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, $d((3,6)) = d([3,6])$.
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$\begingroup$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.
$\endgroup$ 9 $\begingroup$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.
$\endgroup$ $\begingroup$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 $|b-a|=b-a$
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