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There is a surjective continuous map $[0;1]\rightarrow [0;1]^2$ ("space filling curve"). Using such a map one can easily get space filling curves for all finite dimensional cubes. So my question is: Is there a space filling curve of the Hilbert cube $[0;1]\rightarrow [0;1]^\mathbb{N}$ ?

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I believe this is an old theorem that appears in many point-set topology textbooks, that every compact connected, locally-connected metric space is the continuous image of $[0,1]$. Willard or Dugunji should have it. –  Ryan Budney Aug 9 '10 at 8:41

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up vote 10 down vote accepted

Ah, it's called the "Hahn–Mazurkiewicz theorem".


And it apparently appears in the Willard text.

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This also appears in Laskovitch's Conjecture And Proof, where it is used to construct a family of continuous functions $f_1, f2, \dots$ with the property that for any sequence $\alpha_i$ of reals taken from the interval $[0,1]$, there exists a point $x \in [0,1]$ with $f_i(x) = \alpha_i$ for all $i$. –  Nick Salter Aug 19 '10 at 19:23

Well there is indeed a "simple" construction of such a space filling curve.

Let $\gamma:[0;1]\rightarrow [0;1]^2$ be a space filling curve. Then one can obtain a space filling curve for $[0;1]^3$ by postcomposing with $id_\mathbb{R}\times \gamma$. Then one can postcompose with $id_{\mathbb{R}^2}\times \gamma$ and so on. Note that the first coordinates didn't change in the last step.

Putting all this together we get a map

$f:[0;1]\rightarrow [0;1]^\mathbb{N} \qquad t\mapsto (pr_1\circ \gamma \circ (pr_2\circ \gamma)^{n-1}(t))_{n\in \mathbb{N}}, $ where $pr_i$ denote the obvious projections. This map can be seen as the infinite composition of the maps above. By the definition of the product topology this map is continuous.

Especially if we postcompose $f$ with the projection on the first $n$ coordinates, we just get a space filling curve (see above). Let us show, that a arbitrary element $x=(x_i)_{i\in \mathbb{N}}\in [0;1]^\mathbb{N}$ lies in the Image of $f$. We already know, that for each $n$ there is a element $y^n$ in Im$(f)$ agreeing with $x$ in the first $n$ coordinates.

As $[0;1]$ is compact, Im$(f)$ is compact and hence closed ($[0;1]^\mathbb{N}$ is Hausdorff).

And $\lim_{n\to\infty}y^n=x$. Hence $x\in$ Im$(f)$. So $f$ is a continuous surjective map $[0;1]\rightarrow [0;1]^\mathbb{N}$.

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