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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.

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  • $\begingroup$ There is no problem with Rudin's arguments. Although simpler arguments can be used. $\endgroup$ – Oliver Diaz May 16 at 7:07
  • $\begingroup$ Yes to the final question: the topology on $\Bbb R$ is the one induced by the standard distance and also the standard order. This is the default for $\Bbb R$ and can be assumed throughout the book. $\endgroup$ – Henno Brandsma May 16 at 7:11
  • $\begingroup$ When you refer to "the topology on $\mathbb{R}$," is this to say that any alternate distance function we define on $\mathbb{R}$ induces the same topology? $\endgroup$ – JeremyS May 16 at 7:28
  • $\begingroup$ Not any metric, there are other ones too. But they are never used in a analysis but can serve as counterexamples in a more general course, e.g. $\endgroup$ – Henno Brandsma May 16 at 8:53
  • $\begingroup$ So in other words, there's an implicit footnote in the question to assume that we're using the standard topology on $\mathbb{R}$? I'm hoping it doesn't sacrifice generality to use this distance function. I'm hoping there's a "more general proof" of sorts. $\endgroup$ – JeremyS May 16 at 16:41
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No, there is no typo in Rudin. He is assuming that $y\notin E$. Since $y\in\overline E$, every interval $(y-x,y+x)$ contains an element of $E$. That element cannot be $y$ itself, since rudin is assuming that $y\notin E$. And, since $y=\sup E$, no element of $(y,y+x)$ can belong to $E$. So, yes, there is some element of $(y-x,y)$ which belongs to $E$.

And, yes, it's the standard distance that is used here.

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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.

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$x=y$ is not possible because $y \notin E$ and$x \in E$. Hence we have strict inequality $x<y$. Rudin is using the standard metric on $\mathbb R$.

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