Erdős Problem #1071

Erdős Problem 1071

Statement and convention.

Both questions have affirmative answers: there is a finite maximal collection of pairwise disjoint open unit segments in the closed unit square, and there is a region admitting a countably infinite maximal such collection. Here a segment is a rotated and translated copy of $(0,1)$. An endpoint of one segment may lie in the interior of another: the endpoint itself is absent from the former. The stronger convention allowing closed segments to meet only at common endpoints is a different question.

Five interior segments.

Set \[ c=\cos15^\circ=\frac{\sqrt6+\sqrt2}{4},\qquad s=\sin15^\circ=\frac{\sqrt6-\sqrt2}{4}. \] Let $O=(0,0)$, $V=(c,s)$, $W=(s,c)$, and write $\sigma(x,y)=(y,x)$. The triangle $OVW$ is equilateral of side one, since $c^2+s^2=1$ and $c-s=1/\sqrt2$. We construct a unit chord $XY$ through $V$, with \[ X=(x,0),\qquad Y=(1,y),\qquad y=\frac{s(1-x)}{c-x}. \tag{1} \] There is a choice with \[ .95<x<.955,\qquad .8<y<1,\qquad (1-x)^2+y^2=1. \tag{2} \] Here are explicit existence bounds. The radical formulas give $.965<c<.966$ and $.258<s<.260$. At $x=.95$, equation $1$ gives \[ .8<y<\frac{13}{15}, \qquad (1-x)^2+y^2<1. \] At $x_0=(c-s)/(1-s)$ it gives $y=1$ and hence $(1-x_0)^2+y^2>1$. The same bounds give $.95<x_0<.955<c$. The function $y(x)$ is continuous and strictly increasing on this interval, so the intermediate value theorem supplies $2$. Moreover $X,V,Y$ are collinear, with $V$ strictly between $X,Y$.

Take the following five open segments: \[ OV,\quad OW,\quad VW,\quad XY,\quad \sigma X\,\sigma Y. \tag{3} \] All have length one and lie in the open square. They are pairwise disjoint. The chord $XY$ meets the first and third supporting sides only at their omitted endpoint $V$, and is separated from $OW$ by the bounds $x>.95>s$; the reflected statements hold for the other chord. The two chords themselves do not meet in the square: their supporting lines meet on the diagonal at coordinate \[ z=\frac{xy}{x+y-1}>1, \] because $x+y-1>0$ and $xy-(x+y-1)=(1-x)(1-y)>0$.

Why these segments block the interior.

Their arrangement partitions the square into the following six closed polygonal cells, with disjoint interiors: \[ \begin{array}{lll} \operatorname{conv}(O,V,W),& \operatorname{conv}(O,V,X),& \operatorname{conv}(O,W,\sigma X),\\ \operatorname{conv}(X,(1,0),Y),& \operatorname{conv}(\sigma X,(0,1),\sigma Y),& \operatorname{conv}(V,Y,(1,1),\sigma Y,W). \end{array} \tag{4} \] This partition follows directly by first cutting along the two disjoint boundary-to-boundary chords, then along the three equilateral-triangle sides in the remaining region. The only vertices inside the open square are $V,W$; crucially, $V$ belongs to the open chord $XY$, and $W$ to its reflected chord. Thus every boundary between cells inside the open square is covered by $3$, including its interior vertices.

Each cell has diameter one, and every pair at distance one consists of endpoints of a segment already in $3$. For the first cell, these are its three sides. For the next four triangular cells, respectively the unique longest sides are $OV,OW,XY,\sigma X\,\sigma Y$: their other sides are strictly shorter than one by $2$ and the bounds on $c,s$. For the last pentagon the only unit vertex pair is $V,W$. All other vertex pairs are strictly closer: the coordinate bounds give, for the pairs involving $V$ or $W$, squared distances at most \[ .17^2+.75^2<1; \] the pairs among $Y,(1,1),\sigma Y$ have squared distances at most $2(.2)^2<1$. These bounds also cover the reflected pairs.

We use the elementary fact that the diameter of a finite convex hull is attained between vertices. More precisely, any unit-distance pair in a polygon whose vertex distances are all at most one must itself be a pair of vertices: write both points as convex combinations of vertices, apply the triangle inequality to their difference, and use strict convexity of the Euclidean norm in the equality case. Equality forces all contributing differences to be the same unit vector, so neither convex combination can use two different vertices.

