| = p\}$. I would like to know large $Z(f,\aleph_0)$ can be?
I conjecture that we can find $f$ with $|Z(f,\aleph_0)|=2^{\aleph_0}$.
Note that I can find f with $Z(f,2^{\aleph_0}) = \aleph_0$ which I think makes the conjecture plausible:
We know that the zero set of a continuous function on $[0,1]$ can be any closed subset as we can take $f$ to be the distance to the closed set. If we let the closed set be the cantor ternary set $C$ on $[0,1]$ we can create a function on $[0,1]$ with uncountably infinite zero set. By taking suitable scalings of the function on intervals $[\frac{1}{2^{2n+1}}, \frac{1}{2^{2n}}]$ with ramps in between we can find a function $f$ s.t. $Z(f)$ is countably infinite. Hence we have $Z(f,2^{\aleph_0}) = \aleph_0$.
Best Answer
How about Devil's staircase (a.k.a. Cantor function) $C(x)$, but with every horizontal segment replaced by a rescaled zigzag $Z(x)$? By a zigzag I mean, for example: $$ Z(x) = \tfrac{2}{\pi} \arcsin(\sin(2\pi x)) .$$ The function $f$ is formally defined as: $$ f(x) = \begin{cases} C(x) & \text{if $x$ has no $1$ in ternary expansion,} \\ C(x) + 2^{-n} Z(3^n(x - a_{n,k})) & \text{if $x \in [a_{n,k}, a_{n,k}+3^{-n}]$,} \end{cases} $$ where $a_{n,k}$, $k = 1, 2, \ldots, 2^n$, is the enumeration of left endpoints of maximal line segments of length exactly $3^{-n}$ on which $C(x)$ is constant.
For every $y_0$ which is not a dyadic rational there is exactly one $x$ with no $1$ in ternary expansion such that $f(x) = C(x) = y_0$, and for each $n = 1, 2, \ldots$ the line $y = y_0$ intersects exactly one zigzag of $f$ of horizontal length $3^{-n}$.
EDIT: To clarify the definition of $f$, write $$ x = \sum_{n = 1}^\infty \frac{x_n}{3^n} $$ for the ternary expansion of $x$, and let $K \in \{1, 2, \ldots, \infty\}$ be the position of first digit $1$. The Cantor function $C(x)$ is equal to $$ C(x) = \sum_{n = 1}^K \frac{\lceil x_n/2 \rceil}{2^n} . $$ The function $f$ is defined by $$ f(x) = \begin{cases} C(x) & \text{if $K = \infty$,} \\ C(x) + 2^{-K} Z\biggl(\sum_{n = 1}^\infty \dfrac{x_{K+n}}{3^n}\biggr) & \text{otherwise.} \end{cases} $$
Here is the plot of $f$:
To see that $f$ is continuous, it is enough to observe that $f - C$ is an infinite series of continuous "zigzag" functions with disjoint supports and decreasing supremum norms. The series thus converges uniformly, and consequently $f - C$ is continuous.
Regarding the level sets: Suppose that $y$ is not a dyadic rational, with binary digits $y_n$: $$ y = \sum_{n = 1}^\infty \frac{y_n}{2^n} . $$ Then $C(x) = y$ has exactly one solution (namely: an $x$ with $x_n = 2 y_n$), and this will also be a solution of $f(x) = y$. All other solutions $x$ necessarily have $K < \infty$. For such an $x$, we have $|f(x) - C(x)| \le 2^{-K}$, that is, $|y - C(x)| \le 2^{-K}$. Since $y$ is not a dyadic rational, it follows that $x_n = 2 y_n$ for $n = 1, 2, \ldots, K - 1$, and of course $x_K = 1$. Therefore, $$ 2^{-K} Z\biggl(\sum_{n = 1}^\infty \frac{x_{K+n}}{3^n}\biggr) = f(x) - C(x) = y - C(x) = \frac{y_K - 1}{2^K} + \sum_{n = K + 1}^\infty \frac{y_n}{2^n} . $$ The above equation clearly has exactly two solutions ($Z$ is essentially two-to-one). It follows that $f(x) = y$ has one solution with $K = \infty$ and two solutions corresponding to every finite $K$.
By the way, level sets corresponding to dyadic rationals are countable, too, by a very similar argument.