I read in the book 'One two three infinity' by George Gamow that the cardinality of sets of all plane curves is greater than the number of real numbers and so it can be thought of as a strictly larger infinite crdinality $\aleph_2$. I wanted to know the reasoning or proof of this.
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1If Gamow really wrote that it was $\aleph_2$, he was being careless. But see https://math.stackexchange.com/questions/2078067/proof-that-the-set-of-all-possible-curves-is-of-cardinality-aleph-2 also https://math.stackexchange.com/questions/411643/the-number-of-curves-in-the-plane-has-cardinality-2-aleph-0 – Gerry Myerson Sep 12 '17 at 07:39
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Usually when talking about curves in a plane, we mean continuous functions $\phi:\mathbb{R}\to\mathbb{R}^2$ and the cardinality of this set is $2^{\aleph_0}$. The proof is as follows
Each continuous function $\phi:[0,1]\to\mathbb{R}^2$ is determined by its action on $\mathbb{Q}\cap[0,1]$. Now, the set $\mathbb{Q}\cap[0,1]$ is countable, so for each $x$ that belongs to it, there are at most $|R^2|=2^\aleph_0$ choices $\phi$ can take. Therefore, there are at most $(2^\aleph_0)^\aleph_0=2^{\aleph_0\cdot\aleph_0}=2^\aleph_0$ such functions. In other words, there are $$|\mathbb{R^2}|^{|\mathbb{Q}\cap[0,1]|}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}$$ such functions.
So, the upper bound on the number of such functions $\phi$ is $2^{\aleph_0}$. The lower bound is obtained by considering all constant functions to the plane; there is $2^{\aleph_0}$ of them (as many as the reals$. This means
$$2^{\aleph_0}\le |\phi:[0,1]\to\mathbb{R}^2|\le 2^{\aleph_0}$$ $$\implies$$ $$|\phi:[0,1]\to\mathbb{R}^2|=2^{\aleph_0}$$
Now, if by curve in a plane, what is meant is an arbitrary collection of points in a plane, than the calculation proceeds as follows $$(2^{\aleph_0})^{2^{\aleph_0}}=2^{\aleph_0\cdot 2^{\aleph_0}}=2^{2^{\aleph_0}},$$ The above is strictly larger than $2^{\aleph_0}$, by Cantors theorem.
Žiga Sajovic
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