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

Question

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

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

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. $s \notin O$.

Answer 1

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.

Answer 2

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.