Does this "in between" space exist? Let


*

*$C=C([0,1])$ be the space of continuous functions from $[0,1]$ to $\mathbb{R}$ (or $\mathbb{C}$)

*$B=B([0,1])$ be the space of bounded funciton from $[0,1]$ to $\mathbb{R}$ (or $\mathbb{C}$)

*$l_\infty$ the space of bounded sequences of real (or complex numbers)

*$\{q_n\}_{n\in\mathbb{N}}$ an enumeration of the rational numbers $\mathbb{Q}\cap[0,1]$
The function
$$\begin{align}
 \varphi:B & \longrightarrow l_\infty\\
f & \longmapsto \varphi(f)=(f(q_1),f(q_2),\dots,f(q_n),\dots)
\end{align}$$
is onto but clearly not one-to-one. However, $\varphi$ restricted to $C$ is one-to-one because $\{q_n\}_{n\in\mathbb{N}}$ is dense in $[0,1]$ and the functions are continuous. But when restricted to $C$, $\varphi$ is not onto because something like $(0,0,0,1,1,1,1,1,1,1,1,\dots)$ is not the image of any continuous function.
So is there any intermediate space $C\le X\le B$ such that the restriction $\varphi\vert_{X}$ is bijective?
I have thought vaguely of some things that may work, like making the function of $(x_n)_{n\in\mathbb{N}}$ with some limit process but now i don´t have time to check a lot of those ideas. There should be some difficulties with the fact that $\{q_n\}_{n\in\mathbb{N}}$ has zero measure in $[0,1]$ so maybe instead of $B$ we should be using $L^\infty([0,1])$. Does someone have an idea or a complete solution? Also $\varphi$ is (i strongly think but cant check now) continuous and even norm preserving on $C$ but obviously not on $B$. Maybe we shouldn't be thinking of a subspace of $B$ but rather of $l_\infty$?
Also, how does the enumeration change $\varphi$?
Thanks!
edit: maybe we are looking for bounded functions continuous only on the irrational numbers?
 A: We do this with ultrafilters.  Thus it is very "non-constructive".  
For each point $x \in [0,1]$ choose an ultrafilter $\mathcal U_x$ on $\mathbb N$ such that
$$
\lim_{n,\mathcal U_x} q_n = x ,
$$
that is, the limit of $q_n$ according to $\mathcal U_x$ is $x$.  Another restriction: if $x$ is rational, say $x=q_m$, then $\mathcal U_x$ must be chosen as the fixed ultrafilter at $m$.
Write $\mathfrak U$ for this system $(\mathcal U_x)_{x \in [0,1]}$.  
Definition.  A function $f : [0,1] \to \mathbb R$ is said to be $\mathfrak U$-continuous iff: for all $x \in [0,1]$,
$$
f(x) = \lim_{n,\mathcal U_x} f(q_n) .
$$
Our space $X$ is the set of all bounded $\mathfrak U$-continuous functions.  $C \subset X \subset B$.  
What about extension?  Existence:  Let $f : \mathbb Q\cap[0,1] \to \mathbb R$ be bounded.  Define $g : [0,1] \to \mathbb R$ by:
$$
\text{for all $x \in [0,1]$, let}\quad g(x) = \lim_{n,\mathcal U_x} f(q_n) .
\tag{1}$$
For $x$ rational, $\mathcal U_x$ is fixed, so we get $g(x)=f(x)$.  That is, $g$ is an extension of $f$.  Next, $g$ defined in this way is $\mathfrak U$-continuous.  Indeed, let $x \in [0,1]$.  For all $n \in \mathbb N$, the ultrafilter $\mathcal U_{q_n}$ is fixed at $q_n$.  Then
$$
\lim_{n,\mathcal U_x} g(r_n)
=\lim_{n,\mathcal U_x} \;\lim_{m,\mathcal U_{r_n}} f(r_m)
=\lim_{n,\mathcal U_x} f(r_n) = g(x) .
$$
So $g$ is $\mathfrak U$-continuous.  
Injective:  Let $f_1, f_2$ be two bounded functions on $\mathbb Q\cap[0,1]$.  Suppose $f_1 \ne f_2$.  Then there is $m$ such that $f_1(q_m) \ne f_2(q_m)$.  Let the extensions as defined be $g_1, g_2$.  We have $g_1(r_m) = f_1(r_m) \ne f_2(r_m) = g_2(r_m)$, so $g_1 \ne g_2$.
Surjective:  Let $h \in X$ be a bounded $\mathfrak U$-continuous function.  Let $f$ be the restriction to $\mathbb Q \cap [0,1]$.  We claim the extension $g$ defined by $(1)$ is $h$.  Indeed, for any $x \in [0,1]$,
$$
h(x) = \lim_{n,\mathcal U_x} h(r_n)= \lim_{n,\mathcal U_x} f(r_n) = g(x) .
$$
