Supremum of closed sets I have been trying to understand the proof for the fact that the supremum of a closed set of real numbers belongs the set itself, as given by Theorem 2.28 in "Principles of mathematical analysis" by Walter rudin. The proof in the textbook goes as follows:
Assume the supremum y $\notin$ E. For every $h>0$ there is a point $x  \in E$ such that $y-h \leq x \leq y$, since y is the supremum. I was able to understand up until this point.
Then the argument is that every neighborhood of y contains a point $x \in E$ and $x \neq y$ since $y \notin E$. Thus y is a limit point of E but $\notin E$, thus contradictory to the assumption that E is closed. My rough interpretation of the statement is that since every neighbourhood of y is non empty and any given element of the neighbourhood can be written as $y-h$ for some $h>0$, it should contain  an element $x \in E$, thus making it the limit point.
Now, I am confused if this would be valid for any metric. In other words is $y-h \leq x \leq y$ equivalent to saying $d(y,y-h) < r \implies d(y,x) < r$ for any metric?
 A: In the statement of the theorem, y is given as the supremum.  In the proof, it is assumed that y $\notin$ E.  But it is still the supremum and, therefore, an upper bound of E.  Thus for every x $\in$ E, x $\le$ y. 
If you take an arbitrary radius, h $\gt$ 0, y - h $\lt$ y.  An x in E can be found in this neighborhood of y, since E is non-empty.  Otherwise, y - h would be an upper bound of E, because there would be no points of E greater than y - h.  But y - h cannot be an upper bound, since it is less than y, the supremum.
Because h was arbitrarily chosen, every neighborhood of y contains a point of E.  Thus y is a limit point of E by definition.  This implies that y $\in\overline E$, the closure of E.  If E is closed, E = $\overline E$, thus y $\in$ E.
As far as the metric goes, the metric used is the normal metric in $\mathbb{R}$, namely the absolute value of the difference between real numbers.  In this case,
d(x,y) $\lt$ h is equivalent to |x - y| $\lt$ h, where d is the normal metric on $\mathbb{R}$.  This implies that y-h < x < y, for those points x $\lt$ y within distance h of y.
A: A closed set includes all its limit points.
Also in $\mathbb{R}$ Sup of any set is limit of a sequence of its elements so is a limit point of the set, then it must be in the set.
