# 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 $\mathbb{R}$?

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Are cutpoints preserved by continuity alone? I don't think so; the map fromI= [0,1] to $S^1$ seems to be a counter. Otherwise, one could maybe use the fact that compact metric spaces are the continuous image of the Cantor space C , and try to make the triangle f:C-->X and f: C-->Y commute. –  gary Oct 22 '11 at 4:51
You are right, cut points won't help, and in fact, the result is actually true (see my edit). –  Iian Smythe Oct 22 '11 at 6:02

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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I may be just missing something, but could you explain why that $f_3$ maps onto $[0,1]^3$? –  Iian Smythe Oct 22 '11 at 14:10
Let $(x,y,z)$ in $[0,1]^3$. There is $p \in [0,1]$ with $u(p)=y, v(p)=z$. There is $t \in [0,1]$ with $u(t)=x, v(t)=p$. Compute $f_3(t)$. –  GEdgar Oct 22 '11 at 17:39
Thanks, I see it now. –  Iian Smythe Oct 22 '11 at 17:52

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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It seems a nice argument. but I did not understand the contradiction. $x_n$ does not belongs to $\pi_n f([-n,n])$ but it could be an element of $\pi_n f([-k(n),k(n)])$ for some big $k(n)$. So if this happens we don't get the contradiction. can you manage to include this analysis in your argument ? Or it is there and I did not saw it ? –  Leandro Oct 22 '11 at 7:13
@Leandro: It doesn’t matter: the contradiction is still there. Here are a few more details. For $n\in\omega$ let $K_n=f[[-n,n]]$; $\pi_n[K_n]$ is compact, so we pick $x_n\in\mathbb{R}\setminus \pi_n[K_n]\ne\varnothing$. Let $x=\langle x_k:k\in\omega\rangle\in\mathbb{R}^\omega$. For each $n\in\omega$ we have $\pi_n(x)=x_n\notin\pi_n[K_n]$, so $x\notin K_n$. Hence $x\notin\bigcup_nK_n=\mathbb{R}^\omega$, a contradiction. It’s a kind of diagonal argument: we use the $n$-th coordinate of $x$ to avoid $K_n$. –  Brian M. Scott Oct 22 '11 at 7:47
@BrianM.Scott thanks for post this. In fact, the answer's idea it is correct but this argument let more precise how to get the contradiction –  Leandro Oct 22 '11 at 8:31
Thanks, this definitely does the trick. –  Iian Smythe Oct 22 '11 at 14:07