Can a set of non self-intersection points of a space-filling curve contain an arc? Consider a continuous surjection $f:[0,1]\to[0,1]\times[0,1]$.
It can be proved that set of self-intersection points must be dense.
In the Hilbert curve, the set of self-intersections are points (a,b) such that either a or b can be written as $\frac{m}{2^k}$ for some integers $k≥1$ and $1≤m<2^k$ (see this explanation).
So the set of self-intersections is dense and of measure $0$, but you cannot draw any vertical or horizontal line without intersecting it.
This leads me to the question: can there exist, for some space-filling curve, an arc (homeomorphic to a non-degenerate closed interval) 
$\tau\subset [0,1]\times[0,1]$ such that $\forall_{t_1,t_2\in [0,1]} f(t_1) = f(t_2) \in \tau \implies t_1 = t_2$? That is, such that $\tau$ includes only non-self-intersection points?
 A: There is no such arc.
Suppose $f$ has no points of self-intersection in $\tau$. Take a point $x$ in $\tau$, and let $x_1,x_2,\ldots$ be any sequence of distinct points in $\tau$ converging to $x$. For each $x_i$, there is a unique $t_i\in[0,1]$ such that $f(t_i)=x_i$. The sequence $t_1,t_2,\ldots$ is an infinite subset of a compact set, so it has an accumulation point $t$. For any subsequence $t_{i_1},t_{i_2},\ldots$ which converges to $t$, we have $x_{i_n}=f(t_{i_n})\to f(t)=x$, so we must have $f(t)=x\in \tau$, so the sequence $t_1,t_2,\ldots$ has a unique accumulation point and is therefore convergent to $t$. This shows that $f^{-1}(x_n)$ converges to $x$ for all sequences $x_1,x_2,\ldots$ of points in $\tau$ which converge to $x$, that is, $f^{-1}|_\tau$ is continuous. (Here I've used that sequential continuity is equivalent to continuity; that's fine because all our spaces here are first countable.) Since $\tau$ is connected, $f^{-1}(\tau)$ must also be connected and is thus a closed interval $I$. We have that $[0,1]\setminus I^\circ$ is closed and whose image under $f$ is $[0,1]^2$ minus the relative interior of $\tau$, but this is not possible, because $f$ is continuous and maps closed sets to closed sets.
A: No, there can be no such arc.
Suppose for the sake of contradiction that $g$ is an arc in $[0,1]^2$ that contains no self-intersections of $f$.
Let $g \colon [0,1] \to A$ be a homeomorphism.
Then $A$ is compact and connected.
Since $A$ is compact and $[0,1]^2$ is a Hausdorff space, $A$ is closed.
Since $f$ is continuous, $B = f^{-1}[A]$ is also closed, and therefore compact.
Now the restriction of $f$ to $B$, $f \restriction B$, is a continuous bijection.
A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so $B$ is homeomorphic to $A$.
Thus $B$ is also connected.
A closed, connected set of real numbers is a closed interval, so $B$ is a closed interval $[p, q]$.
Let $D = ([0,1] \setminus B) \cup \{ \min B, \max B \} = [0,p] \cup [q, 1]$.
Then $D$ is obviously compact, but $f(D)$ is not:
$f(D)$ is the square with the arc $A$ removed and then just two points put back, so any other element of $A$ is a limit point of $f(D)$.
