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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 $\mathbb{R}$?

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    You are right, cut points won't help, and in fact, the result is actually true (see my edit).2011-10-22

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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) . $

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    @GEdgar: I understand the proof why functions $f_n$ are onto, but I don't see why $f_∞$ should be. I can only see that the range contains a sequence with any initial segment by the property of $f_n$ and the fact that $v$ is onto. So it has dense image, and now I see that it is onto because of compactness. But in the answer it seems that it follows directly from the property of $f_n$.2016-06-26
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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.

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    Thanks, this definitely does the trick.2011-10-22