# 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.

• There is no problem with Rudin's arguments. Although simpler arguments can be used. – Oliver Diaz May 16 at 7:07
• 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. – Henno Brandsma May 16 at 7:11
• 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? – JeremyS May 16 at 7:28
• 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. – Henno Brandsma May 16 at 8:53
• 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. – JeremyS May 16 at 16:41

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.

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.

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