An open unit segment contained in the open square and avoiding $3$ must lie in a single cell interior, since it is connected and cannot cross a covered cell boundary. Its closure would be a unit segment in that cell. By the preceding diameter analysis it would have to coincide with one of the existing sides, contradicting avoidance. Therefore $3$ blocks every unit segment in the open square.

The finite maximal family in the closed square.

Add the four open boundary sides of the square to $3$, obtaining nine disjoint open unit segments. An open segment in the closed square either lies on a boundary side or has all its points in the open square. Indeed, if an interior point of a straight segment has a coordinate equal to $0$ or $1$, convexity forces that coordinate to be constant along the segment. In the boundary case a unit segment equals the corresponding open side; in the other case the preceding argument blocks it. No further disjoint open unit segment can therefore be added.

A countably infinite maximal family in a convex region.

Take the closed half-strip \[ R=[0,\infty)\times[0,1]. \] In each unit square $[j,j+1]\times[0,1]$, $j=0,1,2,\ldots$, put a translated copy of the nine-segment construction. Retain each shared vertical boundary segment only once. The result is a countably infinite pairwise disjoint family of unit segments. Any proposed further open unit segment in $R$ lying on $y=0$ or $y=1$ intersects the existing horizontal boundary segments, which leave only the integer vertices uncovered. Otherwise every point of the proposed segment has $0<y<1$. It cannot cross any line $x=j$, because the entire portion with $0<y<1$ is an existing vertical segment. Connectedness therefore confines it to a single unit square, where the finite construction already blocks it. The infinite family is maximal. The region used is connected and convex; it is not necessary for the second question to use a bounded region.

Attribution and assessment.

The finite construction reconstructs the equilateral-triangle example recorded with the problem, using the coordinates also present in the public formalization. The half-strip deduction proves the second question as stated. Alexeev and Howe have stronger constructions with a countably infinite maximal family inside the unit square itself; that stronger assertion is not needed here. Sources checked 24 September 2026: https://www.erdosproblems.com/1071 and the public proof source https://github.com/plby/lean-proofs/blob/main/src/latest/ErdosProblems/Erdos1071.lean. The geometric argument above is supplied directly; the Lean source was inspected but not locally compiled. No novelty is claimed.

Completion estimate.

COMPLETION: 100%

FORMAL RESTATEMENT

Question 1: Does there exist a finite set $\mathcal{S}$ of unit line segments contained in the unit square $[0,1]^2$ such that no two segments in $\mathcal{S}$ intersect and $\mathcal{S}$ is maximal with respect to this property (i.e. no additional unit segment contained in $[0,1]^2$ can be added while preserving pairwise disjointness)?

Question 2: Does there exist a region $R\subset\mathbb{R}^2$ such that $R$ admits a maximal set of pairwise disjoint unit segments that is countably infinite?

Ambiguities to fix.

• "Line segment" could mean closed $[0,1]$-segment or open $(0,1)$-segment; this affects whether two segments sharing an endpoint are considered intersecting.
• "Region" might mean an arbitrary subset, or might implicitly mean a connected set with nonempty interior.

The problem file states that the answer to Question 1 is "yes" and refers to examples in [Er87b]. I will therefore take as the intended meaning the one under which such examples exist (typically: open segments, or equivalently closed segments allowed to meet only at endpoints).

QUICK LITERATURE/CONTEXT CHECK

The problem file asserts: "The answer to the first question is yes. There are two examples in Erdos [Er87b]." No status is given for Question 2. I will not import any details of those examples; instead I record elementary reductions and a trivial resolution of Question 2 under a weak interpretation of "region".

ATTACK PLAN

$1$ Clarify the open-vs-closed segment model and show the equivalence between "open segments disjoint" and "closed segments intersect only at endpoints". $2$ For Question 2, construct an explicit region $R$ that forces maximal families to be countable (or build a region in which a chosen countable family is maximal) under the weakest reading. $3$ Explain precisely what additional constraints on $R$ would make Question 2 nontrivial.

WORK

Lemma 1071.1 (open-segment model equals endpoint-touching model). Let $s_1,s_2$ be two unit segments in the plane. Let $s_i^{\circ}$ be the corresponding open segments (endpoints removed) and $\overline{s_i}$ the corresponding closed segments (endpoints included). Then \[ s_1^{\circ}\cap s_2^{\circ}=\emptyset \quad\Longleftrightarrow\quad \overline{s_1}\cap \overline{s_2}\subseteq \partial\overline{s_1}\cap\partial\overline{s_2} \] i.e. the open segments are disjoint iff the closed segments intersect, if at all, only at endpoints.

