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A subset of ${\mathbb R}^2$ is open in the usual topology iff it is a (not necessarily disjoint) union of open disks. It is well-known that the plane is connected, so that there is no nontrivial partition of ${\mathbb R}^2$ into open sets.

Now say that a subset of ${\mathbb R}^2$ is weakly open iff it is a (not necessarily disjoint) union of "open disks in dimension 1", i.e. a union of open segments). With this definition, there is an obvious nontrivial partition of ${\mathbb R}^2$ into weakly open subsets : if $\cal D$ is the set of all lines parallel to some direction, any nontrivial partition of $\cal D$ will yield a partition of ${\mathbb R}^2$ into weakly open subsets. Are there any other such partitions ?

UPDATE 06/11/2012 16:30 Since the original question was quickly answered, I ask now : what about partitions into closed, weakly open subsets ?

Sub-question : if a subset is both closed and weakly open, does it necessarily contain a straight line ?

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5 Answers 5

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Lemma: Let $A$ be a closed, weakly open set, $p$, $q$ and $r$ three distinct points, $p \notin A$, and suppose the open line segments $pq$ and $pr$ are disjoint from $A$. Then the open triangle $T$ with vertices $p$,$q$,$r$ is disjoint from $A$. Moreover, the open segment $qr$ is either disjoint from $A$ or contained in $A$.

Proof: If $T$ intersects $A$, then there is a point $y$ of $T \cap A$ that is as far as possible from the line $L$ through $qr$. $y$ must be contained in a maximal open segment $S$ contained in $A$, and since no point of $S$ can be farther from $L$ than $y$, $S$ must be parallel to $L$. But since $S$ can't intersect the open segments $pq$ and $qr$, it must have an endpoint $y' \in T$. This endpoint is also in $A$, and also maximizes the distance from $L$, but any open segment containing $y'$ and contained in $T$ must not be parallel to $L$, contradiction.

If some point $z$ of the open segment $qr$ is in $A$, it is contained in an open segment contained in $A$, but such a segment can't intersect $T$ so it must be on the line $L$. But if not all of $qr$ is in $A$, that segment has an endpoint $z' \in qr$, and then an open segment containing $z'$ and contained in $A$ must not be on $L$, contradiction. That concludes the proof of the Lemma.

Theorem: Suppose $A$ and $B$ are disjoint nonempty, closed, weakly open sets. Then there is some line $L$ disjoint from $A \cup B$, such that both $A$ and $B$ contain translates of $L$.

Proof: Along a line segment from a point of $A$ to a point of $B$ there must be an open interval disjoint from $A \cup B$ that has one endpoint in $A$ and the other in $B$. Let $p$ be a member of such an interval. Let $C_A$ be the set of points $s$ of the unit circle $C$ such that for some $t \in (0,\infty)$, $p + t s \in A$ and for all $t' \in (0,t)$, $p + t' s \notin B$. Similarly define $C_B$, interchanging $A$ and $B$. It is not hard to show that $C_A$ and $C_B$ are open and nonempty. So there must be $s_1$ and $s_2$ in $C$ that are neither in $C_A$ nor $C_B$, dividing $C$ into arcs $C_1$ and $C_2$ where $C_A$ intersects $C_1$ and $C_B$ intersects $C_2$. But by Lemma 1, neither of those arcs can subtend an angle less than $\pi$. So both arcs must subtend $\pi$, i.e. $s_2 = -s_1$, and the line $L$ through $p$ in direction $s_1$ is disjoint from $A \cup B$.

Now let $q$ be a point of $A$ such that the open interval $pq$ is disjoint from $A$, and let $L'$ be the line through $q$ parallel to $L$. Let $U$ be the open strip between $L$ and $L'$. $U$ is the union of the open triangles with vertices $p$, $q$ and points $r$ of $L$. Since the open segments $pq$ and $pr$ are disjoint from $A$, the Lemma says $U$ is disjoint from $A$. Now there is an open segment containing $q$ and contained in $A$, and since this segment can't intersect $U$ it must be contained in $L'$. Applying the second part of the Lemma to a triangle with vertices $p$, $q$ and another point of $L'$, we see that $L' \subseteq A$.

