I'm given the assignment to prove that $\operatorname{diam}\bar A = \operatorname{diam}A$, where $\operatorname{diam}A=\sup\{\rho(a,b): a,b \in A\}.$
How can they be equal if $\bar A= \partial A \cup A$:
Say $x\in \partial A$, then does it not follow that $x > a, \; x> b$, thus $ x \notin A$ ?
On
We don't necessarily have any ordering among the points of the metric space (consider e.g. $\Bbb R^n$), so your statements $x>a$ and $x>b$ are meaningless.
To prove the claim, take any $a, b\in\bar A$ and consider sequences $a_n, b_n\in A$ with $a_n\to a, \ b_n\to b$, and note that $\rho(a_n, b_n) \to\rho(a, b) $.
On
Since $A\subseteq\bar{A}$, it is obvious that $\operatorname{diam}A\le\operatorname{diam}\bar{A}$.
If $\operatorname{diam}A=\infty$, there is nothing to prove, so we can assume $\operatorname{diam}A$ is finite.
Suppose $\operatorname{diam}\bar{A}>\operatorname{diam}A$. Then there are $x,y\in\bar A$ such that $\rho(x,y)>\operatorname{diam}A$.
Let $\varepsilon=(\rho(x,y)-\operatorname{diam}A)/2$. Then, since $x,y\in\bar{A}$, there are $a,b\in A$ such that $\rho(a,x)<\varepsilon$ and $\rho(b,y)<\varepsilon$. By applying the triangle inequality we have $$ \rho(x,y) \le\rho(x,a)+\rho(a,y) \le\rho(x,a)+\rho(a,b)+\rho(b,y)< \varepsilon+\operatorname{diam}A+\varepsilon<\rho(x,y) $$
On
You want to show $$\operatorname{diam}\bar A = \operatorname{diam}A$$
Obviously $$\operatorname{diam}A \le\operatorname{diam}\bar A $$
On the other hand if $$\operatorname{diam}A <\operatorname{diam}\bar A $$ we have two points in the $\bar A$ whose distance is greater than $\operatorname{diam}A$
At least one of these points must be in $\partial A $
That implies that the point in the boundary is included in an open ball which does not intersect A.
That contradicts the definition of $\partial A $
Consider the intervals $(0,2)$ and $[0,2]$. One is open -- not containing either $0$ or $2$. The other is closed -- containing both $0$ and $2$.
The length of both these intervals is $2-0 = 2.$