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 $|a-s|<epsilon$, 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

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