TITLE: Is $[0,1]^\omega$ a continuous image of $[0,1]$?
QUESTION [9 upvotes]: 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 $\mathbb{R}$?

REPLY [2 votes]: 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:
$$
f_\infty(t) = \big(u(t),u(v(t)),u(v(v(t))),u(v(v(v(t)))),\dots\big) .
$$