I was interested in putting a complete metric on the space of Jordan curves. Say, just planar Jordan curves contained in $B(\bar{0}, 2) \backslash B(\bar{0}, 1)$ which separates $\bar{0}$ and infinity.

As we can see, the standard Hausdorff metric does not do the job, for example we can have a Cauchy sequence of Jordan curves with thinner and thinner 'fingers', the limit is a circle with a spike, hence not a Jordan curve. (see Figure below)

Hence in some sense, the metric needs to not only 'see' the curve but also what it encloses. I looked a bit into the literature and didn't find anything...Does anyone know such a metric? Any ideas?

An attempt of mine in constructing the metric:

First we can consider the plane as in the Riemann sphere, so our space is merely the set of Jordan curves that separates the unit disk around the North pole and that around the South pole. For each curve we have (up to a Mobius automorphism of the disc) two Riemann maps from the standard unit disc to the North and South hemisphere separated by the curve.

Define the distance between two curves to be the minimum distance between the pair of 'North hemisphere Riemann maps', plus that of the South hemisphere. It's easy to check that this does give a metric. In our 'finger' case above, the $ \{ C_n \} $ is not Cauchy in our metric (because Riemann map will map less and less of the disc inside the finger so there is a subsequence of distance finger-length apart).

I strongly believe this metric is indeed complete, but can't find a good argument.

What can be done is, given a Cauchy sequence of Jordan curves under the above metric, pick a pair of distance minimizing Riemann map for each curve (such thing always exist by a simple compactness argument) then apply Carathéodory, all Riemann maps extends to the homeo on the boundary, so we can take the pointwise limit of the boundary map. Do the same for both North and South hemisphere. All we need to show is this limit is still injective. (I think it should be because if not, either North or South will have a infinitely thin 'neck', so the Riemann maps on that half will not be Cauchy).

I ran into many topological difficulties when trying to make an argument for this (mainly because we know nothing about the limiting curve and it can be really complicated)...Any ideas on how one might go about that would be greatly appreciated. :-P

Figure for my comment on professor Thurston's response:

---A simple case to illustrate that the two Riemann maps does not match up on the equator. i.e. radial lines in the North hemisphere does not 'want' to pass through the thin tube and land on the tip, but radial lines in the South hemisphere does not care. Hence the pre-image of the black arc will be very short in the Northern disc but normal in the Southern disc. However, there of course exists unique homeo $h$ on the boundary to glue them up. I do not know how would that give a homeo from the sphere to itself, though.