Let $S=[0,1]\times [0,1]$ be the closed unit square. Suppose we label its four edges in cyclic order as $E_1,E_2,E_3,E_4$ so that $E_1$ is parallel to $E_3$ and $E_2$ is parallel to $E_4$.

Now, choose some partition $\{A,B\}$ of $S$. Must at least one of the following alternatives hold?

  1. There is a quasicomponent of $A$ intersecting both $E_1$ and $E_3$.

  2. There is a quasicomponent of $B$ intersecting both $E_2$ and $E_4$.

This statement is analogous to the (discrete) statement that a game of hex always has a winner. It seems that the continuous statement I'm asking about is intuitively correct (especially if we were to add a requirement that $A$ is closed or otherwise well-behaved), but it seems unclear how to handle the generality of this statement. Moreover, things do seem to break down: this question shows that both conditions could hold, as well as that neither could hold if we replace "quasiconnected" by "path connected". The comments give a counterexample for "connected" as well.

I'd primarily like to apply this in the case where $A$ and $B$ are reasonably well behaved (something comparable with $A$ being closed), but I wasn't able to quickly cook up any counterexamples even if we try to bring the axiom of choice into this, so I've been thinking that maybe such assumptions are unnecessary.

  • $\begingroup$ You need more than just $A$ closed, I think. Let $A=E_2$ and $B=S\setminus A$, and neither condition holds. $\endgroup$
    – Steve Kass
    Jun 18, 2018 at 20:57
  • $\begingroup$ @SteveKass $A$ would connect $E_1$ and $E_3$ in that case. (I'm considering each edge to be a closed set - i.e. $E_1$ and $E_3$ each share a point with $E_2$. I can edit that to be more explicit, if it's ambiguous) $\endgroup$ Jun 18, 2018 at 21:00
  • $\begingroup$ @Steve Kass wouldn't $E_4$ be a connected component intersecting $E_1$ and $E_3$ in your example? $\endgroup$ Jun 18, 2018 at 21:01
  • $\begingroup$ Ah, you’re right. (Though I think you mean $E_2$, not $E_4$. In my example, $E_4$ is not part of $A$, but $E_2$ is. I don’t think you need to be more explicit, and these comments should keep others from making the mistake I made. $\endgroup$
    – Steve Kass
    Jun 18, 2018 at 21:05
  • 2
    $\begingroup$ Let $Q_a$ denote the boundary (i.e. union of the four closed edges) of the square $[a,1-a]\times[a,1-a]$. Define $A=(\bigcup_{q\in[0,\frac12]\cap\mathbb Q} Q_a)\setminus\{(0,0),(0,1),(1,0),(1,1)\}$ and $B=S\setminus A$. I believe this is a counterexample: the components of the two sets are the sets $Q_a$ for $a\in(0,\frac12]$, the open edges of $S$ and the vertices of $S$. The statement might still be true if you replace "components" by "quasicomponents". $\endgroup$
    – Dejan Govc
    Jun 18, 2018 at 21:11

1 Answer 1


Here is a counterexample. Begin by partitioning the square $S$ into parallel diagonal line segments $$D_r=\{(x,y)\in S\mid y = x+r\}$$ for $r\in[-1,1]$. Split $D_0$ into two parts $D_0'=\{(0,0)\}$ and $D_0''=D_0\setminus D_0'$. The desired partition will consist of unions of some of these sets. Let $P=[-1,1]\setminus\mathbb Q$ and $Q=\left([-1,0)\cup(0,1]\right)\cap\mathbb Q$ and define $$A=D_0'\cup\bigcup_{r\in P}D_r,\qquad B=D_0''\cup\bigcup_{r\in Q}D_r.$$ These sets have the following properties, which should be straightforward to verify:

  1. $A\cup B=S$ and $A\cap B=\emptyset$,
  2. the (quasi)components of $A$ are the singleton $D_0'$ and the closed segments $D_r$ with $r\in P$,
  3. the (quasi)components of $B$ are the "half-open" diagonal segment $D_0''$ and the closed segments $D_r$ with $r\in Q$.

In particular, none of the (quasi)components connect a pair of opposite edges.

For slightly nicer $A$ and $B$, you can take $Q=\{\pm\frac1n\mid n\in\mathbb N\}$ and $P=([-1,0)\cup(0,1])\setminus Q$ instead. In that case both $A$ and $B$ are $F_\sigma$ as well as $G_\delta$ sets. In fact, $A$ is an open set plus a point and $B$ is a closed set minus a point.


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