How do I show that $2a$ is the diameter of $A=(-a,a)$? proof. Suppose that $A = (-a,a) = (-a,0]\cup [0,a)$ and let $x \in [0,a)$ and $y\in (-a,0]$ . Then $x + \epsilon = a = \sup A$ for some $\epsilon > 0$ and $y-\beta = -a = \inf A$ for some $\beta >0$. So,
$2a=a-(-a)=x+\epsilon -(y-\beta)=|x-y| + (\epsilon + \beta)$. Then 
$2a \ge |x-y|$.
Would I have to show that $2a$ is the supremum of $|x-y|$ by also showing the cases when $x,y \in (-a,0]$ , and $x,y \in [0,a)$
 A: For any $x,y \in A$ we have $|x-y| < 2a$. [If you to prove this claim more rigorously: Suppose without loss of generality that $x \le y$. Then $-a < x \le y < a$ so $y-x < a - (-a) = 2a$.] This shows $\text{diameter}(A) \le 2a$.
To show $\text{diameter}(A) = 2a$, we need to show $\sup_{x,y \in A} |x-y| = 2a$.
For any fixed $\epsilon > 0$, can you find $x,y \in A$ such that $|x-y| \ge 2a - \epsilon$?
A: First you can prove that the diameter is $\le 2a$ like this. Take any $x,y \in (-a,a)$. Now consider two cases. 


*

*If $x \le y$ then $-a < x \le y < a$ and so $|x-y|=y-x<a-(-a)=2a$.

*If $y \le x$ then $-a < y \le x < a$ and so $|x-y|=x-y<a-(-a)=2a$.
Next you can prove that the diameter is $\ge 2a$ like this. For any $\epsilon \in (0,a)$ we have $-a < -a+\frac{\epsilon}{2} < a - \frac{\epsilon}{2} < a$ and so the diameter is $\ge (a - \frac{\epsilon}{2}) - (-a + \frac{\epsilon}{2}) = 2a - \epsilon$. Since the diameter is $\ge 2a-\epsilon$ for all $\epsilon>0$ it follows that the diameter is $\ge 2a$.
A: You could do, but there are far easier ways. For example, simple note that for $x>y\in A$, we have $|x-y| \leq |a-x|+|x-y|+|y-(-a)| = |a-(-a)| = 2a$, so $2a$ is an upper bound for the diameter of $A$, and also that for any $\varepsilon > 0$ with $\delta := \min(\varepsilon,a)$, if we take $x = a - \frac{\delta}{2}$ and $y = \frac{\delta}{2}-a$, we have $x > y \in A$, and $|x-y| = |2a-\delta| \geq 2a - \delta\geq 2a-\varepsilon$, so $2a$ is also a lower bound for the diameter of $A$, hence is equal to the diameter of $A$.
