Can I cover a square with many line segments?

Question

Not sure If I've chosen the tags correctly.

Anyway, is it possible to obtain a unit square with enough line segments oriented vertically, placed next to each other? We know that a unit square has area $1$. I don't know anything about measure theory, except for the fact that measure is used to assign a number representing 'size' of a given set.

My intuition is that a line segment (of length $1$) has area $0$, then no matter how many line segments we take, their total area will be $0$. There exists no point at which adding yet one more line segmnets increases the area, so we will never reach area $1$ or a unit square. However, I've heard that in the limit, by taking infinitely many line segments, we will eventually have posiitve area. How can it be? If I take union of $100$ empty sets (buckets with apples let's say), I still have an empty set, so even with infinitely many empty buckets I don't have more that I had before.

How is the area defined anyway? What does it mean for a square to have area $1$ if this is supposed to be possible to obtain a square from line seqments?

Answer 1

The measure is $\sigma$-subadditive: in other words, we impose $$meas\left(\bigcup_{n\in\Bbb N}A_n\right)\le\sum_{n\in\Bbb N}meas(A_n).$$

However, to cover a square with line segments you will need more than countable number of segments, so this case is not covered by the properties of measure.

Answer 2

Obviously, the square $[0,1]$ is the union of uncountably infinitely many line segments, since $$[0,1] = \bigcup_{x\in[0,1]}(\{x\}\times[0,1])$$

A more interesting question is "can I do it with fewer line segments?"

What you are referring to as "area" is mathematically formalized into measure theory, as you correctly suspect. The "standard" measure of sets in $\mathbb R^n$ (the measure in which $[0,a]^n$ has a "volume" of $a^n$) is called the Lebesgue measure.

In this measure, all lines in $\mathbb R^2$ have a measure of $0$ (i.e., their "surface area is $0$").

An important fact from measure theory is this:

If $A_1,A_2,\dots$ are disjoint measurable sets with measures $m(A_1), m(A_2),\dots$, then the measure of ther union is the sum of their measures, i.e. $$m\left(\bigcup_{i=1}^\infty\right)=\sum_{i=1}^\infty m(A_i)$$

From this, and some measure theory facts, it also follows that if $A_1,\dots,$ are not disjoint, then their measure is smaller or equal to the sum of their measures.

Furthermore, any finite union of lines is still a measurable set in $\mathbb R^2$ and, because of what I wrote above, its measure is less than or equal to the sum of the individual measures, which is $0$, so the measure of any finite union of lines must be $0$. In fact, the same is true for countably infinite many lines, so you can see that the square (having measure $1$) cannot be the union of a finite or countably infinite number of lines (because that union has measure $0$).

That said, there do exist certain ways in which you can construct a continuous surjective mapping from $[0,1]$ to $[0,1]^2$, using so called "space filling curves".

In summary:

• Can the unit square be written as a countable union of straight lines? No.
• Can the unit square be written as a union of straight lines? Yes, but it must be a union of uncountably many lines.
• Can the unit square be covered with a curve? Yes.

Answer 3

Consider the square $[0,1]\times[0,1]$ in $\mathbb R^2$. That is, the square consists of all points whose coordinates $(x,y)$ satisfy $0\leq x \leq 1$ and $0\leq y \leq 1$.

Now consider the coordinates of any particular point in the square; that is, take any point $(x_0,y_0) \in [0,1]\times[0,1]$. Consider the line segment $L$ described by the equations $y = y_0$, $0\leq x \leq1$. Clearly $(x_0,y_0) \in L$.

To do this without coordinates, choose one side of the square as a reference segment. Then for any point in the square, construct the line through that point parallel to the reference segment, and take the line segment defined by the intersection of that line with the square (that is, all points on the line that are not outside the square).

Every point in the square belongs to such a line segment. So if we take the set of all such line segments, it covers the square.

There is no contradiction between this and your statement that at no point will adding another line create positive measure. Even in the limit of a process that adds lines one at a time, you still have only a countable number of lines. The set of line segments defined by the procedures above is an uncountably infinite set of line segments, which is an entirely different thing. It is much the same question as how a line segment can have positive measure when individual points have zero measure.