# Proof of the fact that the closure of a set always contains its supremum and an open set cannot contain its supremum

I would like to ask, if my proof checks out and is completely sound.

Exercise 3.2.4 from Stephen Abbot's Understanding Analysis. $$\newcommand{\absval}{\left\lvert #1 \right\rvert}$$

Let $$A$$ be non-empty subset of $$\mathbf{R}$$ and bounded above, so that $$s = \sup A$$ exists. Let $$\bar{A} = A \cup L$$ be the closure of $$A$$.

(a) Show that $$s \in \bar{A}$$.

(b) Can an open set contain its supremum?

My Attempt.

(a) $$\bar{A} = A \cup L$$ is the closure of $$A$$ and contains the limits points of $$A$$. We proceed by contradiction. Assume that $$s \notin \bar{A}$$ and is not a limit point of $$A$$.

Since, $$s$$ is the supremum for $$A$$, looking at the definition of least upper bound, it must satisfy two properties: (i) $$s$$ is an upper bound for $$A$$. (ii) Given any small arbitrary, but fixed positive real $$\epsilon > 0$$, $$(s - \epsilon)$$ should not be an upper bound for $$a$$.

From (ii), it follows that, given any $$\epsilon > 0$$, there exists $$t \in A$$, such that $$s - \epsilon < t$$. Thus, \begin{align*} \absval{t - s} < \epsilon \end{align*}

Thus, every $$\epsilon$$-neighbourhood of $$s$$, $$V_\epsilon(s)$$ intersects $$A$$ in points other than $$s$$. So, $$s$$ is the limit point of $$A$$. Therefore, $$s \in \bar{A}$$, which contradicts our initial assumption. Hence, our initial assumption is false.

(b) An open set cannot contain its supremum. We proceed by contradiction. Let $$O$$ be an open set. Assume that $$s \in O$$.

Since $$O$$ is an open set, for all points $$x$$ belonging to $$O$$, there exists an $$\epsilon$$-neighbourhood $$V_\epsilon(x)$$ that is contained in $$O$$. In particular, $$V_\epsilon(s) \subseteq O$$. So, if \begin{align*} s - \epsilon < t < s + \epsilon \end{align*}

then $$t \in O$$, for some $$\epsilon$$. But, that implies, for some $$\epsilon > 0$$, we must have \begin{align*} s < t < s + \epsilon \end{align*}

$$t \in O$$. So, $$s$$ is not an upper bound for $$O$$. This is a contradiction. Our initial assumption must be false. $$s \notin O$$.

• Looks good to me :) Jan 5, 2021 at 21:13
• @jlammy: It’s not quite correct: what if $A=(0,1)\cup\{2\}$, for instance? Jan 5, 2021 at 21:14
• @BrianM.Scott Is that an open set? The set $\{2\}$ is closed. Jan 5, 2021 at 21:17
• @Apoorv: $A$ is any non-empty subset of $\Bbb R$ that is bounded above. Jan 5, 2021 at 21:18
• @jlammy: I’m talking about (a). Jan 5, 2021 at 21:19

The argument for (a) isn’t quite correct, because $$s$$ need not actually be a limit point of $$A$$. For instance, let $$A=(0,1)\cup\{2\}$$; then $$s=2$$, and for any positive $$\epsilon\le 1$$ the open interval $$(s-\epsilon,s)$$ is disjoint from $$A$$. And as a minor point, you don’t need to argue by contradiction.

If $$s\in A$$, then certainly $$s\in\operatorname{cl}A$$, so suppose that $$s\notin A$$. Let $$\epsilon>0$$; then $$s-\epsilon< s$$, so $$s-\epsilon$$ is not an upper bound for $$A$$, and therefore $$A\cap(s-\epsilon,s]\ne\varnothing$$. Moreover, $$s\notin A$$, so $$A\cap(s-\epsilon,s)\ne\varnothing$$. Thus, for each $$\epsilon>0$$ there is an $$a\in A$$ such that $$|a-s|<\epsilon$$, so $$s$$ is a limit point of $$A$$, and therefore $$s\in\operatorname{cl}A$$.

(Note that while there is absolutely nothing wrong with including extra detail, and it can be a good idea when you’re still learning, it’s really not necessary to say more by way of justifying the various steps than I did above.)

The argument for (b) is fine.

• I am having some trouble following your proof through to the end. If $A = (0,1) \cup \{2\}$ then $s = 2$ and say I pick $\epsilon = 0.5$, wouldn't $(s-\epsilon,\epsilon)$ be $(1.5,0.5)$? That's an odd interval. How could $s - \epsilon$ be smaller than $\epsilon$? Jan 5, 2021 at 22:07
• @Quasar There $s=2 \in A$ so that part of the proof does not apply, as we are in the case $s \notin A$. Jan 5, 2021 at 22:09
• @Quasar: That’s an obvious typo, which I’ll fix as soon as I post this. Since $s=2$, $(s-\epsilon,s)=(1.5,2)$. Jan 5, 2021 at 22:13
• @HennoBrandsma, okay got it. By the way, I think he means $s - \epsilon < s$ (it's a typo). Jan 5, 2021 at 22:13

Depending on your definition of open and closed sets in $$\mathbb{R}$$, and depending on what previous theorems you have been given:

A boundary point $$x$$ for a non-empty set $$A$$ is a point such that in any open interval around $$x$$, no matter how small, there will be at least one point in the interval that is in $$A$$ and one point in the interval that is not in $$A$$.

Given any non-empty set $$A$$ that is bounded above, the supremum (i.e. least upper bound) of $$A$$ is a boundary point of $$A$$.

Any non-empty set that contains one of its boundary points can not be an open set.

Any non-empty closed set must contain all of its boundary points. This last assertion is a consequence of defining a non-empty set as closed if and only if its complement is open.