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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?

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Are you willing to assume compactness or at least closedness? Otherwise I'd be worried about some kind of continuum-hypothesis-based weirdness. –  user31373 Jul 17 '12 at 1:54
    
I'm not sure. Could you please elaborate? –  MJD Jul 17 '12 at 1:58
    
@MarkDominus If you assume closedness, then the cardinality of your set (which is clearly not finite) is either $\aleph_0$ or $\mathfrak c$, independent of the CH. I think Leonid's concern is that otherwise transfinite induction might produce something with properties dependent on cardinality, which would in turn depend on CH. –  Alex Becker Jul 17 '12 at 2:03
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@AlexBecker: Surely his set cannot be $\aleph_0$ as he wants every sufficiently small circle around any point to intersect it in 3 points? There are as many small circles as the continuum has points. –  us2012 Jul 17 '12 at 2:12
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@MarkDominus I was thinking of various Sierpinski decompositions which rely on CH. I have a vague suspicion that one could construct a non-explicit, non-connected everywhere 3-valent set in such a way. But I don't really know. –  user31373 Jul 17 '12 at 2:16
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2 Answers

Here are the details of the transfinite induction argument:

Well-order the set of points in $\mathbb{R}^d$ and let $p_\alpha$ denote the point at ordinal index $\alpha$. Then define:

$S_{<\alpha} = \bigcup_{\beta \lt \alpha}{S_\beta}$

$S_0 = \{p_0\}$

$S_\alpha = S_{<\alpha}$ when there is a $(d-1)$-sphere centered at $p_\alpha$ intersecting $S_{<\alpha}$ at more than $n$ points.

$S_\alpha = S_{<\alpha} \cup \{p_\alpha\}$ otherwise.

It can be seen that $\bigcup_{\alpha}{S_\alpha}$ is everywhere-$n$-valent.

This is actually a little bit stronger since every point has $n$ neighbors at every distance, not just arbitrarily small ones. It works for any $n$ and also any $d$, but I don't know exactly which metric spaces.

I don't know if there is any concrete example. Maybe it is worth considering the valency of plane fractals like the Koch snowflake; perhaps there are concrete examples of everywhere-$2 \cdot n$-valent curves that are wiggly enough to enter and exit arbitrarily small circles multiple times. Because of the Jordan curve theorem, this approach seems less promising for finding everywhere-odd-valency sets.

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How does this make $\bigcup_{\alpha} S_\alpha$ everywhere $n$-valent? Certainly if $p_\alpha \in S_\alpha$, there are no more than $n$ points of $S_\alpha$ intersecting any sphere centred at $p_\alpha$. But then you may put $p_\beta$ in $S_\beta$ with $\beta > \alpha$, and this will be on some sphere centred at $p_\alpha$ that may have had $n$ points in $S_\alpha$. –  Robert Israel Jul 17 '12 at 7:08
    
Maybe you want to also have $S_\alpha = S_{<\alpha}$ if there is a sphere containing $p_\alpha$ centred at some point of $S_{<\alpha}$ that has $n$ points of $S_{<\alpha}$. But I see no reason to think you'll get any points at all on any particular sphere. –  Robert Israel Jul 17 '12 at 7:12
    
You are right, my construction is missing something. –  Dan Brumleve Jul 17 '12 at 8:36
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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:

  1. $S_\alpha$ contains exactly three points on every circle $C_\beta$ for $\beta \le \alpha$.
  2. $S_\alpha$ does not contain more than three points on any circle.
  3. $\text{card}(S_\alpha) \le 3\, \text{card}(\alpha)$

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)).

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