# The length of an open set.

Our textbook states:

The length l(I) of an interval I is defined to be the difference of the endpoints of I if I is bounded, and infinity if I is unbounded. Lenght is an example of a set function, that is, a function that assosiates an extended real number to each set in a collection of sets. In the case of length, the domain is the collection of all intervals. In this chapter we extend the set function length to a large collection of sets fo real numbers. For instance, the "length" of an open set will be the sum of the lengths of the countable number of open intervals of which it is composed."

I don't understand this last sentence.

1) How can there be a "countable number of open intervals" in an open interval? For example, if we have $(0,2)$, I can choose any two numbers x,y such that $0 < x,y <2$ and create an open interval from them, right? So I don't understand what they mean by a "countable number of open intervals".

2) The text is telling us the the "'length' of an open set will be the sum of the lengths of the countable number of open intervals of which it is composed". But how can we know the length of those "countable number of open intervals of which it is composed.

I was wondering if anybody could give me an example in order to clarify what this means...

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The author must be thinking of the open set as a countable disjoint union of open intervals. – user108903 Jan 22 '13 at 19:38
The last sentence that you quoted is about the length of an open set, not necessarily an interval. If you already know the lengths of intervals, then, given any open set $U$, it is the union of countably many disjoint open intervals (this should be proved in the textbook), and you define the length of $U$ to be the sum of the lengths of those intervals. – Andreas Blass Jan 22 '13 at 19:39

1) The phrase "of which it is composed" seems to be intended to mean that you write the open set $U$ as $$\tag1 U=\bigcup_{i\in S}(a_i,b_i)$$ Where $(a_i,b_i)\cap (a_j,b_j)=\emptyset$ for $i\ne j$.

2) In (1) above, the length of $(a_i,b_i)$ is $b_i-a_i$. Then the author wants to define $$l(U)=\sum_{i\in S} (b_i-a_i).$$ Of course, one needs to show that this s well defined

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But if U is a set (v,w), then why don't we just say that its length is w-v instead of saying $l(U)=\sum_{i\in S} (b_i-a_i)$? In other words, why do we need to break U up into pieces to take the sum of their lengths, instead of just taking the length of U itself? Is that because v and w are not included in the set (since its open)? But, can't we also say that $l(U)=\sum_{i\in S} (b_i-a_i)$ also does not represent the true length of U, since there is not miminum or maximum point to start from or end at (since U is open)? – user58289 Jan 22 '13 at 20:03
@Artus You seem to be assuming that these are all single intervals. Some are not: $(1,2)\cup(3,4)$ is an open set. It makes more sense to give it length $2$ rather than $3$, since it doesn't contain the middle section. A single open interval on its own would be given a length in the way you describe. – Robert Mastragostino Jan 22 '13 at 20:39

The author means to say that an open set in the reals is a countable set of disjoint open intervals. For example, the open set $(2n,2n+1): n \in \mathbb{N}$ is open, and is made up of a countable number of open intervals.

The author says that the 'total length' is the sum of these 'smaller lengths,' which I think is the most intuitive way of thinking about it.

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