# Proving that the closure of a set contains the $\inf$ and $\sup$

I came across the following problem about closures:

If $A$ is a bounded nonempty subset of $\mathbb{R}$, prove that $\sup A \in \overline{A}$ and $\inf A \in \overline{A}$.

Proof. By hypothesis, $A$ satisfies the least upper bound property (and the greatest lower bound property). So $\sup A$ and $\inf A$ exist. Now we know that $$\overline{A} = \{x \in \mathbb{R}: a_n \to x \ \text{for some sequence} \ (a_n) \ \text{in} \ A \}$$

Moreover, $x \in \overline{A} \Longleftrightarrow (\forall \epsilon >0) \ \exists a \in A \ni |x-a| < \epsilon$. If $\sup A$ and $\inf A$ are in $A$, then we are done. So suppose they are not. Let $x_1 = \sup A$ and $x_2 = \inf A$. By definition, for every $\epsilon >0$, there exists $x_A \in A$ such that $x_A \leq x_2 +\epsilon$. Likewise, there exists $x_B \in A$ such that $x_1- \epsilon \leq x_B$. So we get $$x_A-x_2 \leq \epsilon$$ and $$x_1-x_B \leq \epsilon$$

It seems like we would be done if we put absolute value signs on those quantities. Because then we can approximate the supremum and infimum as closely as we like with members of $A$.

How would I get the absolute value signs on the above two quantities?

• You can indeed replace $x - x_2$ by $|x - x_2|$, since $x - x_2 \geqq 0$ for each $x \in A$ by definition of the infinum. Similarly for the other statement. – Dylan Moreland Jun 27 '11 at 5:39
• Try to construct a sequence of elements of $A$ that converges to $x_1$, and then try to construct a (possibly different) sequence that converges to $x_2$. Hint. Start by considering $\epsilon_1=1$; then $\epsilon_2=\frac{1}{2}$; then $\epsilon_3=\frac{1}{3}$; and so on, in order to construct each sequence. – Arturo Magidin Jun 27 '11 at 5:41

## 2 Answers

You are extremely close.

$|x_A-x_2|\leq \varepsilon$ is equivalent to $-\varepsilon\leq x_A-x_2\leq \varepsilon$, so you'd be done with the inf part if you could show that $-\varepsilon\leq x_A-x_2$, or in other words, $x_2\leq x_A+\varepsilon$. Can you see why that is true?

Similarly, show that $-\varepsilon\leq x_1-x_B$, or, $x_B\leq x_1+\varepsilon$.

(A pedantic point about notation: Matching $A$ with $2$ and $B$ with $1$ may catch a reader off guard. An even more pedantic point: The "$A$" used as a subscript is the same symbol used to denote the original set, which might also be best avoided.)

Generally speaking, if you are trying to prove something about sups, you will often need to use both key components of their definitions.

1. Every number less than the sup is not an upper bound. This is the part that you used.
2. The sup is an upper bound. This is the part that you hadn't yet used.

The analogous statements hold for infs.

Once you have the essentials down, it's also worth noting that your assumption that the inf and sup are not in the set is superfluous. That is, whether or not $\sup A$ and $\inf A$ are in $A$, the existence of $x_A$ and $x_B$ as you stated is guaranteed by the definition of inf and sup.

The following answer is given in your notation:

We do not need absolute value signs on these inequalities since $0\leq x_2-x$ and $0\leq x-x_1$. The reason is that $x_2=\sup A$ and since $x\in A$, it follows that $x\leq x_2$, i.e., $x_2-x\geq 0$. Similarly, since $x_1=\inf A$ and since $x\in A$, it follows that $x_1\leq x$, i.e., $x-x_1\geq 0$.

The following exercises are relevant to your question:

(1) Let $A=(0,1)$. Determine:

(a) $\sup A$,

(b) $\inf A$,

(c) a sequence $\{x_n\}$ of elements of $A$ that converges to $\sup A$, and

(d) a sequence $\{y_n\}$ of elements of $A$ that converges to $\inf A$.

(2) Let $A=[0,1]$. Determine:

(a) $\sup A$,

(b) $\inf A$,

(c) a sequence $\{x_n\}$ of elements of $A$ that converges to $\sup A$, and

(d) a sequence $\{y_n\}$ of elements of $A$ that converges to $\inf A$.

(3) You know that if $A$ is a closed and bounded subset of $\mathbb{R}$, then $\inf A, \sup A\in A$. Prove that if $A$ is a connected (bounded) subset of $\mathbb{R}$, then the converse is also true, that is, if $A$ is connected and if $\inf A, \sup A\in A$, then $A$ is a closed set.

(4) Is the converse of the result of your question true in general? More precisely, if $A\subseteq \mathbb{R}$ is a bounded set and if $\inf A,\sup A\in A$, then can we always conclude that $A$ is a closed set?

(5) Prove that if $A$ is a bounded subset of $\mathbb{R}$ and if $x=\sup A$, we can choose a sequence of elements of $A$ converging as fast as desired to $x$, that is, if $M$ is a positive real number, then we can choose a sequence $\{x_n\}$ of elements of $A$ such that $\Sigma_{n=1}^{\infty} (x-x_n) < M$.

I hope this helps!