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This an example of a definition of an interval as given by K.G. Binmore: Mathematical Analysis: A Straightforward Approach.

A (real) interval is a subset I of the real numbers such that:

$\forall x, y \in I: \forall z \in \mathbb{R} : ({x \le z \le y \implies z \in I})$

I'm wondering how does this definition fit together with open intervals where endpoints $x, y$ are not included in the interval. Does it still hold that $x, y \in I$?

Furthermore, how should I understand $[3,2] = \emptyset$ which is a closed interval? Shouldn't closed intervals by definition contain endpoints and thus the set is not empty?

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Open intervals are still fine by this definition. In less compact language, the definition states that an interval $I$ is composed of all the numbers such that if you pick two numbers $x$ and $y$ in the interval, then every $z$ between them (including endpoints) is also in the interval. So, if you have an open interval, meaning the endpoints are not included, then you can't take either of the endpoints as your $x$ and $y$.

As for the set $[3,2]$, remember that $$[3,2] = \{x\in \mathbb{R} : 3 \leq x \leq 2\}$$ by definition of the notation $[a,b]$. However, in this case, as you've pointed out, $[3,2] = \emptyset$. However, we can still think of it as the empty interval, because it is vacuously true that if you pick any two numbers $x,y$ in the empty set then every number in between those two numbers are also in the empty set (because no two such number exist in the first place).

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    $\begingroup$ Thank you for such a clear explanation! $\endgroup$ – Treex Dec 3 '20 at 12:43

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