Question about bounded interval $\left[0,0\right]$ I'm studying intervals (tangentially, they came up in the awesome book How to Design Programs) and was wondering if the interval $\left[0,0\right]$ had only one element and why.  It makes sense to me that the interval $\left[0,1\right]$ would have infinite real numbers as elements.  And $\left[0,0.1\right]$ would have infinite real elements (and so on...). But $\left[0,0\right]$ seems like it should have only one element in it, but I don't understand how it could be proven.
Sorry if the question is simplistic!
Fond Regards!
 A: $[0, 0]$ does indeed have only one element, $0$: this is because, by definition, $[0, 0]=\{x: 0\le x\le 0\}$, and by trichotomy (https://en.wikipedia.org/wiki/Trichotomy_(mathematics)) $0$ is the only number satisfying that condition. 

Interesting aside: although the interval $[0, a]$ approaches $[0, 0]$ as $a$ approaches $0$ (in a certain sense), the cardinality does "jump" - basically, what this means is that cardinality is not a continuous function on the space of sets of real numbers, although this takes a LOT of work to formalize.
The measure of size that is continuous(er) is, well, measure - Lebesgue measure, that is, which is roughly speaking a generalization of the length of an interval https://en.wikipedia.org/wiki/Lebesgue_measure. On the plus side it has many nice properties, including a kind of continuity; on the downside it is very abstract, and working with it requires serious thought (real analysis). For example, even proving that the measure of $[0, a]$ is $a$ requires some work.
A: It just comes down to definitions. The definition of $[a,b]$ is 
$$
[a,b] = \{x\in \mathbb{R} | a\leq x \leq b\}$$Thinking about $[0,0]$ makes it clear that $0$ is the only element satisfing $0\leq x \leq 0$. For any two distinct reals, $a < b$, there are infinitely many reals satisfy $a\leq x \leq b$.