Corollary: In a partition of ${\mathbb R}^2$ into closed, weakly open sets, the boundary of each member of the partition consists of parallel straight lines.

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Thanks, I'll fix it. –  Robert Israel Jun 14 '12 at 7:22
    
I think you mean to pick $y$ in *closure*$(T) \cap A$, since $T$ itself is open. –  hardmath Jun 14 '12 at 12:00
    
@RobertIsrael : thanks very much for your answer. How do you deduce from lemma 1 that "neither of those arcs can subtend an angle less than $\pi$" ? –  Ewan Delanoy Jun 14 '12 at 14:54
    
@hardmath: note that the edges $pq$ and $pr$ don't intersect $A$, so if $y$ is in the closure of $T$ (and is not on $L$) it is in $T$. –  Robert Israel Jun 14 '12 at 16:51
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@EwanDelanoy: because $s_1 \notin C_A$. –  Robert Israel Jun 14 '12 at 19:41

Your definition gives the plane the discrete topology: each singleton is the intersection of two open segments. Thus, every partition of the plane qualifies.

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I was wrong to ask the question in a topological setting, where it is indeed trivial as you explain. I edited my original post to make the question more meaningful. –  Ewan Delanoy Jun 11 '12 at 10:25

The "weakly open" intervals and lines need not all be parallel, to suggest alternative examples. For example take a countable set of lines parallel to the x-axis, $y = k$ for $k$ each integer, and fill in the regions between each consecutive such pair with unit length open intervals parallel to the y-axis.

With a little ingenuity it seems the lines can further be broken up in such a way that the partition involves only finite length intervals. For each "even" line $y = 2k$, remove the points $(2m,2k)$ and combine them with the intervals above and below that parallel the y-axis. Similarly with "odd" lines $y = 2k+1$ using points $(2m+1,2k+1)$. Thus no interval in the partition will have length greater than two.


Added: Here's another example of a closed "weakly open" set that contains no infinite segment. Any bounded convex polygon, whether including or excluding its interior, with extremal points (vertices) removed is a disjoint union of finite length open segments. For simplicity we take a triangle without vertices and countably many translates of it, positioned so that the missing vertices coincide with midpoints of neighboring edges:

plane triangulation

Note that if the triangles were equilateral and without interior, then all the constituent open segments would have equal length.

As a replacement conjecture, it seems that closed "weakly open" planar subsets are unbounded, even if only finite segments are contained in one.

The definition of "weakly open" sets seems to need tightening. Were it not clarified by the "open segment" phrase that follows, the "union of open disks in lower dimension" might invite single points as (homeomorphic to) open disks in $\mathbb{R}^0$. Another aspect is whether "union" or "disjoint union" was intended here. So far all the examples adduced satisfy the stricter qualification of a disjoint union of open segments, and I'm not sure if the two approaches are equivalent.

However the biggest improvement would come from choosing different terminology. "Weakly open" is inapt because these sets do not provide an open basis other than for the discrete topology, as Brian M. Scott pointed out. My misgivings about this terminology are enlarged by the juxtaposition of (topologically) closed with "weakly open".

