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What is the cardinality of the set of all simple closed curves in $R^2$? Furthermore, what resources are there which present a proof, if any, of said cardinality?

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Do you want to place any restrictions on the curves, like requiring them to be smooth or piecewise-smooth or (Riemannian) embeddings, or are you only concerned with the topological embeddings of $S^1$ into $\mathbb{R}^2$? For example, one can probably relatively easily characterize the space of circles in $\mathbb{R}^2$. It is even easier to classify topological embeddings $S^1\hookrightarrow\mathbb{R}^2$ up to homotopy. It is probably not so easy to tackle the general topological or smooth problem. More guidance would help determine which answer you want. – Neal Dec 7 '11 at 2:25
@Neal I am sorry for being so general, but the only restriction is that the curves are Jordan curves in $R^2$. I would prefer not to make the restriction that they are smooth or piecewise-smooth. – analysisj Dec 7 '11 at 2:33
@Neal: It doesn’t matter: the answer in all of those cases is $2^\omega$. – Brian M. Scott Dec 7 '11 at 2:45
I misinterpreted the question. My bad. PS- The cardinality of homotopy classes of simple closed curves in $\mathbb{R}^2$ is not $2^\omega$. ;) – Neal Dec 7 '11 at 3:13
up vote 5 down vote accepted

A simple closed curve in $\mathbb{R}^2$ is the image of $S^1$ under a continuous map. $S^1$ has a countable dense subset $D$ (for instance, the set of points with polar coordinates of the form $\langle 1, q\pi\rangle$, where $q$ is a rational number in the interval $[0,2)$). A continuous function from $S^1$ to $\mathbb{R}^2$ is uniquely determined by its restriction to $D$. That is, if $f,g:S^1\to\mathbb{R}^2$ are continuous functions such that $f(x)=g(x)$ for every $x\in D$, then $f=g$. Thus, there are at most as many embeddings of $S^1$ into $\mathbb{R}^2$ as there are functions from $D$ into $\mathbb{R}^2$. $|\mathbb{R}^2|=2^\omega$, and $|D|=\omega$, so there are $$|\mathbb{R}^2|^{|D|}=(2^\omega)^\omega=2^\omega$$ functions from $D$ into $\mathbb{R}^2$. This shows that there are at most $2^\omega$ simple closed curves in $\mathbb{R}^2$.

On the other hand, it’s easy to construct $2^\omega$ distinct simple closed curves on $\mathbb{R}^2$: the circles centred at the origin form such a family, for instance. Thus, there are exactly $2^\omega$ simple closed curves in the plane.

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So then is the continuum equinumerous with this set? – analysisj Dec 7 '11 at 3:06
Yes: $\mathfrak{c}=2^\omega$. – Brian M. Scott Dec 7 '11 at 3:22
I think @analysisj also meant to say "thank you". :) – Bruno Joyal Dec 7 '11 at 5:39
@BrianM.Scott Thank you! – analysisj Dec 7 '11 at 11:18

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