Why are there no wild arcs in the plane? On math.stackexchange it was asked whether all arcs in the plane are ambient-isotopic. I suggested that one could prove this by appealing to the Schönflies theorem, which you can do as long as you can extend your Jordan arc to a Jordan curve. That is, as long as you can extend an embedding of an interval to an embedding of a circle. However, I have to admit I don't readily see how to do this, and there are examples which show this is a subtle question. E.g. you can take an arc which spirals infinitely into a point. So my question is how one can show an embedding of an arc into $\mathbb R^2$ extends to an embedding of a circle into $\mathbb R^2$, or, failing that, if someone knows some other proof that all planar Jordan arcs are standard.
Edit: I want to highlight Bill Thurston's elegant answer (in a comment) to the question of whether you can extend an arc to a circle, even though I accepted his other answer using Caratheodory's theorem. Namely, assume your arc runs from $0$ to $\infty$ in $\mathbb C\cup\{\infty\}$. Then take the pre-image under the double branched cover $z\mapsto z^2$. The original arc can be identified with one of its two preimages, while the other preimage fills it in to a circle. Then one can apply the Schönflies theorem. 
 A: By stereographic projection we can assume that the given Jordan arc
lies on a sphere, and that its two ends are at opposite poles, $N$
and $S$. Now project the arc onto a cylinder that touches the sphere
at the equator, so that $N$ and $S$ go to infinity on opposite ends
of the cylinder.
Unrolling the cylinder, we now have our arc on a strip of the plane,
with the ends of the arc at opposite ends of the strip. Now we can
inverse stereographically project the arc back onto the sphere, so
that both ends go to the point of projection. We now have a closed
Jordan  curve on the sphere, and we can apply the Schoenflies theorem.
Edit. Ryan Budney has pointed out the flaw in this argument, so I
withdraw it as it stands. Thanks for the amendment, Ryan.
A: The desired result appears in M.H.A. Newman's classic book, Elements of the Topology of Plane Sets of Points (2nd ed., 1951), as Theorem 14.5 in Chapter VI, on pg. 164:
Theorem 14.5: Every simple arc in $X^2$ is an arc of a simple closed curve in $X^2$.
$X^2$ is Newman's notation for a space that is either the "open" plane, $R^2$, or the "closed" plane, $R^2 \cup \{\infty\}$.
N.B. The proof of Theorem 14.5 does not involve Schoenflies' Theorem. Also, Newman's definition of an arc is such that it has ends.
A: One way to show all arcs are tame is to apply the Riemann mapping theorem to the complement of the arc on $S^2$. Caratheodory proved that whenever the complement of a simply-connected domain is locally connected, then the Riemann mapping extends to a continuous map of the disk to the plane. One half of the disk parametrizes the arc.
Another (related) method is to take the stereographic projection of the arc to a sphere
in $\mathbb E^3$, and form its convex hull.  If the interior of the ball is interpreted as the projective model for hyperbolic 3-space, then the boundary of the convex hull intersect the interior of the ball is a developable surface, i.e. its path metric (that is, distance between points is the minimum arc length of a path connecting them) is a complete metric locally isometric to the path-metric of the hyperbolic plane. Since it's simply-connected, it's globally isometric to the hyperbolic plane.  For reasons parallel to Caratheodory's the map extends continuously to a map of the closed disk (2-dimensional projective disk model of 
the hyperbolic plane, together with its circle at infinity) to the closed ball.
