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This is Exercise 6.59 from Goldrei's Classic Set Theory.

(a) Show that any two points in $\Bbb R^2\setminus \Bbb Q^2$ of the plane can be joined by a circular arc which lies entirely in this subset. (This is a throwaway remark of Cantor's!)

(b) How many such arcs are there? (That is, is the set of all such arcs joining the two points equinumerous with a well-known set like $\Bbb N$ or $\Bbb R$?)

My attempt:

(a) Suppose we are given $2$ points $a,b$ in $\Bbb R^2\setminus \Bbb Q^2$. Given any point $c$, $c$ is the center of a circle containing $a,b$ iff $c$ lies on the perpendicular bisector of the line joining $a,b$. Denote by $C_c$ the circle containing $a,b$ with center $c$. There are at most two rational points on a circle with non-rational center. Suppose it is the case that any circle with a non-rational center and containing $a,b$ has at least $1$ rational point. Then given any point non-rational $c$ on the perpendicular bisector of the line joining $a,b$, we may it to $(q,q)$ in the case that $C_c$ has exactly one rational point $q$ and to $(p,q)$ in the case that $C_c$ has exactly two rational points $p,q$, with ($x$-coordinate of $p$< $x$-coordinate of $q$) or ($x$-coordinate of $p$=$x$-coordinate of $q$ and $y$-coordinate of $p$< $y$-coordinate of $q$). This map is injective since if $2$ circles joining $a,b$ have the same rational point(s), then they are identical. It follows that the non-rational points on the perpendicular bisector is countable, a contradiction. Therefore, there exists a non-rational point $k$ such that $C_k$ has center $k$ and containing $a,b$ and no rational points. Any arc with end points $a,b$ on $C_k$ satisfies the requirement of the problem.

(b) Given any arc satisfying the requirement in (a), we may map it to the unique point of intersection of the arc and the perpendicular bisector of the line joining $a,b$. This map is injective because any $2$ distinct arc satisfying (a) must have distinct points of intersection with the perpendicular bisector of the line joining $a,b$. Hence, there are at most $|\Bbb R^2|=|\Bbb R|$ such arcs.

By a similar argument in (a), there are at least uncountably many arcs satisfying (a), but it doesn't show that the cardinality of the set of those arcs is at least $|\Bbb R|$. My guess is that there are $|\Bbb R|$ such arcs but I don't know how to prove it.

19021605
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  • Surely the points on the perpendicular bisector that you exclude are countable – Mark Bennet Sep 07 '24 at 08:37
  • I know but the required set $S$ of arcs being uncountable doesn't imply that $|\Bbb R|\le |S|$. – 19021605 Sep 07 '24 at 09:23
  • You are deducting a countable set from a continuum set. – Mark Bennet Sep 07 '24 at 09:49
  • I get what you mean. By https://math.stackexchange.com/questions/30282/does-k-aleph-0-mathfrakc-imply-k-mathfrakc-without-the-axiom-of-choic ,we deduce the cardinality of the set of arcs satisfying (a) is continuum. But none of the proofs on that page uses only the techniques that I have learnt while reading up to the page of this exercise. The first $2$ proofs have set theory concepts that I haven't learnt and the last proof uses the fact that any countable set is well-ordered which isn't proved even in later chapters. I'm guessing there's a solution to this exercise without that result. – 19021605 Sep 07 '24 at 12:49

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