Let $(f_k)$ be a sequence of continuous functions $[0,1]\to\mathbb R$ satisfying these two conditions:
- It converges uniformly to a limit $f$.
- For every $k$ the zero set $f^{-1}_k(0)$ is uncountable.
Does it follow that $f^{-1}(0)$ is uncountable?
This is not true if the compact interval is replaced with the whole real line. Just let $C$ be a compact uncountable set and define $f_k(x)=\tanh(d(x-k,C))$. The zero sets are $k+C$ but the limit function is $f\equiv1$. I imagine compactness could help, but I have found no proof or counterexample. Perhaps the sets $f^{-1}_k(0)$ can be shown to converge in a suitable way as $k\to\infty$.
No; define $$f_k(x)=\begin{cases} 0 &:0\leq x\leq 1/k\\ x-\frac{1}{k} &: 1/k\leq x\leq 1\end{cases}.$$
This converges uniformly to $f(x)=x$, for every $k$, the set $f_k^{-1}(0)$ is uncountable, but $f^{-1}(0)=\{0\}$.