A strangely connected subset of $\Bbb R^2$ Let $S\subset{\Bbb R}^2$ (or any metric space, but we'll stick with $\Bbb R^2$) and let $x\in S$.  Suppose that all sufficiently small circles centered at $x$ intersect $S$ at exactly $n$ points; if this is the case then say that the valence of $x$ is $n$.  For example, if $S=[0,1]\times\{0\}$, every point of $S$ has valence 2, except $\langle0,0\rangle$ and $\langle1,0\rangle$, which have valence 1.
This is a typical pattern, where there is an uncountable number of 2-valent points and a finite, possibly empty set of points with other valences. In another typical pattern, for example ${\Bbb Z}^2$, every point is 0-valent; in another, for example a disc, none of the points has a well-defined valence.
Is there a nonempty subset of $\Bbb R^2$ in which every point is 3-valent? I think yes, one could be constructed using a typical transfinite induction argument, although I have not worked out the details. But what I really want is  an example of such a set that can be exhibited concretely.
What is it about $\Bbb R^2$ that everywhere 2-valent sets are well-behaved, but 
everywhere 3-valent sets are crazy? Is there some space we could use instead of $\Bbb R^2$ in which the opposite would be true?

I claim there is a set $S \subseteq {\mathbb R}^2$ that contains exactly three points in every circle.
Well-order all circles by the first ordinal of cardinality $\mathfrak c$ as $C_\alpha, \alpha  < \mathfrak c$. By transfinite induction I'll construct sets $S_\alpha$ with 
$S_\alpha \subseteq S_\beta$ for $\alpha < \beta$, and take
$S = \bigcup_{\alpha < {\mathfrak c}} S_\alpha$.  These will have the following properties:


*

*$S_\alpha$ contains exactly three points on every circle $C_\beta$ for $\beta \le \alpha$.

*$S_\alpha$ does not contain more than three points on any circle.

*$\text{card}(S_\alpha) \le 3\, \text{card}(\alpha)$ 


We begin with $S_1$ consisting of any three points on $C_1$.
Now given $S_{<\alpha} = \bigcup_{\beta < \alpha} S_\beta$, consider the circle $C_\alpha$.
Let $k$ be the cardinality of $C_\alpha \cap S_{<\alpha}$.  By property (2), $k \le 3$.  If $k = 3$, take $S_\alpha =  S_{<\alpha}$.
Otherwise we need to add in $3-k$ points.  Note that there are fewer than ${\mathfrak c}$ circles determined by triples of points in $S_{<\alpha}$, all of which are different from $C_\alpha$, and so there are fewer than $\mathfrak c$ points of $C_\alpha$ that are
on such circles.  Since $C_\alpha$ has $\mathfrak c$ points, we can add in a point $a$ of $C_\alpha$ that is not on any of those circles.  If $k \le 1$, we need a second point $b$ not to be on the circles determined by triples in $S_{<\alpha} \cup \{a\}$, and if $k=0$ a third point $c$ not on the circles determined by triples in $S_{<\alpha} \cup \{a,b\}$.  Again this can be done, and it is easy to see that properties (1,2,3) are satisfied.
Finally,  any circle $C_\alpha$ contains exactly three points of $S_\alpha$, and no 
more than three points of $S$ (if it contained more than three points of $S$, it would have more than three in some $S_\beta$, contradicting property (2)).  
