Is there a continuous surjection from the reals to the continuous functions? $\require{begingroup}\begingroup
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It is known that the set of real numbers has the same cardinality as the set of continuous $\R$ to $\R$ functions---and therefore a bijection exists between them. My question is whether a continuous bijection exists between them, or at least a continuous surjection from $\R$ to $C(\R)$, where $\R$ has the usual metric and $C(\R)$ has whatever metric works.
The reason this interests me is I have always felt that one of the reasons differential equations are hard to solve is because we don't have a big enough "vocabulary" to describe all of the functions in continuous function space. This is in contrast to the space of real numbers where we do have a robust method to express any real number to arbitrary precision: decimal notation. And historically, decimals have allowed us to solve many previously intractable problems like, say, computing cube roots---something the ancient Greeks could never do.
If a continuous surjection exists between real numbers and continuous functions, we could then use decimal notation to describe any continuous function to arbitrary precision, instead of being restricted to compositions of elementary functions or their extensions. We might even be able to express functional operations in terms of arithmetic operations. And maybe this could (but I'm only speculating) give us a way to solve difficult differential (or other functional) equations in much in the same way decimals allowed us to compute cube roots.

But I'm going out on a limb now. Even if such a surjection exists, I have no idea how much it would actually help in solving such problems. But I'm very curious about that, so if anyone has any additional insights, please tell!
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 A: 
Is there any metric on $C(\mathbb{R})$ that allows for a continuous surjection from $\mathbb{R}$ to $C(\mathbb{R})$.

Of course there is, for a boring reason. Let $F\colon C(\mathbb{R})\to \mathbb R$ be any bijection (the sets have the same cardinality). Define a metric on $C(\mathbb{R})$ by $d(f,g) = |F(f) - F(g)|$. Then $F^{-1}:\mathbb{R}\to C(\mathbb{R})$ is not only a continuous surjection, but an isometric bijection.
Totally useless, of course. 
Here is something of more interest. Suppose $(V, d)$ is an infinite-dimensional vector space with a metric $d$ that makes it a complete metric space. Then there is no continuous surjection from $\mathbb R$ onto $V$. 
Indeed, if $f:\mathbb{R}\to V$ is such a surjection, then $V$ is $\sigma$-compact:
$$
V = \bigcup_{n=1}^\infty K_n,\quad \text{where } K_n=f([-n,n])
$$
But a compact subset of $V$ has empty interior, since every locally compact Hausdorff topological vector space is finite-dimensional. So we have written $V$ as the union of countably many closed subsets with empty interior, contradicting the Baire category theorem.  
In particular, the above applies to $C([0,1])$ with the uniform metric, or to $C(\mathbb R)$ with some natural metrics like 
$$
d(f,g)=\sum_{k=1}^\infty 2^{-n}\frac{\|f-g\|_n}{1+\|f-g\|_n} \quad\text{where } 
\|f-g\|_n = \sup_{[-n,n]}|f-g|$$
