Peano Curve

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 $|p_{k+1}(t) - p_k(t)| \le \mathrm{diam}\, Q_j^{(k)} = \sqrt{2}\, 3^{-k}$.

Sum the tail: for $m > n$, the triangle inequality gives $|p_m(t) - p_n(t)| \le \sum_{j=n}^{m-1} |p_{j+1}(t) - p_j(t)| \le \sum_{j=n}^{m-1} \sqrt{2}\, 3^{-j} = \sqrt{2}\, 3^{-n}\, \frac{1 - 3^{-(m-n)}}{1 - 1/3} \le \frac{3\sqrt{2}}{2}\, 3^{-n}$. The right side tends to 0 as $n \to \infty$, independently of $t$ and $m$, so $(p_k)$ is uniformly Cauchy. Since $\mathbb{R}^2$ is complete, $p_k(t)$ converges for each $t$, and uniform Cauchyness makes the convergence uniform — 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.

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 is at most $\frac{3\sqrt{2}}{2}\, 3^{-k_j}$ by the uniform tail estimate proved in peano-uniform-limit ($|p_m - p_n| \le \frac{3\sqrt{2}}{2}\, 3^{-n}$ for $m > n$; let $m \to \infty$), and the second is at most $\sqrt{2}\, 3^{-k_j}$ because both points lie in the same $Q^{(k_j)}$ of diameter $\sqrt{2}\, 3^{-k_j}$. Both tend 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.

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 — a closed subset of a compact set is compact (for $K = [0,1]$ this is Heine–Borel) — 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.

In contrast, $[0,1]^2$ minus the corresponding point $r := f(c)$ is path-connected, shown here with explicit formulas.

Take two points $p, q \ne r$. If $p = q$, the constant path $\gamma(t) \equiv p$ works. If $p \ne q$, look at the line $L(t) = (1-t)p + tq$ ($t \in [0,1]$). If $r \notin L([0,1])$, then $\gamma = L$ is the desired path.

If $r \in L([0,1])$, there is a unique $t_0 \in (0,1)$ with $r = L(t_0)$ (since $r \ne p, q$). Set the midpoint $m = \frac{p+q}{2}$, direction $d = q - p$, unit normal $n = \frac{(-d_2, d_1)}{|d|}$, and pick the detour point $s = m + \varepsilon n$ ($\varepsilon \ne 0$, chosen explicitly below). The path is two segments joined: $\gamma(t) = (1-2t)p + 2ts$ ($0 \le t \le \frac12$), $\gamma(t) = (2-2t)s + (2t-1)q$ ($\frac12 \le t \le 1$).

Check that $r$ is avoided. Points $x$ on the line $pq$ satisfy $n \cdot (x - m) = 0$ (since $n \perp d$), and $r$ lies on it, so $n \cdot (r - m) = 0$. For a point $(1-u)p + us = (1-\frac{u}{2})p + \frac{u}{2}q + u\varepsilon n$ on $[p,s]$, $n \cdot (x - m) = u\varepsilon$; equaling $r$ forces $u\varepsilon = 0$, i.e. $u = 0$ — only the point $p \ne r$.

For a point $(1-u)s + uq = \frac{1-u}{2}p + \frac{1+u}{2}q + (1-u)\varepsilon n$ on $[s,q]$, $n \cdot (y - m) = (1-u)\varepsilon$, so $(1-u)\varepsilon = 0$ gives $u = 1$ — only the point $q \ne r$. Hence $\gamma$ never passes through $r$.

Explicit choice of $\varepsilon$ (forcing $s \in [0,1]^2$): $m$ cannot be a corner (a corner $m$ would give $p = q = m$, contradicting $p \ne q$). Put $\delta = \min(m_1, 1-m_1, m_2, 1-m_2) \ge 0$. If $\delta > 0$ ($m$ interior), take $\varepsilon = \delta/2$: $|s - m| = \delta/2 < \delta$, so $s$ is interior.

If $\delta = 0$, then $m$ lies on an edge but is not a corner. For instance with $m = (m_1, 0)$, $m_1 \in (0,1)$ (the other three edges are identical via symmetries like $(x,y) \mapsto (x,1-y)$), put $\eta = \frac12\min(m_1, 1-m_1) > 0$ and take $s = m + \eta n$ if $n_2 \ge 0$, $s = m - \eta n$ if $n_2 < 0$. Then $s_2 = \eta|n_2| \in [0,\eta] \subset [0,1]$ and $|s_1 - m_1| \le \eta \le \frac12\min(m_1, 1-m_1)$, so $s_1 \in (0,1)$ — i.e. $s \in [0,1]^2$. With $s \in [0,1]^2$, convexity keeps both segments $[p,s]$, $[s,q]$ inside $[0,1]^2$.

Thus any two points of $[0,1]^2 \setminus \{r\}$ are joined by an explicit path avoiding $r$ — path-connected, 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.