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I sure understand your criticisms, but what terminology would you suggest instead of "weakly open" ? –  Ewan Delanoy Jun 14 '12 at 4:20
    
I updated the OP in order to clarify the points you mentioned. –  Ewan Delanoy Jun 14 '12 at 4:21
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"Closed weakly open subsets are unbounded" is actually a fact that's easy enough to prove. If such a set $A$ were bounded, then $b={\sf sup}(y, (x,y) \in A)$ would be finite. Let $D$ be the horizontal line $\lbrace (x,y) | y=b \rbrace$, then $A$ is below $D$. And $A'=A\cap D$ is nonempty since $A$ is closed. For any point $p \in A'$, there is an open segment $s$ contained in $A$ and containing $p$, and $s$ is necessarily horizontal. So we see that $A'$ is nonempty and both open and closed in $D$, showing $A'=D$ : then $A$ contains a straight line and is therefore unbounded, qed. –  Ewan Delanoy Jun 14 '12 at 4:34
    
Ewan Delanoy: Your technique appears to prove that for closed weakly open $A$ and line $L$ disjoint from $A$, $A$ contains a translate of $L$. This looks similar to what @RobertIsrael proves in his post. As far as novel terminology goes I'm prepared to suggest Delanopen ! –  hardmath Jun 14 '12 at 7:18
    
I'll look at your suggestion on translates of a bounding line. If it turns out be right, it will lead to a simpler proof than RobertIsrael's. –  Ewan Delanoy Jun 14 '12 at 14:57

Here is a counterexample to your subquestion.

I don't know why the image isn't working. Anyway, take a square and extend the left side upward indefinitely, extend the top to the right, extend the right side downward and extend the bottom side to the left. This is a weakly open closed set with no line. (Although it does have 4 rays.)

enter image description here

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Thanks for your answer. It can be modified to produce a "saw-like" example, without half lines (take an infinite path made of equally sized horizontal and vertical segments, and add small squares at each corner to ensure the set is weakly open). –  Ewan Delanoy Jun 13 '12 at 14:05
    
By the way, the image works on my computer (but not on yours it seems, how ironic!) –  Ewan Delanoy Jun 13 '12 at 14:09
    
Taken as four finite segments, it appears the design above can be stacked in staggered rows and columns in an unbounded way that nevertheless contains no lines (or half-lines). –  hardmath Jun 13 '12 at 14:09
    
@Ewan: It's working now. Not sure what happened earlier. –  Grumpy Parsnip Jun 13 '12 at 19:18
    
+1 This is nice. I was trying to construct something like the closure of the union of blue tiles in a Penrose tiling here. But can't prove that it works, and this is simpler to boot. –  Jyrki Lahtonen Jun 13 '12 at 19:50

Here's another example of a closed weakly open subset that doesn't contain a straight line.

Start with the plane $\Bbb R^2$ and then for each Eisenstein integer, i.e. a point in $\Bbb R^2$ of the form $a(1,0)+b(\cos\frac{2\pi i}{3},\sin\frac{2\pi i}{3})$, where $a,b\in\Bbb Z$, remove an open disk $D_{a,b}$ centered at that point with diameter $1$. Here's a picture:

enter image description here

The resulting set $A=\Bbb R^2\setminus\bigcup_{a,b\in\Bbb Z}D_{a,b}$ is closed of course, since it is a complement of a union of open disks. But it is also weakly open: to see this, it is enough to find for each $x\in A$ an open segment $S_x\subseteq A$ that contains $x$. (As then $A=\bigcup_{x\in A}S_x$ follows.)

If $x$ lies in he interior of $A$, then there is an open set $U\subseteq A$ such that $x\in A$, so such a segment does exist. If $x$ lies in the boundary of $A$, it must lie in either one or two of the boundary circles $C_{a,b} = \partial D_{a,b}$. In either of these two cases, it is possible to choose an open segment that is tangent to this circle at $x$ and small enough to not intersect any of the open disks.

Showing that this set does not contain any straight lines amounts to showing that every such line will have to intersect one of the disks. To do this, suppose there is a line that interesects no such disk. Such a line will have to contain one of the points where two circles kiss (i.e. an element of some set $C_{a_1,b_1}\cap C_{a_2,b_2}$), since removing all such points from $A$ disconnects $A$ into bounded components. The line will have to be tangent to both such circles and cut a third one in half, thus (after filling in all the details) proving the claim.

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