Find the cardinality of the set of all convex subsets of $\mathbb{R}^2.$
Is it true that the cardinality of the set in $1$ is at most that of the real numbers? If not, then the following only gives a rough lower bound on that cardinality. A convex subset $C$ of $\mathbb{R}^2$ is one that satisfies that for all $a,b \in C, t\in[0,1], ta + (1-t)b\in C.$ One would need to construct a distinct convex subset of $\mathbb{R}^2$ for each positive real number, and intuitively this can be done by taking a convex subset and rescaling it for each positive real number (if $A$ is convex and $k >0,$ then $kA$ is convex because for all $a,b \in kA, a = kx, b= ky$ for some $x,y\in C$ so for any $t\in [0,1], ta + (1-t)b = t(kx) + (1-t)(ky) = k(tx + (1-t)y) \in C$. Then wouldn't this describe an injective function from $\mathbb{R}^+$ to the set of convex subsets? And if so, the cardinality of the set of convex subsets is at least $|\mathbb{R}|$. Since the cardinality is also at most $|\mathbb{R}|$, it is equal to $|\mathbb{R}|.$
Best Answer
No, the cardinality of the family of convex sets of $\Bbb R^2$ (and of $\Bbb R^n$ for any $n\ge2$) is $\beth_2=2^{\lvert \Bbb R\rvert}$. In order to see this, call $S^1=\{x\in\Bbb R^2\,:\, \lVert x\rVert=1\}$ the unit sphere and $B_1=\{x\in\Bbb R^2\,:\,\lVert x\rVert<1\}$. Notice that, for all $H\subseteq S^1$, $B_1\cup H$ is convex, essentially because a segment with endpoints on $S^1$ is entirely contained in $B_1$ except at most its own two endpoints.
So the map $H\mapsto B_1\cup H$ is an injection of $\mathcal P(S^1)\cong\mathcal P(\Bbb R)$ into the set of convex subsets.
Clearly, the cardinality of convex subsets of $\Bbb R^1$ is $\lvert \Bbb R\rvert$.
For $n\ge1$, the cardinality of closed convex subsets of $\Bbb R^n$ is $\lvert \Bbb R\rvert$, essentially because the cardinality of the closed sets is.