Are Darboux Functions closed under uniform limit? I suspect there exists a sequence of functions $f_n : [0,1] \to \mathbb{R}$ s.t $f_n$ converges uniformly to $f$, $f_n$ Darboux but $f$ not Darboux. However, I haven't found any examples. What are some examples?
 A: Let us denote $f$ the Conway base 13 function, defined on $[0,1]$, which is a Darboux function and maps every interval onto $\mathbb{R}$.
Therefore $f^{-1}(\{0\})$ is dense in $[0,1]$. Now we take $E = \{ a_k \mid k\in \mathbb{N} \}$ a countable subset of $f^{-1}(\{0\})$ which is dense in $[0,1]$. 
(to construct $E$, you can use that for all rational $q \in [0,1]$, there exists a sequence $(\alpha_{q,n}) \in \big(f^{-1}(\{0\})\big)^{\mathbb{N}}$ converging towards the rational $q$, and we can define $E$ as the union of all the $\alpha_{q,n}$. $E$ is obviously dense, and it is countable because $\mathbb{Q}\cap[0,1]$ is countable)
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The idea is now to construct $(f_n)$ such that $f_n^{-1}(\{0\})$ "shrinks".
With this $E$, you can define a sequence of function as follows : for $n \in \mathbb{N}$, $f_n : [0,1] \longrightarrow\mathbb{R}$ and 
$f_n\ \colon\ 
x\longmapsto\begin{cases}
2^{-k}&\text{if $x=a_k\ $ for $k\le n$}\\
0&\text{if $x=a_k\ $ for $k>n$}\\
1&\text{if $x \in f^{-1}(\{0\})\backslash E$} \\
f(x)&\text{otherwise}
\end{cases}$
Using that $E$ is dense, $f_n$ still maps every interval onto $\mathbb{R}$.
Moreover, the sequence $(f_n)$ converges to some function $F$ and $||f_n-F||_{\infty} \leqslant \frac{1}{2^{n+1}}$. Therefore $(f_n)$ is a sequence of Darboux functions which converges uniformly to $F$.
Yet, for all $a \in \mathbb{R}^*,\ F^{-1}(\{a\})$ is dense in $[0,1]$, but $F^{-1}(\{0\})$ is empty. Hence $F$ has both positive and negative images, but never vanishes, and therefore is not Darboux.
