Is the length of a segment between 0 and 1 exactly 1?

What is the length of the line segment between points A and B on a number line, where A = 0 and B = 1? Is it exactly 1? Perhaps I am thinking about it in an incorrect manner, but it seems to me that 1 is distance from A to B, not taking into account that infinitesimally small width of point A and B. Or is it that the widths of A and B are so infinitely small that they're essentially 0?

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There are several ways to "measure" sets in $\mathbb{R}$, but in the traditional sense it's just as you say: the points are so small that they measure $0$. That kind of stuff is studied in Measure Theory. – Git Gud Jan 20 '13 at 21:58
The width of the point A is indeed infinitesimal. However, zero is the the only infinitesimal in the real numbers. – Hurkyl Jan 20 '13 at 22:06

Not just essentially $0$: they are $0$, as the linear size of a set of real numbers is normally measured. The closed interval $[1,1]$ is precisely the set $\{1\}$ whose only member is the number $1$, so the ‘width’ of the point $1$, if you want to call it that, is the length of the closed interval $[1,1]$. But for $a\le b$ the length of the interval $[a,b]$ is $b-a$, so the length of the interval $[1,1]$ is $1-1=0$.

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The notion of length is inherently $1$-dimensional. As such the length of any point is precisely equal to $0$. Similarly, when considering area, which is inherently a 2-dimensional creature, the area of any line segment is precisely $0$. So, the area of a square $[0,1]^2$ is precisely $1$ regardless if you measure the sides or not. Similarly in ever higher dimension.

Curiously (as I don't know if this is at all useful, and this is certainly not how it is done in measure theory) for simple shapes you can measure the total length (or area or volume) as a linear combination of measures of various dimensions and this gives plausible results in computations (at least some times). For instance, let us introduce a notion of the content of a single point (a $0$-dimensional notion of volume). The content of any single point is $1P$ (the 'P' reminds us that this is the content of a point).

Now, if we agree that the length of $(0,1)$ is $1L$ (the 'L' reminds us that we are now measuring a line) then we can agree that the total measure of $[0,1]$ is 1L+2P. Now if we agree that the area of the square $(0,1)^2$ is $1R$ ('R' reminds us of rectangles) then we may agree that the total measure of $[0,1]^2$ is $1S+4L+4P$ (since we added four segments like $(0,1)$ and four points to the square $(0,1)^2$ in order to get $[0,1]^2$. We can also compute the total measure of $[0,1]^2$ as the square of the total measure of $[0,1]$, that is $(1L+2P)^2$. Arguing formally, the latter is equal to $1L^2+4PL+4P^2$ and since (arguing formally) $L^2=R$ and $P^2=P$ and $PL=L$ we obtain the same result.

Again, please note that as far as I know this (non-measure theoretic) point of view is just a curiosity.

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I guess the fact that $\ell([a,b]) = b-a,\,a<b$ is intuitive enough for you. Now, let us assume that $\ell((a,b))$ exists. Intuitively, if our set is bigger, its length would be bigger, so $\ell((a,b)) \leq \ell([a,b]) = b-a$. On the other hand, $[a+\frac{1}{n},b-\frac{1}{n}] \subset (a,b)$ for all sufficiently large $n$, and therefore, $\ell((a,b)) \geq \ell([a+\frac{1}{n},b-\frac{1}{n}]) = b-a-\frac{2}{n}$. Since $n$ can be chosen arbitrarily large, we obtain $\ell((a,b)) \geq b-a$. Combining the two inequalities, we obtain $\ell((a,b)) = b-a$. To formalize these arguments, we need the "Lebesgue measure theory."

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The usual way to measure things, $[0,1]$ and $(0,1)$ have the same length.

So, let's look at a new way to measure things!

We want the interval $[a,b)$ to have length $b-a$, and $(a,b)$ to have one fewer point and $[a,b]$ to have one extra point. I chose this arrangement so that taking the unions of adjacent intervals is sane: we want the length of $[a,b)$ plus the length of $[b,c)$ to be the length of $[a,c)$.

(exercise: work things out if you'd rather the length of $[a,b]$ plus the length of $[b,c]$ to be the length of $[a,c]$... or show that things won't behave reasonably if you do)

Well, we could define new numbers for the length of $[0,1]$ and $(0,1)$, such as $1^+$ and $1^-$ respectively. Then the length of $[0,2)$ could also be computed as the length of $[0,1] \cup (1,2)$, i.e. $1^+ + 1^- = 2$.

What if we want to start measuring disjoint intervals? What should the length of $(0,1) \cup [2,3]$ be? Or $(5,6) \cup (3,4)$? Note that these have to add up to the length of $(0,1) \cup [2,4) \cup (5,6)$!

We could leave some sums undefined. Or we could introduce an infinitesimal $\epsilon$. We're allowed to add or subtract it in (and this means we can multiply it by integers), but not multiply it by real numbers or by itself. We can now record the length of $(0,1) \cup [2,3]$ as $(1 - \epsilon) + (1 + \epsilon) = 2$, and of $(5,6) \cup (3,4)$ as $(1 - \epsilon) + (1 - \epsilon) = 2 - 2\epsilon$. And these correctly add up to the length of $(0,1) \cup [2,4) \cup (5,6)$ which is $(1 - \epsilon) + 2 + (1 - \epsilon) = 4 - 2 \epsilon$

Note the length of a point $[x,x]$ is $0 + \epsilon$.

With care, we can prove it all works out, so we can talk about the lengths of unions of intervals using numbers of the form $a + b \epsilon$ where $a$ is a non-negative real number and $b$ is an integer (bonus exercise: work out a theory of directed intervals that can have negative lengths). However, I don't actually have any idea for how to use these numbers, other than simply to say that we can talk about lengths of intervals in a way that makes individual points have nonzero infinitesimal length.

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I'd be interested in a source for your claim that "it all works out", not because I doubt your word, but because I'd like to follow up. – MJD Jan 20 '13 at 22:59
@MJD: Admittedly, I don't have a reference; I've merely worked through enough detail to see how it goes. The main idea is to consider formal disjoint unions (or even formal sums) of finitely many intervals; the only obstacle to well-definedness of length then is when two intervals disjoint union to a larger interval. Trying to extend to infinite unions has surprises, though; the length of $(1,2) \cup (1/2,1) \cup (1/4,1/2) \cup (1/8,1/4) \cup \cdots$ is not defined because the coefficient on $\epsilon$ is $\sum_{i=0}^{\infty} -1$. – Hurkyl Jan 20 '13 at 23:33