# What is the length of an open interval and how do you prove it? [closed]

If an open interval does not contain its endpoints, why does its length is the same as a closed interval with the same endpoints? For example, $d((3,6)) = d([3,6])$.

## closed as off-topic by Shailesh, Saad, Xander Henderson, Ethan Bolker, timurMay 29 '18 at 13:04

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• the distance between any two points $a,b$ on the number line is $|b-a|$ this is very intrinsic to calculus. If you know that $b>a$ you can drop the absolute values $|b-a|=b-a$ – N8tron May 28 '18 at 23:07
• It's by definition – Jeffery Opoku-Mensah May 28 '18 at 23:11
• @N8tron I know what distance is. I didn't know that single points have no length as Eric Wofsey mentioned. So I did not understand why an open interval has the same length of a closed one. – Mor Haham May 28 '18 at 23:46
• What's the area of a rectangle under $f(x)=1$ on the interval $[a,b]$ in terms of calculus? What's the area under the same $f$ on $(a,b)$ in terms of calculus? The whole idea is lines don't have area so they don't affect area. Points don't have length – N8tron May 28 '18 at 23:53
• it is a convention that the distance from a point to itself is 0. But, there are generalization of distances (called partial metrics) where the distance from a point to itself may be positive. While you probably need not worry about that, here are a couple of links: nyaspubs.onlinelibrary.wiley.com/doi/abs/10.1111/… and dcs.warwick.ac.uk/pmetric/monthly708-718.pdf – Mirko May 28 '18 at 23:59

The length of an interval $(a,b)$ (or $[a,b]$, or $(a,b]$, or $[a,b)$) is defined to be $b-a$. This is just a definition, so it requires no proof. Intuitively, it should make sense: if you think of the interval as being a "stick" cut out from the number line, then it would be $b-a$ units long. We don't care whether the interval contains its endpoints because a single point has no length.

The following intuition may help. Suppose you have a ruler, with lengths marked in centimeters along it. If you want to measure a length of $c$ centimeters, you would usually just measure from the start of the ruler to the point marked $c$. However, you could also measure from a point marked $a$ to a point marked $b$, as long as $b-a=c$. The reason is that you could just shift the ruler forwards by $a$ centimeters, so the point that was marked $a$ is now at the start of the ruler and the point that was marked $b$ is now marked $b-a$.

A vast generalization of this notion of "length of an interval" is Lebesgue measure on $\mathbb{R}$, which is a way of defining the "length" of much more complicated sets than just an interval. In the context of Lebesgue measure, depending on your definitions, it may be a theorem that the length of $(a,b)$ is $b-a$. But in calculus or basic analysis, this is usually just taken as a definition.

• it does not make sense because, for example if you have $(3,6)$ so $3,6\notin(3,6)$. – Mor Haham May 28 '18 at 23:16
• I don't understand your objection. – Eric Wofsey May 28 '18 at 23:21
• It does make sense because, by definition, an open interval does not contain its endpoints. – marty cohen May 28 '18 at 23:22
• A single point (or two points) has no length. – Eric Wofsey May 28 '18 at 23:32
• @MorHaham: Well, what do you expect people to think when you don't tell them the reasoning behind your question? – Eric Wofsey May 28 '18 at 23:55

This question is more related to measure theory than topology. The most standard measure on the collection of sets generated by open intervals is Lebesgue measure where $\mu([a,b])=b-a$. You can associate a measure with any right continuous function F such that $\mu_F([a,b])=F(b)-F(a)$. There's no correct answer for the measure of an interval, there's lots of different ways to do it.

Here's an attempt of intuitively justifying it to you. Think about the distance between a point and the origin.

• In $\mathbb{R}^3$ take $P(x,y,z)$ has distance $\sqrt{x^2+y^2+z^2}$
• In $\mathbb{R}^2$ (the $xy$--plane) take $P(x,y)$ has distance $\sqrt{x^2+y^2}$
• In $\mathbb{R}^1$ (the number line ) take $P(x)$ has distance $\sqrt{x^2}=|x|$

So on the number line the distance between any point $P(x)$ and the origin is $|x|$. Now if you want to find the distnce between two points $P(x)$ and $Q(y)$ on the number line first consider their difference $R(x-y)$ it will have the same distance from $0$ as $P$ has from $Q$ so $|b-a|$ is the distance between $a$ and $b$ on the number line. When $b>a$ you can drop the absolute values $|b-a|=b-a$