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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.2012-07-17
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    I'm not sure. Could you please elaborate?2012-07-17
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    @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.2012-07-17
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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.2012-07-17
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    @us2012 Good point, I was not thinking.2012-07-17
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    @MarkDominus I was thinking of various [Sierpinski decompositions](http://mathoverflow.net/questions/50952/can-sierpinskis-anisotropic-bicolouring-of-the-plane-assuming-the-continuum-hyp) 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.2012-07-17
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    @Leonid I know that construction. That's just the sort of thing I had in mind when I imagined that a 3-valent set could be constructed via transfinite induction. But I don't really know either.2012-07-17
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    I have deleted the part of my question that asks if an everywhere 3-valent set is necessarily connected, since that seems unlikely. Probably I should have said *locally* connected, but I will have to think about it a bit more.2012-07-17
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    I have not accepted an answer here because I am still hoping for answers to the questions of the last paragraph.2012-10-26

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