Peano Curve | Real Analysis

A continuous map

[0,1] \to [0,1]^2

can fill the whole square — recursively repeating the 9-subsquare snake yields uniformly convergent approximations with a surjective continuous limit. Injectivity is impossible (Netto's theorem, proved in full).

Examples & Exercises

Show that the order- $k$ approximants $p_k: [0,1] \to [0,1]^2$ (the recursive subdivision of the snake polyline through the 9 subsquares) are uniformly Cauchy, and prove completely that the uniform limit $p$ is continuous.

\|p_{k+1} - p_k\|_\infty \le \frac{\sqrt{2}}{3^k} \to 0

Order

k = 1

: 9 segments snake through the 9 subsquare centers — this pattern repeats recursively inside each subsquare.

Show proof

Each $p_k$ is a polyline, hence continuous. The parametrization is: $p_k$ traverses each interval $[(j-1)/9^k,\, j/9^k]$ inside the $j$ -th level- $k$ subsquare $Q_j^{(k)}$ (side $3^{-k}$ ), in snake order.
$p_{k+1}$ refines $p_k$ : on $[(j-1)/9^k,\, j/9^k]$ it traverses the 9 level- $(k+1)$ subsquares inside the same $Q_j^{(k)}$ (the recursion applies the same pattern within each $Q_j$ ). Hence for every $t$ , $p_k(t)$ and $p_{k+1}(t)$ both lie in $Q_j^{(k)}$ , so $|p_{k+1}(t) - p_k(t)| \le \mathrm{diam}\, Q_j^{(k)} = \sqrt{2}\, 3^{-k}$ . Thus $(p_k)$ is uniformly Cauchy and converges uniformly — call the limit $p$ .
A uniform limit of continuous functions is continuous: from $|p(s) - p(t)| \le |p(s) - p_k(s)| + |p_k(s) - p_k(t)| + |p_k(t) - p(t)|$ , make the first and third terms small by uniform convergence, then the middle term small by continuity of the fixed $p_k$ (the $\varepsilon/3$ argument). Hence $p: [0,1] \to [0,1]^2$ is a continuous limit curve.

Prove completely that the limit curve $p: [0,1] \to [0,1]^2$ from the problem above is surjective.

p([0,1]) = [0,1]^2

Order

k = 2

: 81 segments visit every cell of the

9 \times 9

grid — the square fills up as the order grows.

Show proof

Take any $y \in [0,1]^2$ . At each level $k$ , $y$ lies in some level- $k$ subsquare $Q^{(k)}$ (side $3^{-k}$ ). Since $p_k$ visits every level- $k$ subsquare, there is $t_k$ with $p_k(t_k) \in Q^{(k)}$ .
Since $(t_k)$ lies in $[0,1]$ , compactness gives a convergent subsequence $t_{k_j} \to t$ . By continuity, $p(t_{k_j}) \to p(t)$ . But $|p(t_{k_j}) - y| \le |p(t_{k_j}) - p_{k_j}(t_{k_j})| + |p_{k_j}(t_{k_j}) - y|$ : the first term tends to 0 by uniform convergence, and the second because both points lie in the same $Q^{(k_j)}$ — each is at most $\sqrt{2}\, 3^{-k_j} \to 0$ . Hence $p(t) = y$ .
Since $y$ was arbitrary, $p$ passes through every point of $[0,1]^2$ — a continuous surjection $p: [0,1] \to [0,1]^2$ . That a 1-dimensional interval can "fill" a 2-dimensional square was Peano's 1890 discovery.

Prove completely that no continuous bijection $[0,1] \to [0,1]^2$ exists (Netto's theorem). Hence the Peano curve above cannot be injective.

\text{no continuous bijection } [0,1] \to [0,1]^2

Order

k = 3

: 729 segments — the curve crosses itself countless times, visibly ruling out injectivity.

Show proof

Two lemmas first. (A) The continuous image of a compact set is compact: if $K$ is compact and $f$ continuous, an open cover $\{U_i\}$ of $f(K)$ pulls back to an open cover $\{f^{-1}(U_i)\}$ of $K$ , which has a finite subcover whose images cover $f(K)$ . (B) A compact subset of a Hausdorff space is closed: if $K \subseteq Y$ is compact and $y \notin K$ , each $x \in K$ has disjoint neighborhoods $U_x \ni x$ , $V_x \ni y$ (Hausdorff); finitely many $U_{x_i}$ cover $K$ , so $\bigcap V_{x_i}$ is a neighborhood of $y$ missing $K$ .
Netto's theorem: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Indeed, if $F \subseteq K$ is closed then $F$ is compact, so $f(F)$ is compact by (A) and closed in $Y$ by (B) — i.e. $f$ is a closed map. A closed bijection is a homeomorphism ( $(f^{-1})^{-1}(F) = f(F)$ is closed, so $f^{-1}$ is continuous).
If $f: [0,1] \to [0,1]^2$ were a continuous bijection, it would be a homeomorphism. Removing an interior point $c \in (0,1)$ disconnects the interval: $[0,1] \setminus \{c\} = [0,c) \cup (c,1]$ is a separation into disjoint nonempty opens. But $[0,1]^2$ minus a point is path-connected (route around the missing point), hence connected. A homeomorphism restricts to a homeomorphism of the punctured spaces — contradiction. So no continuous bijection exists, and the Peano surjection cannot be injective.