How "big" can the boundary support of a continuous function be? I was advised to place this part of my previous question (https://math.stackexchange.com/a/3083158/512018) into a new question, namely: 
Considering the factor $\partial \{x \in \mathbb R: f(x)\neq 0\}$, does this mean that the set $\overline{\{x \in \mathbb R: f(x)\neq 0\}}$ can contain points $y \in \mathbb R$ such $f(y)=0$ but these sets of points need to be countable? Or can they be uncountably infinite?
 A: The zero set in the support of a continuous function can have any cardinality.  I think that it is not too hard to find examples for finite cardinalities:  it is relatively simple to construct a function which is continuous on $\mathbb{R}$ and supported on an interval—a "tent" or "triangle" function does the job quite nicely, e.g. the function
$$ f(x) = \left( 1-|x| \right) \chi_{[-1,1]} $$
is continuous on $\mathbb{R}$ and supported on $[-1,1]$.  The boundary of the support is the two element set $\{-1,1\}$, which has cardinality 2.  More generally, it might be helpful to think of this function as the function
$$ d(x,E) \chi_{[-1,1]}, $$
where $E = \{-1, 1\}$ and $d(x,E) = \inf\{ |x-y| : y\in E\}$.  By extension, if $E$ is any finite set with $\min(E) = a$ and $\max(E) = b$, then the function
$$ f(x) = d(x,E) \chi_{[a,b]} $$
is supported on $[a,b]$ and is zero on $E$.  In this case, since $E$ is finite, closing the set $\{x : f(x) \ne 0\}$ will "put back" the points of $E$ (where $f$ is zero) into the support.  That is,
$$ \operatorname{supp}(f) = \overline{[a,b]\setminus E} = [a,b]. $$
This same process does not work in general with countably infinite sets.  For example, this same construction does not work to give a continuous function supported on $[0,1]$ which is zero on the rational numbers—a continuous function which is zero on the rationals must be zero everywhere (by a density argument).  However, if we take $E = \{ \frac{1}{n} : n\in\mathbb{N} \}$, then the function
$$ f(x) = d(x,E)\chi_{[0,1]} $$
will be continuous with
$$ \operatorname{supp}(f) = \overline{\{x : f(x) \ne 0\}} = \overline{[0,1]\setminus \{\tfrac{1}{n} : n\in\mathbb{N} \} } = [0,1],$$
which gives an example of a compactly supported function with countably many zeroes in the support.
Obtaining a continuous function with uncountably many zeros in the support is a little trickier, but still doable.  In the above example, it is the lack of density of the zero-set which makes things happen, so the goal is to play the same game with a discrete, uncountable set.  So, let $C$ be the usual ternary Cantor set.  It is not too hard to show that $C$ is (1) uncountable and (2) nowhere dense.  Then the function
$$ f(x) = d(x,C) \chi_{[0,1]} $$
is continuous, nonzero on $[0,1] \setminus C$, and zero on $C$.  Since $C$ is nowhere dense in $[0,1]$, it follows that
$$ \operatorname{supp}(f)
= \overline{[0,1]\setminus C}
= [0,1], $$
which gives an example of a continuous function supported on $[0,1]$ with an uncountable set of zeroes.

As an added bonus, we can actually construct a function which is not only continuous, but smooth on $[0,1]$ which possesses the Cantor set as a zero set.  Define the "bump function"
$$ \eta(x) := \exp\left( -\frac{1}{1-x^2} \right). $$
Observe that this function is smooth on $\mathbb{R}$ (it has derivatives of all orders) and supported on $[-1,1]$.  If $[a,b]$ is any interval, then we can construct a smooth bump function on $[a,b]$ by translating and scaling $\eta$:
$$ \eta_{a,b} := (b-a)\eta\left( \frac{2}{b-a}x - \frac{a+b}{b-a} \right). $$
The factor of $(b-a)$ out in front is there mostly to ensure that the derivatives don't get too out of control.  To get something supported on the complement of the Cantor set, we can use the construction of the Cantor set to give us intervals on which to put these bump functions:
At the $n$-th stage of the construction of the Cantor set, $2^{n-1}$ open intervals of length $1/3^n$ are removed from the open interval.  Specifically:
\begin{align}
\text{Stage 1:} &&& I_{1,1} = (a_{1,1}, b_{1,1}) := \left( \tfrac{1}{3}, \tfrac{2}{3} \right) \\
\text{Stage 2:} &&& I_{2,1} = (a_{2,1}, b_{2,1}) := \left( \tfrac{1}{9}, \tfrac{2}{9} \right) \\
&&& I_{2,2} = (a_{2,2}, b_{2,2}) := \left( \tfrac{7}{9}, \tfrac{8}{9} \right) \\
\text{Stage 3:} &&& I_{3,1} = (a_{3,1}, b_{3,1}) := \left( \tfrac{1}{27}, \tfrac{2}{27} \right) \\
&&& I_{3,2} = (a_{3,2}, b_{3,2}) := \left( \tfrac{7}{27}, \tfrac{8}{27} \right) \\
&&& I_{3,3} = (a_{3,3}, b_{3,3}) := \left( \tfrac{19}{27}, \tfrac{20}{27} \right) \\
&&& I_{3,4} = (a_{3,4}, b_{3,4}) := \left( \tfrac{25}{27}, \tfrac{26}{27} \right) \\
\text{and so on...}
\end{align}
Then define the function
$$ f(x) := \sum_{m=1}^{\infty} \sum_{n=1}^{2^n} \eta_{(a_{m,n},b_{m,n})}(x). $$
It can be shown (though it is a bit of work to do so) that $f$ is smooth (infinitely differentiable), supported on the entire unit interval $[0,1]$, and zero exactly on the Cantor set.  I've used Desmos to give an idea of what this function looks like, at least up to the fourth stage of the construction:

To engage in some rampant speculation, this last example is, I think, somewhat enlightening:  it gives an idea of how to generalize to higher dimensions using sums of smooth bump functions defined for open sets with "interesting" boundaries (the complement of the Cantor set is such an open set with an "interesting" boundary).
Moreover, it seems plausible that this construction could be used to find functions with large zero sets in terms of measure, instead of in terms of cardinality.  For example, consider a smooth function supported on the complement of a "fat" Cantor set in $\mathbb{R}$.  Such a fat Cantor set still has the necessary topological features to push through the construction, but has positive Lebesgue measure.  Hence a zero set needn't even be a null-set!
Finally, even more generally, this construction can be souped up to discuss smooth functions on manifolds with "large" zero sets, where "large" is appropriately defined for whatever setting you are interested in (measure and cardinality being, for example).
