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

Mathematics Asked on February 23, 2021

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

Exercise 3.2.4 from Stephen Abbot’s Understanding Analysis.
$$newcommand{absval}[1]{leftlvert #1 rightrvert}$$

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. $$s notin O$$.

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 $$epsilonle 1$$ the open interval $$(s-epsilon,s)$$ is disjoint from $$A$$. And as a minor point, you don’t need to argue by contradiction.

If $$sin A$$, then certainly $$sinoperatorname{cl}A$$, so suppose that $$snotin A$$. Let $$epsilon>0$$; then $$s-epsilon< s$$, so $$s-epsilon$$ is not an upper bound for $$A$$, and therefore $$Acap(s-epsilon,s]nevarnothing$$. Moreover, $$snotin A$$, so $$Acap(s-epsilon,s)nevarnothing$$. Thus, for each $$epsilon>0$$ there is an $$ain A$$ such that $$|a-s|, so $$s$$ is a limit point of $$A$$, and therefore $$sinoperatorname{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.

Correct answer by Brian M. Scott on February 23, 2021

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.

Answered by user2661923 on February 23, 2021