Contents

Idea

The original “Peano (space-filling) curve” is a surjective continuous function $I \to I^2$, from the closed interval $I = [0, 1]$ to the product with itself, the square. The existence of such an entity (due to Peano) came as a surprise.

One may characterize exactly which Hausdorff spaces arise as the continuous images of a unit interval. These are called Peano spaces.

One can similarly show that there is a continuous surjection $\mathbb{R}^1 \to \mathbb{R}^2$ from the real line to the plane (both regarded as Euclidean spaces equipped with their metric topology), and similarly characterize which spaces arise as continuous images of the real line. These are sometimes called $\sigma$-Peano spaces.

Notice that, while of course there is also an injection $\mathbb{R} \to \mathbb{R}^2$, there is no homeomorphism between these two spaces, or generally between Euclidean spaces of differing dimension. This is the statement of topological invariance of dimension.

Construction

There are many constructions of space-filling curves, but one of the quickest is due to Lebesgue and is closely connected with the Cantor-Lebesgue function.

Cantor space (following the “middle thirds” construction) can be described as the subspace $C$ of $[0, 1]$ consisting of points whose base-$3$ representation $.a_1 a_2 a_3 \ldots = \sum_{i=1}^{\infty} \frac{a_i}{3^i}$ has $a_i \in \{0, 2\}$ for all $i$ (no $1$'s). Define a function $\phi: C \to I$ by

$\phi\left(\sum_{i=1}^\infty \frac{a_i}{3^i}\right) = \sum_{i=1}^\infty \frac{a_i/2}{2^i}$

(i.e., replace $2$‘s in the base $3$ representation by $1$’s and reinterpret the sequence as representing a number in base $2$). Notice that $\phi$ maps the two endpoints of any one of the open intervals removed during the middle-thirds construction to the same point, e.g., for the first middle third $(\frac1{3}, \frac{2}{3})$ we have

$\phi(.02222222\ldots_3) = .01111111\ldots_2 = 10000000\ldots_2 = \phi(.20000000\ldots_3).$

In any case, it is very easy to see that $\phi: C \to I$ is a continuous surjective map.

Now: $C$ is homeomorphic to the product space $2^\mathbb{N}$, a countable product of copies of the discrete space $2 = \{0, 1\}$. Of course we also have a bijection $\mathbb{N} \cong \mathbb{N} + \mathbb{N}$, inducing a homeomorphism

$2^\mathbb{N} \cong 2^{\mathbb{N} + \mathbb{N}} \cong 2^\mathbb{N} \times 2^\mathbb{N}$

and hence a “pairing function” $pair: C \cong C \times C$ that is a homeomorphism (see Jonsson-Tarski algebra). We use this to construct a continuous surjection

$C \stackrel{pair}{\to} C \times C \stackrel{\phi \times \phi}{\to} I \times I,$

denoted say $g: C \to I \times I$, and Lebesgue’s idea is to extend $g$ to a continuous function $f: I \to I \times I$ by linear interpolation: if $x \in I$ belongs to one of the open intervals $(a, b)$ removed during the middle thirds construction, say $x = t a + (1 - t)b$ for some $t \in (0, 1)$, then define

$f(x) = t g(a) + (1 - t)g(b).$
Proposition

The function $f: I \to I \times I$ thus defined is surjective and continuous.

Proof

Surjectivity follows from the fact that its restriction $g: C \to I \times I$ is surjective.

Obviously for each open interval removed in the middle thirds construction, $f$ is continuous at each interior point $x$ (being locally an affine map there), and so it remains to check that $f$ is continuous at each point of $C$. So let $a \in C$, and let us prove that $f(x)$ approaches $f(a)$ as $x$ approaches $a$ from the right; a similar argument will prove continuity from the left. This is obvious if $a$ is the left endpoint of one of the removed open intervals, again because $f$ is affine to the immediate right of $a$. If not, then $a$ is a limit from the right of points of $C$. Now $g: C \to I \times I$ is continuous, so given $\epsilon \gt 0$ there exists $\delta \gt 0$ such that ${|g(x) - g(a)|} \lt \epsilon$ whenever $x \in [a, a + \delta)$ and $x \in C$. What if $x \notin C$? Shrink $\delta$ a little more, and assume $a + \delta$ is a right-hand endpoint of a removed open interval, and consider the case where $x \in [a, a + \delta)$ and $x \notin C$, say $x \in (b, c)$ where $(b, c)$ is a removed open interval and $b, c \in (a, a + \delta]$. Then we get the same $\epsilon$-bound as before: putting $x = t b + (1 - t)c$, we have

$\array{ {|f(x) - f(a)|} & = & {\left|[t g(b) + (1 - t)g(c)] - g(a)\right|} \\ & = & {\left|t(g(b) - g(a)) + (1 - t)(g(c) - g(a))\right|} \\ & \leq & t{\left|g(b) - g(a)\right|} + (1 - t){\left|g(c) - g(a)\right|} \\ & \lt & t\epsilon + (1 - t)\epsilon = \epsilon }$

which completes the demonstration.

The same method can be used to exhibit a space-filling curve $I \to I^S$ for any set $S$ of finite or countable cardinality. Note that in the case where $S$ is a singleton, where we extend the surjection $C \to I$ to $I \to I$ by linear interpolation, we get the Cantor-Lebesgue function.

Hahn-Mazurkiewicz theorem

The eponymous theorem may be stated as follows:

Theorem

A Hausdorff space $X$ admits a continuous surjection $f: I \to X$ from the closed interval $I$ if and only if it is a connected, locally connected compact metrizable space.

(N.B. According to the nLab, connected spaces are nonempty!)

The “only if” half is relatively easy; see here for some details. The “if” half is rather more involved, but Willard’s General Topology contains a proof. A space $X$ satisfying the stated conditions is called a Peano space.

Given this characterization, it is not difficult to characterize which spaces are continuous images of $\mathbb{R}$:

Theorem

A path-connected Hausdorff space $X$ admits a continuous surjection $\mathbb{R} \to X$ if and only if it is a $\sigma$-Peano space, i.e., a countable union $\bigcup_{n: \mathbb{N}} A_n$ of Peano spaces.

Proof

The “only if” half being fairly obvious, the “if” part may be proved as follows. Since there are continuous surjections $[0, \infty) \to \mathbb{R}$ and $\mathbb{R} \to [0, \infty)$, it suffices to show that a $\sigma$-Peano space admits a continuous surjection from $[0, \infty)$. For each $n \in \mathbb{N}$ choose a continuous surjection $f_n: [2 n, 2 n + 1] \to A_n$. Then for each $n$ choose a path $g_n: [2 n + 1, 2 n + 2] \to X$ such that $g_n(2 n + 1) = f_n(2 n + 1)$ and $g_n(2 n + 2) = f_{n+1}(2 n + 2)$. Then the $f_n$ and $g_n$ paste together to form a continuous surjection $[0, \infty) \to X$.

An example of such a space is the Warsaw circle.

References

The proof of the Hahn-Mazurkiewicz theorem is given in section 31 (page 219ff) within chapter 8 of Willard’s classic text: