Rudin's Theorem 2.28 Theorem 2.28 in Rudin states:

Let $E$ be a nonempty set of real numbers which is bounded above. Let $y = \sup E$. Then $y \in \overline{E}$. Hence $y \in E$ if $E$ is closed.

I don't think I follow Rudin's proof completely. His proof is:

If $y \in E$ then $y \in \overline{E}$. Assume $y \not \in E$. For every $h > 0$ there exists then a point $x \in E$ such that $y - h < x < y$, for otherwise $y - h$ would be an upper bound of $E$. Thus $y$ is a limit point of $E$. Hence $y \in \overline{E}$.

I don't think the assertion $y - x < x < y$ is fully correct, i.e., the second inequality should be non-strict. Given $h > 0$, $y - h < y$, so it can't be an upper bound of $E$, as that would contradict the fact that $y$ is the supremum of $E$. So there must exist $x \in E$ such that $y - h < x$. But $x \leq y$ since the supremum is an upper bound, so $y - h < x \leq y$. The second inequality needn't be strict, e.g., we could have $E = \{1\}$, so $\sup E = 1 \in E$.
Am I correct that this is a typo in Rudin?
From this point, how does one reason that $y$ is a limit point of $E$? If the above assertion in Rudin's proof were true, then we have $x \in (y-h, y+h)$. I suppose this is still true even if the second inequality is non-strict, but this presupposes that $d(x,y) := |x-y|$. I know this is the standard distance on $\mathbb{R}$, but is it necessary that we use it? Should I assume this is the distance on $\mathbb{R}$ unless otherwise specified? I assume there are other valid and equivalent notions.

In this part of the proof Rudin assumes $ y \notin E$. This implies in particular that $y \neq x$ (as $x \in E$). We know $x \le y$ from the sup-condition, plus $x \neq y$ is exactly $x < y$ as claimed. There is no typo. It's correct.