Proof. ($\Rightarrow$) If $s_1^{\circ}\cap s_2^{\circ}=\emptyset$ but $\overline{s_1}\cap\overline{s_2}$ contains a point $x$ that is not an endpoint of $s_1$ or not an endpoint of $s_2$, then $x$ lies in the interior of at least one of the closed segments. If $x$ lies in the interior of both, then $x\in s_1^{\circ}\cap s_2^{\circ}$, contradiction. If $x$ lies in the interior of $\overline{s_1}$ but is an endpoint of $\overline{s_2}$, then $x\notin s_2^{\circ}$ so this does not contradict disjointness; in this case the intersection point is an endpoint. Symmetrically, any intersection can only occur at endpoints.

($\Leftarrow$) If $\overline{s_1}\cap\overline{s_2}$ is contained in endpoints, then in particular there is no point that lies in the interior of both segments, so $s_1^{\circ}\cap s_2^{\circ}=\emptyset$. $\square$

Lemma 1071.2 (Question 2 is trivial for disconnected/degenerate "regions"). If "region" is interpreted as an arbitrary subset of $\mathbb{R}^2$ with no connectedness or interior requirement, then the answer to Question 2 is yes.

Proof (explicit construction). Let $R$ be the disjoint union of countably many translated copies of the unit square, \[ R := \bigsqcup_{n\ge 1} \left([0,1]^2 + (3n,0)\right). \] The components are pairwise at distance at least $2$, so any unit segment contained in $R$ lies entirely within a single component. In each component, choose a maximal family of disjoint unit segments (existence follows from Zorn's lemma: order families by inclusion; any chain has a union that is still pairwise disjoint). By Question 1 (as stated in the problem file) we may even choose a finite maximal family in each component. Let $\mathcal{S}$ be the union over all components. Then $\mathcal{S}$ is a countable union of finite sets, hence countable; and it is maximal in $R$ because any additional unit segment would lie in some component, contradicting maximality in that component. Thus $R$ admits a countably infinite maximal family. $\square$

FAST REALITY CHECK (sanity).

• With $\mathcal{S}=\emptyset$ in the unit square, the family is not maximal (one can add a unit segment).
• Lemma 1071.2 shows Question 2 depends strongly on what constraints are placed on $R$; without connectedness/interior constraints it has a straightforward affirmative construction.

VERIFICATION

Lemma 1071.1 is a direct set-theoretic unpacking of "open segment" vs "closed segment". Lemma 1071.2 is verified by the distance-separation condition between components, ensuring any unit segment lies in one component, and by the Zorn maximality argument inside each component.

FINAL

UNRESOLVED.

(i) Strongest proved partial result: Question 2 has a trivial affirmative answer if "region" is allowed to be disconnected (Lemma 1071.2). The distinction between open segments and allowing endpoint-touching is formally equivalent (Lemma 1071.1). The problem file asserts that Question 1 has an affirmative answer via explicit constructions.

(ii) First gap (crisp statement): Formulate and resolve Question 2 under a nontrivial notion of "region" (e.g. $R$ connected with nonempty interior, or $R$ convex/bounded), and under a fixed convention about whether endpoints may coincide.

(iii) Top 3 next moves (concrete lemmas/computations):

1. Pin down the intended definitions by extracting from [Er87b] whether segments are open or whether endpoint-touching is allowed; then restate both questions with that convention.
2. For connected regions $R$ with interior, attempt to construct a "corridor" region that forces any unit segment to lie in one of countably many "cells", yielding a countable maximal family.
3. Conversely, attempt to prove an impossibility result: under natural regularity assumptions on $R$ (e.g. convex), show that every maximal family must be uncountable or must have accumulation structure incompatible with countability.

(iv) Minimal counterexample structure: For a negative answer to Question 2 under a strengthened definition of region, one would need to show that in every such region $R$ any maximal family of disjoint unit segments is either finite or uncountable (or that any countable disjoint family can always be extended). For a positive answer, one needs an explicit region $R$ and a countably infinite disjoint family $\mathcal{S}$ such that every unit segment contained in $R$ intersects at least one member of $\mathcal{S}$.