The identity I was asked to prove is as follows:
Let $(X, d)$ be a metric space and $Y \subset{X}$. Prove that $diam(Y) = diam(\bar{Y})$ where $diam(Y) = sup\{ d(x,y)\ |\ x,y \in Y \}$
My proof is this:
$Y \subset \bar{Y} \implies diam(Y) \le diam(\bar{Y})$
Therefore, to prove $diam(Y) = diam(\bar{Y})$ it suffices to show that assuming $diam(Y) \lt diam(\bar{Y})$ leads to contradiction.
Suppose $diam(Y) \lt diam(\bar{Y})$. Then $\exists\ a \in \bar{Y}$ such that $B(a,r)\ \cap\ Y = \emptyset$ for some $r \gt 0$. However, $\bar{Y}$ consists of all points adherent to Y, or
$\bar{Y} = \{x \in X\ |\ B(x,r)\ \cap\ Y \neq \emptyset, \forall\ r \gt 0 \}$
Thus, $\bar{Y}$ must contain a point which is not adherent to $Y$, which is a contradiction of its definition.
Thus, $diam(Y) = diam(\bar{Y})$
$\blacksquare$
Is this proof correct? or are there faults\inconsistencies in it?
Best Answer
I'd they the general proof is correct. The only step you could make a bit more precise is
To me, it is not immediately clear why this is true (it is -- but it requires some intermediate steps (at least for me)).
If you want to stick closer to the definitions here, I would probably start by saying that $\mathrm{diam}(Y)<\mathrm{diam}(\bar Y)$ implies that there exist some $x,y \in \bar Y$ with $d(x,y) > \mathrm{diam}(Y)$ and then deduce a contradiction from there.