Is the area of a line really $0$

Question

Let's take a square with side $A$. The area of this is defined as $A\times A$.
The way I explained this to myself is by reapeatedly deviding the square till I reached a single line.
Then stacking up lines one on top of the other $A$ times.

But if the area of a line was really $0$ no matter how many lines were stacked up, they would always have an area of $0$.(would exist in one dimension only)

Another way to see this is defining a function $V(F)$ which gives the volume of two dimensional figure $F$.
If this volume became $0$ , no matter how many such Figures I stack up on each other they would never have a volume.

When this is combined with physics it makes a little more sense.
Even if I were to chop up a cube infinitely I could never reduce its height to less than the Planck length. This means that the volume would be $Area \times ℓP$ (Planck length)

So the question is for a three dimensional object to exist must every two dimensional object have a non zero volume?

Answer 1

You've touched on one of the reasons why a careful axiomatization of measure theory is important. How you answer your question depends on how you are defining your notion of "volume"/"area" (what we generally call "measure").

• Let's say in your setup you give line segments of length $1$ a nonzero volume $c$. And let's suppose that your notion of volume is additive (in the sense that the volume of the union of disjoint regions is equal to the sum of the volumes of the individual regions). One might think of a $1 \times 1$ square as stacking infinitely many (actually, an uncountable infinitude of) copies of this line segment. Then your $1 \times 1$ square would have infinite volume since each segment has volume $c$ and you are combining infinitely many of these segments.
• On the other hand, if we say that these line segments have zero volume, then this seems to suggest that the $1 \times 1$ square has zero volume as well, since it is made up of infinitely many segments each having zero volume.
• The "solution" to this paradox that is established in measure theory is to restrict the additivity property to only countable unions. That is, if $E_1, E_2,\ldots$ are a countable collection of disjoint sets, then $\text{volume}\left(\bigcup_{i =1}^\infty E_i \right) = \sum_{i=1}^\infty \text{volume}(E_i)$; we do not require that this property holds for uncountable collections of disjoint sets. Then, in the case of Lebesgue measure in two-dimensional space, we assign "two-dimensional" regions volume in the usual sense ("area"), and we assign "one-dimensional" regions (like line segments) zero volume. The "paradox" in the previous bullet-point does not occur simply because we do not require our notion of volume to be additive across uncountable collections of disjoint sets (like stacking uncountably many line segments to make a square). This is just a description of Lebesgue measure, however; you can assign measures in other non-intuitive ways.