For arbitrary metric spaces, if $\sup E\not\in E$, then must $\sup E$ be a limit point of $E$? Let $E\subset\mathbf R$, and suppose that $\sup E$ exists in $\mathbf R$. It is not too difficult to prove that if $\sup E\not\in E$, then $\sup E$ is a limit point of $E$:

Let $\alpha=\sup E$, and suppose that $\alpha$ is not a limit point of $E$. Then there is a $h>0$ such that $(\alpha-h,\alpha+h)$ does not contain any members of $E$. Hence every member of $E$ must be less than or equal to $\alpha-h$. But this contradicts the fact that $\alpha$ is the least upper bound of $E$.

My question is: can this proof be generalised? More explicitly, suppose that $X$ is both an ordered set and a metric space, and let $E\subset X$. Assuming that $\sup E$ exists in $X$, must it be the case that if $\sup E\not\in E$, then $\sup E$ is a limit point of $E$? In case the answer is negative, is there an extra condition which does guarantee this property?
 A: The answer is quite trivially negative when the order and metric are unrelated, but it is also negative in a suitably good setup, like subspace of order topology.
Consider $X=[0,1)\cup \{2\}$ and $E=[0,1)$ with the Euclidean metric and order. Note that $2$ is the supremum of $E$ but it is not its limit point. This space however is only a subspace of order topology. It is not the order topology for the inherited order (that would be homeomorphic to $[0,1]$ interval).
In general it is true for order topology over total order. In that situation let $s=\sup E\not\in E$. Then for any $x\in E$ there is $y\in E$ such that $x<y<s$. Therefore every open neighbourhood of $s$ contains some element from $E$ and so it is a limit point of $E$.
If the partial order is not total then again there are counterexamples, even for order topology. For example consider $X=\{0,1,2\}$ with $0<2$ and $1<2$, while $0$ and $1$ are unrelated. The induced order topology has only two (nontrivial) open subsets: $\{0,1\}$ and $\{2\}$. So $E=\{0,1\}$ has $2$ as the supremum but it is not its limit point.
