Lecture 2 - Upper Bounds, Bounded Above Sets

Comment on the following sets

$\{1,2,3,4,5\}$	$\{1,2,3,4,5,6\}$
$[-10,5]$	$[1,9]$
$(3,5]$	$[6,10)$
$(-2,5)$	$\{5\}\cup \{1,6,7\}$
$(-\infty,5]$	$[5,\infty)$
$(-\infty,5)$	$(-\infty,5.2)$
$[2,3]\cup [4,5]$	$\mathbb{R}$

1. Understanding Upper Bounds

Definition: Upper Bound

Let $S$ be a non-empty set of real numbers. A number $\alpha\in \mathbb{R}$ is called an upper bound for $S$ if: $$\forall\, x \in S,\quad x \leq \alpha.$$

Examples

For $S = \{1, 2, 3\}$, the numbers $3,\, 4,\, 100$ are all upper bounds.
For $(0, 1)$, the number $1$ is an upper bound even though $1 \notin (0,1)$.

What does it mean for $\alpha\in \mathbb{R}$ to not be an upper bound for $S$?

$\exists\, x \in S$ such that $x > \alpha.$

Let $S$ be a nonempty set. What you mean by $S$ has a maximum.

$\alpha\in \mathbb{R}$ is maxixum for $S$ means that $\alpha$ satisfies two properties.

$\alpha\in S.$
$\alpha$ is an upper bound for $S.$

Let $S$ be a nonempty set and $\alpha \in \mathbb{R}.$ What you mean by $\alpha$ is not maximum of $S.$

Either $\alpha \notin S$ or $\alpha$ is not an upperbound of $S.$

Why $5$ is not maximum of $S=(1,5)?$
Why $5$ is not maximum of $S=(1,6)?$
Q: If maximum of a set $S$ exsits, is it unique?

Is true that every finite set has a maximum?

Yes. (How do you prove? By Mathematical Induction.
$P(n):$ Every set $S$ with $n$ elements has a maximum.
How to show $P(1)$ is true?
Suppose $P(k)$ is true. What do you mean by that? Any set with $k$ elements has a maximum.
Consider a set $S=\{a_1,a_2,\ldots,a_k\}\cup \{a_{k+1}\}.$
By Indction $T=\{a_1,a_2,\ldots,a_k\}$ has maximim say $\beta.$ Recall what do you mean by $\beta$ is amximum of $T.$?
Let $\gamma=\max\{\beta, a_{k+1}\}.$ Show that
1. $\gamma \in S$ and
2. $\gamma$ is an upper bound of $S.$

Is $3.5$ an upper bound for $S = \{1,2,3,4,5\}$?

No, because $4 \in S$ and $4 > 3.5.$ Hence $3.5$ is not maximum of $S.$

Proposition: If $\alpha$ is an upper bound for $S$ and $\beta > \alpha$, then $\beta$ is also an upper bound for $S.$

Proof. Let $x \in S.$ Since $\alpha$ is an upper bound, $x \leq \alpha.$ Since $\beta > \alpha,$ we get $x \leq \alpha < \beta,$ so $x < \beta.$ As $x$ was arbitrary, $\beta$ is an upper bound for $S.$ $\square$

Consequence:

If $U$ is the set of all upper bounds of $S$ and $\alpha \in U,$ then $[\alpha, \infty) \subseteq U.$

Definition: Bounded Above

Let $A$ be a nonempty subset of $\mathbb{R}.$ Then $A$ is bounded above in $\mathbb{R}$ if $A$ has an upper bound, i.e., $\exists \alpha in \mathbb{R}$ such that $\alpha$ is an upper bound of $A.$
$$\exists\, \alpha \in \mathbb{R} \text{ such that } \forall\, x \in A,\quad x \leq \alpha.$$

Which of the following sets are bounded above: $A=[1,2], B=(2,5), C=[1,\infty), D=\{1,2,3\}, \mathbb{R},\mathbb{R}^+.$

$A$ is bouded above by $2$,
$B$ is bounded above by $5,$
$D$ is bounded above by $3,$
Remaining sets are not bounded above.
What do you mean a set $S$ is not bounded above?

For example, $\alpha\in \mathbb{R}$ then $\alpha<\alpha+1\in \mathbb{R}.$
But note, we cannot apply same procedure for $\mathbb{R}^{+}=(0,\infty)$ as $-99\in \mathbb{R}$ but $-99+1=-98\not \in \mathbb{R}^{+}.$ We prove in two different ways.

By cases: If $\alpha \le 0,$ then $1\in\mathbb{R}^{+},$ will work. If $\alpha>0,$ then $\alpha+1\in \mathbb{R}^{+}.$

Without cases: If $\alpha\in \mathbb{R},$ then $|\alpha|+1\in \mathbb{R}^{+}.$

Proposition: Subsets of bounded-above sets are bounded above.

Let $\emptyset \neq A \subseteq B \subseteq \mathbb{R}.$ If $B$ is bounded above, then $A$ is bounded above.

Proof. Since $B$ is bounded above, $\exists\, \alpha \in \mathbb{R}$ such that $\forall\, x \in B,\; x \leq \alpha.$ Let $y \in A.$ Then $y \in B$ (since $A \subseteq B$), so $y \leq \alpha.$ Hence $\alpha$ is an upper bound for $A.$ $\square$

Are you able to prove that if $\emptyset \neq A \subseteq B \subseteq \mathbb{R}$ and $A$ is not bounded above, then $B$ is not bounded above.

Self Assessment Quiz

Q1: Prove that $S=(0,1)$ has no maximum.

Show Hint

Hint: Show that if $\alpha\in S,$ then $\alpha$ is not an upper bound.

Show Answer

Suppose $\alpha\in S,$ then $\frac{\alpha+1}{2}\in S$ and $\alpha<\frac{\alpha+1}{2}.$
We showed that $S$ does contain none of its upperbounds.

Q2: If an upper bound belongs to $S$, then that is the maximum.

Show Hint

Hint: Recall the definition of maximum of a set.

Show Answer

Let $\alpha\in S$ and an upper bound for $S.$ Then by defition $\alpha$ satisifies both the conditions.

If $\beta$ is an upper bound of $S,$ then $\alpha\le \beta$ as $\alpha\in S.$ That is out of all upperbounds $\alpha$ is the least.

Q3: If $\alpha\in S$ is the maximum, then $\alpha-\epsilon$ is not an upper bound for $S,$ where $\epsilon$ is a postive real number.

Show Hint

Hint: Find an element in $S$ bigger than $\alpha-\epsilon.$

Show Answer

It is clear that $\alpha-\epsilon<\alpha\in S.$

Lecture 2: Upper Bounds, Bounded Above

1. Understanding Upper Bounds

Examples