Intervals are [a,b], right?
What is NOT an interval? Is it a single point?
$\displaystyle (1,2) \cup (3,4)$ is not, and neither is $\displaystyle [1,3] \setminus \{2\}$.
An interval is a subset of $\displaystyle \mathbb{R}$ where if you take two numbers from the subset every number between them is in the subset: $\displaystyle I \subseteq \mathbb{R}, \text{ } a, b \in I$, then $\displaystyle c \in I \text{ } \forall \text{ } a \leq c \leq b \in \mathbb{R}$. So as soon as you take one or more of these numbers out you stop having an interval.
Some authors define an interval in $\displaystyle R^1$ to be a non-degenerate( non-singleton) connected set.
So basically there are eight types, four bounded and four unbounded:
$\displaystyle (a,b),~(a,b],~[a,b),~[a,b],~(- \infty ,b), ~(- \infty ,b],~(a, \infty),~\& ~[a, \infty) $
Because it is required that an interval contain at least two points, $\displaystyle \{a\}$ is not an interval.
unions of non-overlapping intervals are not intervals:
$\displaystyle [0, 1]\cup [2, 3]$ is not an interval.
The set of all rational numbers in [0,1] not only is not an interval but contains no interval because, between any two rational numbers, there is an irrational number.
But that's the definition used only by some authors, as you write. Intervals are not at all generally defined that way. It is, in my humble opinion (and I could quote several books in my side), much more natural to define intervals as the connected subsets of $\displaystyle \mathbb{R}$ - without slapping any further condition(s) on top of that basic requirement.
But, as I wrote above, not a few mathematicians would look askance at you if you told them that $\displaystyle \{a\}$ is not an interval...So basically there are eight types, four bounded and four unbounded:
$\displaystyle (a,b),~(a,b],~[a,b),~[a,b],~(- \infty ,b), ~(- \infty ,b],~(a, \infty),~\& ~[a, \infty) $
Because it is required that an interval contain at least two points, $\displaystyle \{a\}$ is not an interval.
Well, Walter Rudin does it repeatedly, in "Principles of Mathematical Analysis" as well as in "Real & Complex Analysis". Jean Dieudonné does it, in "Foundations of Modern Analysis". And if Dieudonné does it I take it for granted that all the other members of the Bourbaki Group would at least go along with that. (I think it would be not particularly useful to you if I mentioned German authors: but being Swiss German, many of my books are in .. you guessed it .. German.)
Wikipedia also defines intervals in such a way that the "interval" [a,a] is just taken to be the singleton set {a}.
Defining intervals this way ("my way", i.e. Rudin's and Dieudonné's way), allows one to say something simple in a simple way: the image of an interval under a continuous function $\displaystyle f:\, \mathbb{R}\rightarrow \mathbb{R}$ is an interval. You could not say the exact same thing as succinctly if you defined interval in such a way as to exclude singleton sets.
Of course, all definitions are arbitrary - but some definitions are more arbitrary than others....
How very odd. Your edditions/translations of Rudin and Dieudonne differ so from mine.
In both 'Big Rudin' and 'Little Rudin' that I have, this definition appears:
“The segment (a,b) is the set of all real numbers, x , such that a<x<b.”
Then he notes that a interval is a segment that includes its endpoints.
To me that rules out {a} being an interval.
Virtually the same statement appears on page 17 of my translation of Dieudonne.
This is odd, don't you think?
Interpretations sometimes, but nowhere coud I find any source requiring any minimal number of points for something to be considered a real interval, and I think the huge majority of mathematicians would agree that {a} = [a,a] is a closed interval, in particular since the usual topology on $\displaystyle \mathbb{R}$ is Hausdorff and thus every singleton is a closed set, but of course: I could be wrong.
Tonio