10. Supremum, infimum and completeness

Supremum

Let $S\subseteq \mathbb{R}$ and $\alpha \in \mathbb{R}$
$a$ is the supremum of $S$ (also written as $\sup S$) if

1. $s\le \alpha$ for all $s\in S$ &nbsp &nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp ($\alpha$ is an upper bound of $S$)
2. If $s\le b$ for all $s\in S$ then $\alpha \le b$ &nbsp&nbsp&nbsp($\alpha$ is the least upper bound of $S$)

If $S$ has a supremum then $\sup S$ is unique

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Completeness axiom for the real numbers

Let $S$ be a

1. Non-empty subset of $\mathbb{R}$
2. Bounded above

Then $S$ has a supremum

Note that non-empty is very important otherwise every real number is an upper bound

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Infimum

Let $S\subseteq \mathbb{R}$ and $b\in \mathbb{R}$
$\alpha$ is the infimum of $S$ if

1. $s\ge \alpha$ for all $s\in S$ &nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp ($\alpha$ is a lower bound of $S$)
2. If $s\ge b$ for all $s\in S$ then $\alpha \ge b$ &nbsp&nbsp&nbsp&nbsp ($\alpha$ is the greatest lower bound of $S$)

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Properties of $\sup$ within subsets

Let $S,T$ be non-empty subsets of $\mathbb{R}$, with $S\subseteq T$ and with $T$ bounded above
Then

1. $S$ is bounded above

2. $\sup S\le \sup T$

Proof

1. Since $T$ is bounded above then suppose it has upper bound $b$
   Then $t\le b$ for all $t\in T$ so we know that $t\le b$ for all $t\in S$
   Hence $b$ is an upper bound for $S$

2. As $S,T$ are non-empty and bounded above so by completeness axiom then $S,T$ each have a supremum
   Note that $\sup T$ is an upper bound for $T$ hence also for $S$ so

$$\sup T\ge \sup S$$

(since $\sup S$ is the least upper bound for $S$)

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Relation between $\inf$ and $\sup$

Let $T\subseteq \mathbb{R}$ be non-empty and bounded below
Let $S=\{-t:t\in T\}$
Then

1. $S$ is non-empty and bounded above

2. Furthermore $\inf T$ exists and $\inf T=-\sup S$

Proof

Since $T$ is non-empty, so is $S$
Let $b$ be a lower bound for $T$, so $t\ge b$ for all $t\in T$
Then $-t\le -b$ for all $t\in T$ so $s\le -b$ for all $s\in S$
Hence $-b$ is an upper bound for $S$ so $S$ is bounded above

As $S$ is non-empty and bounded above so by completeness axiom, it has a supremum
Then $s\le \sup S$ for all $s\in S$ so $t\ge -\sup S$ for all $t\in T$

Hence $-\sup S$ is a lower bound for $T$
If $b$ is a lower bound for $T$ then $-b$ is an upper bound for $S$

Hence $-b\le \sup S$ (as $\sup S$ is the least upper bound)
Therefore $b\le -\sup S$

Hence $-\sup S$ is the greatest lower bound
So $\inf T$ exists and $\inf T=-\sup S$

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Approximation Property

Let $S\subseteq \mathbb{R}$ be non-empty and bounded above
For any $ϵ>0$ there exists $s_e\in S$ such that

$$\sup S-ϵ<s_{ϵ}\le \sup S$$

TLDR: There exists $s_{ϵ}\in S$ that is arbitrarily close to $\sup S$

Proof

Note that by definition of the supremum we have

$$s\le \sup S\text{ for all }s\in S$$

Suppose for a contradiction that

$$\sup S-ϵ\ge s\text{ for all }s\in S$$

Then $\sup S-ϵ$ is an upper bound for $S$ but $\sup S-ϵ<\sup S$
But this is a contradiction so there

$$\text{exists }{s}_{ϵ}\in S\text{ with }\sup S-ϵ<s_{ϵ}$$

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