Uncountable Collections of Arcs In the Plane with Prescribed Properties I was wondering, if for each angle $0 \leq \theta < 2\pi$ we have an uncountable collection $A_\theta$ of pairwise-disjoint, closed, straight line segments with length $1$ and slope $\theta$ in the plane, is it possible for all of the $A_\theta$'s to be immersed in the plane simultaneously?  That is to say, is there a collection of pairwise-disjoint, closed unit segments in the plane such that for each slope, uncountably many have that slope?
I feel like the answer is no.  By covering the plane with a sufficiently small tiling of squares you will find, for each slope, some pair in the covering such that each has uncountably many end points of segments with that slope.  Then you can pick a square 'between' that pair relative to this slope such that its boundary is intersected by uncountably many segments twice.  Another counting procedure then forces the existence of a square such that this happens for uncountably many different angles.
The details of those parts are trivial, but now I get stuck showing that this is impossible.  Namely, that this compact square can't contain uncountably many families of uncountably many segments spanning its perimeter, each of whose members are disjoint.  It seems to be some order-theoretic statement, since we know that each of these uncountable sets (for a given slope) contains an uncountable closed set.  We can assume it is the Cantor Set for all but countably many, lest we get an uncountable number of pairwise-disjoint neighborhoods of the square.
But there is an uncountable collection of pairwise-disjoint Cantor Sets in the unit interval (equivalently, the circle/square perimeter).  So I am wondering, can this proof be saved by some facts about how these are embedded together?
This area of math has been studied some, in the form of decompositions of the plane or ensembles/configurations of various objects in the plane, but I am not familiar with the literature.  Does anybody have a good starting point for the latter viewpoint (i.e. not usc decomposition theory), preferably open access?  I would assume the above question would have been addressed early in its development, but this problem can be generalized in many ways.  Does anybody see some generalizations that this method would still work for?  Haven't bothered to suss them out, so feel free to take the credit haha
Any other cool results of this flavor, please share!
Thanks!
 A: This is only a partial solution and an explicit formulation of the Cantor Set-theoretic statement that needs to be proved.  This is the remaining step for the proof:
Problem: Suppose that $C_\alpha$ is an uncountable collection of pairwise-disjoint copies of the Cantor Set in the unit interval $[0,1] = I$.  Let $C$ be the union of these sets in $I$.  Is it possible to order the indices $\lbrace \alpha \rbrace$ such that if $x \in C_\beta$ and $y_n \in C$ with $y_n \rightarrow x$ from the left, then eventually each $y_n$ is contained only in some $C_\alpha$'s with $\alpha \leq \beta$?  With the same condition imposed for convergence from the right.
The easy way to state this is: Points from these Cantor Sets $C_\alpha$ can converge from the left to some point in another Cantor Set $C_\beta$ if and only if their indices are eventually to the left of $\beta$.
My intuition has been that this is impossible, namely the sets $C_\alpha$ will be so entangled with each other that they might crunch themselves too thin to all fit in the interval, if that order requirement is imposed.  I still haven't been able to complete the proof, but I think it is almost done.  Here are some reductions I have made:
For each $C_\alpha$ we can write it as the complement of a sequence of finite unions of open intervals as per normal, and $C_\alpha$ can be written as the closure of the end points of these intervals.  Thus for each $C_\alpha$ we can distinguish a countable collection of its points $\lbrace x_\alpha^1, x_\alpha^2, \dots \rbrace$ representing its first point (relative to the order on $I$), its last point, then the two end points of the first removed interval, then the four end points of the next level of excised intervals, etc.  And this set is dense in $C_\alpha$.
Now consider the family $\lbrace x_\alpha^1 \rbrace$.  Since the $C_\alpha$'s are pairwise-disjoint each of these points is unique.  The collection satisfies the natural order inherited from $I$, so by lemma 3.2.1 in Whyburn's Analytic Topology (p. 44), we may discard a countable subcollection and be left with a saturated family, i.e one where each point is an $\omega$-accumulation point of both its predecessors and successors relative to the order from $I$.
Since each point is coming from a different $C_\alpha$, after discarding the Cantor Sets associated with the discarded points, this means that the left end point of each $C_\alpha$ is a condensation point from both the left and right by leftmost points of other $C_\beta$.  Doing this for each $n$ we discard at most countably many $C_\alpha$ and now have an uncountable collection of pairwise-disjoint Cantor Sets such that their designated dense sets all $\omega$-accumulate onto each others'.
This means that in particular the excised intervals converge to each other.  But they cannot be equal since the Cantor Sets are disjoint from each other.  It seems like you now get a contradiction, looking at whether these excised intervals are to the left, right, or contain the middle excised interval for some fixed $C$ that a collection's end points are converging onto.  Does anyone see how to finish it from here?
