Is $[0,1]^\omega$ a continuous image of $[0,1]$? Is $[0,1]^\omega$, i.e. $\prod_{n=0}^\infty [0,1]$ with the product topology, a continuous image of $[0,1]$? What if $[0,1]$ is replaced by $\mathbb{R}$?
Edit: It appears that the answer is yes, and follows from the Hahn-Mazurkiewicz Theorem ( http://en.wikipedia.org/wiki/Space-filling_curve#The_Hahn.E2.80.93Mazurkiewicz_theorem ). However, I am still interested in the related question: is $\mathbb{R}^\omega$ a continuous image of $\mathbb{R}$?
 A: For the first question, this case is easier than the generic Hahn-Mazurkiewicz theorem.  Begin with a Peano curve, or space-filling curve, continuous $f : [0,1] \to [0,1]^2$ onto.  Once we have this, we can get a space-filling curve $f_3 : [0,1] \to [0,1]^3$ by fiddling with this:  if $f(t) = (u(t),v(t))$ are the components of $f$, write
$$
f_3(t) = \big(u(t),u(v(t)),v(v(t))\big) .
$$
Check that it is continuous and onto.
Now to do $f_4 : [0,1] \to [0,1]^4$ try this:
$$
f_4(t) = \big(u(t),u(v(t)),u(v(v(t))),v(v(v(t)))\big) .
$$
Once you understand why these work, it is natural to go on to $f_\infty : [0,1] \to [0,1]^\infty$ by:
$$
f_\infty(t) = \big(u(t),u(v(t)),u(v(v(t))),u(v(v(v(t)))),\dots\big) .
$$
A: So if I'm reading correctly you want to find out if there is a continuous (with respect product topology) surjective map $f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}$?
No, there is not. Note that $\mathbb{R}$ is $\sigma$-compact, so write:
$$\mathbb{R} = \bigcup_{n \in \mathbb{N}} [-n,n]$$
Then using the fact that $f$ is surjective we get:
$$\mathbb{R}^{\omega} = \bigcup_{n \in \mathbb{N}} f([-n,n])$$
By continuity of $f$ each $D_n=f([-n,n])$ is a compact subset of $\mathbb{R}^{\omega}$. So the question boils down to whether is possible that $\mathbb{R}^{\omega}$ is $\sigma$-compact with product topology.
No, let $\pi_{n}$ be the standard projection from $\mathbb{R}^{\omega}$ onto $\mathbb{R}$, then $\pi_{n}f([-n,n])$ is a compact subset of $\mathbb{R}$ so bounded. Thus for each $n \in \mathbb{N}$ choose $x_{n} \in \mathbb{R} \setminus \pi_{n}f([-n,n])$ then $x=(x_{n})$ lies in $\mathbb{R}^{\omega}$ but not in $\bigcup_{n \in \mathbb{N}} f([-n,n])$, a contradiction.
