Does there exist a uniformly continuous surjection $f:\Bbb Q^c\to\Bbb Q$?
This question arose to my mind after constructing a continuous surjection from $\Bbb Q^c$ to $\Bbb Q$. Which is a very simple construction, that is $f(x)=\Big\lceil\frac{1}{x}\Big\rceil$ gives a continuous surjection from $\Bbb Q^c\to\Bbb N$, now taking an enumeration of $\Bbb Q$, say $\{q_n\}$ we construct a map $h:\Bbb N\to\Bbb Q$ as $h(n)=q_n$ (clearly continuous and surjective), now $(h\circ f):\Bbb Q^c\to\Bbb Q$ is the required continuous surjection from $\Bbb Q^c$ to $\Bbb Q$
Thanks in advance!
If possible, such $f$ exists. Now, since $\Bbb Q^c$ is dense in $\Bbb R$ and $f$ is a uniformly continuous function, so it can be extended continuously on $\Bbb R$. Let $g:\Bbb R\to\Bbb R$ be such extension.
Now consider the restriction of $g$ on $I_n=[n,n+1]\,\,\forall\,\,n\in\Bbb N$ is continuous. Hence $g(I_n)$ is compact and connected.
Now consider, $g(I_{n_0})=[a,b]$ for some $a\neq b$ for some $n_0\in\Bbb N$
(Note that, there exists such $I_{n_0}$, since not for all $n\in\Bbb N$, $g(I_n)$ can be a singleton.)
Now, $g(I_{n_0})=g(I_{n_0}\,\cap\,\Bbb Q^c)\,\cup\,g(I_{n_0}\,\cap\,\Bbb Q)$ is countable, since $g(I_{n_0}\,\cap\,\Bbb Q^c)\subset\Bbb Q$ and $g(I_{n_0}\,\cap\,\Bbb Q)$ is atmost coutable but $[a,b]$ is uncountable. (Which is a contradiction.)