Does there exist an uncountable set of pairwise disjoint crosses in the plane?

Question

Does there exist an uncountable set of pairwise disjoint crosses in the plane?

We define a cross to be two open line segments with one point of intersection.

My initial thought is yes:

For example pick any real $r$ on the x-axis and consider the line $x=r$. It seems as though we should always be able to find space on this line to draw a cross centred on it (as we can always draw finite sized crosses and move up and down the line to find space).

We can therefore find a bijection between $ \mathbb{R} $ and the set of crosses by mapping the x coordinate of each crosses centre to the cross itself.

Is this correct?

Answer 1

Supposing $C$ is a set of uncountably many disjoint crosses in the plane, let $C_\epsilon$ be the subset of $C$ consisting of crosses of size at least $\epsilon$. By "size", I mean the length of the shortest "leg" of the cross.

Now, since $C$ is uncountable, for some $\epsilon$ we must have $C_\epsilon$ uncountable as well (since $C=\bigcup_{n\in\mathbb{N}}C_{1\over n}$).

Now it's easy to show that any cross of size $\epsilon$ can't have another cross of size $\ge\epsilon$ "too close" - specifically, the centers of two such crosses must be at least $\epsilon$ distance apart. So each $C_\epsilon$ is countable. But this contradicts the fact above.

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You wrote

It seems as though we should always be able to find space on this line to draw a cross centred on it (as we can always draw finite sized crosses and move up and down the line to find space).

If we unpack this, what you're describing - correctly! - is a process to "add one more cross" assuming we've only added finitely many crosses so far. The problem is, this only gets you countably many crosses!

At this point it's a good idea to consider a "maximal" family of crosses - that is, countably many crosses such that no more crosses can be added without overlapping.

This is actually pretty easy to do, if we view a cross as not containing its four "endpoints" (so, two crosses are allowed to "almost touch"); I leave it as an exercise to modify this counterexample so that it works for crosses with endpoints.

Namely, let $L_1$ be the set of points in the plane both of whose coordinates are integers, with the same parity (so $(0, 0)$ and $(1, 1)$ are good, but $(0, 1)$ isn't); and, having defined $L_n$, let $L_{n+1}$ be the set of points in the plane, both of whose coordinates have the form "$k\over {2^{n}}$" for some odd integer $k$. Intuitively, each $L_n$ is a grid of points cutting the plane into squares of size $2^{1-n}$, and the points in $L_{n+1}$ are the centers of those squares.

Let $L=\bigcup L_n$, and note that each $x\in L$ is in exactly one $L_n$; let that $n$ be the rank of $x$, $n=r(x)$. Now consider, for each $x\in L$, the cross centered at $x$, with legs going vertically and horizontally, each of length $2^{r(x)-1}$. The family $C$ of such crosses is maximal.

Answer 2

If $S$ is a set of such crosses, let $S_{\epsilon}\subset S$ be the subset of crosses which smallest side is $\ge \epsilon$. Around each center of such $\epsilon$-cross, there is a square of side $\epsilon$ in which there is no other center of such $\epsilon$-cross. It means that if a rectangle region $R$ of the plane with sides parallel to the axes contains $N$ crosses in $S_{\epsilon}$, it contains $N$ non overlapping squares of measure $\epsilon^2/4$, and the measure of this rectangle is $\ge N \epsilon^2/4$. In particular a rectangle contains only a finite number of elements of $S_{\epsilon}$ and as $\epsilon$ is arbitrary, it contains only a countable number of crosses. By letting the rectangle grow and cover the whole plane, the result follows:

there is no uncountable set of crosses in the plane.