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

Question

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 $l^\infty$?

Answer 1

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 $l^\infty$.