Does there exist a continuous surjection $f:\mathbb{R} → l^{\infty}$

I have asked a question asking for a surjection from $\mathbb{R}$ to $\mathbb{R}^2$. Any thoughts, anyone? Or can someone start me off by offering me a hint how to find a surjection from $\mathbb{R}$ to $l^\infty$?

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What can you say about topological properties of continuous images of $\mathbb{R}$? Connectedness, separability, etc? –  Martin Apr 10 '13 at 23:26
The prior question –  Martin Apr 10 '13 at 23:30
@Martin this is the stuff I do not know, and need to read up on. These stuff appear as "extra questions" for a course I am looking at. The topological stuff was not part of the course. Would you care to point me to something I should know to help? –  Lost1 Apr 10 '13 at 23:41
This is just your prior question with a different title. –  leo Apr 10 '13 at 23:42
It seems like a stretch to solve this without knowing any topology. If instead you know some linear algebra and set theory, you might make an argument via the cardinality of a basis of $\ell^\infty$. But that's not so elementary either. –  Kevin Carlson Apr 11 '13 at 0:26

Yes. No: see Martin's comment below, but there exists a continuous surjection $\mathbb R\to l^\infty$ when $l^\infty$ is equipped with the topology induced by $\mathbb R^{\mathbb N}$.
First of all, given the existence of space-filling curves $f:[0,1]\to[0,1]^2$, it's easy to construct space-filling curves $g:[0,1]\to[0,1]^3$: if we write $f(t)=(x(t),y(t))$ for all $t\in[0,1]$, we may set $g(t)=(x(t),f(x(t)))$. Similarly we can define space-filling curves from $[0,1]$ to $[0,1]^n$ for any $n>0$.
Let's find a space-filling curve $F_1:[0,\infty)\to [-1,1]^{\mathbb N}$ where $[-1,1]^{\mathbb N}=[-1,1]\times[-1,1]\times\cdots$. Let $g_n:[0,1]\to[-1,1]^n$ be a space-filling curve which starts and ends at the origin in $\mathbb R^n$, and let $G_n:[0,1]\to[-1,1]^{\mathbb N}$ simply be the same curves $g_n$ embedded in $\mathbb R^{\mathbb N}$ (so, for instance, $G_3(t)=(g_3(t),0,0,0,\ldots)$). Now a space-filling curve $F_1:[0,\infty)\to[-1,1]^{\mathbb N}$ is given by first walking along $G_1$, then walking along $G_2$, and so on.
Finally, we construct a space-filling curve $F_2:[0,\infty)\to l^\infty$. The space $l^\infty$ can be viewed as the set of bounded vectors in $\mathbb R^{\mathbb N}$. Let $h:[0,\infty)\to\mathbb R^2$ be a space-filling curve (see this post), and write $h(t)=(x(t),y(t))$ for all $t\geq 0$. Set $F_2(t)=|x(t)|\cdot F_1(y(t))$. Here, $F_2(t)$ will be an element of $l^\infty$ with norm $|x(t)|$, and as we vary $t\geq0$, we will get all elements of $l^\infty$.
This surjection is continuous with respect to the weak*-topology (the topology induced by $\mathbb{R}^\mathbb{N}$), but it is not continuous with respect to the norm topology. If nothing else is said $l^\infty$ is usually equipped with the norm topology. The problem is that $l^\infty$ is not norm-separable and continuous images of separable spaces are separable, so there is no norm-continuous surjection. –  Martin Apr 11 '13 at 9:13