Explicit bijection between Jordan curves and real numbers It is my understanding that the set of all Jordan curves and the set of real numbers are of the same cardinality.  So, it follows that there should exist a bijection between them.  Is there a known, explicit bijection between these two sets?  If so, what is it?
 A: Recall that a Jordan curve is a continuous injection from the circle into $\Bbb R^2$. First let us see that indeed there are only $2^{\aleph_0}$ Jordan curves, then we can discuss explicitness.
The unit circle has the same cardinality as the real numbers (why?) and it is a separable space, that is to say that it has a dense countable subset. Recall that a continuous function is uniquely extended from the values it takes on a dense subset. So we can fix $Q$ to be a countable dense subset and two Jordan curves which take the same values when restricted to that set will have to be equal.
It follows, if so, that the collection of all continuous functions from $S^1$ to $\mathbb R^2$ has the cardinality of at most $2^{\aleph_0}$, as demonstrated in the following argument, $$\left|(\Bbb R^2)^Q\right|=|\mathbb R^2|^{|Q|}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\aleph_0}=2^{\aleph_0}=|\mathbb R|.$$
So the collection of Jordan curves cannot have more than $2^{\aleph_0}$ elements; but it also cannot have less. Simply note that if $r\in\mathbb R$ is a positive number then scaling the unit circle to the circle of radius $r$ is a Jordan curve. Therefore we have $$|\mathbb R|\leq|\cal J|\leq|\mathbb R|.$$
(Where $\cal J$ is the set of all Jordan curves.)
By the Cantor-Bernstein theorem it follows that $|\cal J|=|\Bbb R|$. 
But wait a minute, we didn't really write any explicit bijection! Am I just pulling your legs with some mathematical nonsense? Well, no. I'm not actually. The Cantor-Bernstein theorem allows us to generate an explicit bijection from given injections. It is hardly ever a pretty bijection, though. When we think about "explicit functions" we often think in terms of "$x\mapsto x^2+e^{\pi x}$ or something like that, some nice closed term formula. But the truth is that the majority of functions cannot be written like that, or even piece-wise like that.
If one is so inclined then all it is possible to work from the equalities and inequalities the various injections needed and to use the Cantor-Bernstein theorem to write out an explicit bijection, but this function will not be very pretty.
