Proof that the set of all possible curves is of cardinality $\aleph_2$?

Solution 1:

This assertion is incorrect and even if corrected is misleading. Here is what I'm guessing Gamow had in mind. You can describe a (parametrized) "curve" in the plane as a function $f:\mathbb{R}\to\mathbb{R}^2$. The sets $\mathbb{R}$ and $\mathbb{R}^2$ each have cardinality $2^{\aleph_0}$, so the set of all such functions has cardinality $$(2^{\aleph_0})^{2^{\aleph_0}}=2^{\aleph_0\cdot 2^{\aleph_0}}=2^{2^{\aleph_0}}.$$ By Cantor's theorem, this is strictly larger than the cardinality of the plane itself, which is $2^{\aleph_0}$.

However, it is not correct that $2^{2^{\aleph_0}}=\aleph_2$. The standard axioms of set theory actually are not strong enough to determine whether $2^{2^{\aleph_0}}=\aleph_2$: it might be true, or $2^{2^{\aleph_0}}$ could be larger than $\aleph_2$ (possibly much much much larger). Relatedly, it is not correct to say that $2^{\aleph_0}=\aleph_1$ either: it is possible that this is true, but you cannot prove that it is true from the standard axioms.

The statement is further misleading because it is not particularly reasonable to refer to an arbitrary function $\mathbb{R}\to\mathbb{R}^2$ as a "curve". Indeed, by this definition a "curve" is (roughly speaking) just a completely arbitrary collection of points in the plane. Normally when mathematicians use the word "curve", they mean something much more restricted, such as a continuous function $\mathbb{R}\to\mathbb{R}^2$. And in fact, there are only $2^{\aleph_0}$ such continuous functions, so with that definition is is not true that there are more curves than there are points in the plane.