Bounded sets and functions

• Boundedness is a property of metric, not a property of the general topological space

Question

• How is sequence escaping to infinity different from diverging to infinity?
• Negative $-\infty$ case of limit superior?

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Supremum and infimum §

Let $(X,<)$ be an ordered set with subset $S\subset X$. We have the following definitions in $X$:

Definition: Upper and lower bounds

An upper bound for $S$ is an element $x\in X$ where for all $s\in S$, we have $s\le x$. A lower bound for $S$ is an element where we have $x\le s$.

Definition: Bounded above and below

$S$ is bounded above if there exists an upper bound for $S$, and similarly bounded below if there exists a lower bound.

Definition: Least upper bound (supremum) and greatest lower bound (infimum)

If one exists, the least upper bound or supremum of $S$ is an upper bound $x\in X$ for $X$ so that for any other upper bound $x^{\prime }\in X$ for $S$, we have $x\le x^{\prime }$. This is written as $\text{lub}(S)$ or $\text{sup}(S)$.

The greatest lower bound or infimum is written as $\text{glb}(S)$ or $\text{inf}(S)$.

• (Lemma) If $(X,<)$ is an ordered set and $S\subset X$ is a subset with both $x,x^{\prime }$ as its least upper bounds, then we have $x=x^{\prime }$.

Theorem: Elements in a subset of \mathbb R can be arbitrarily close to the supremum and infimum

Suppose $S\subset \mathbb{R}$ is a set of real numbers, and that $x\in \mathbb{R}$ is an upper bound for this set (i.e., for all $s\in S$, we have $s\le x$). Then

$$x=\mathsf{lub}(S)⟺{\forall }_{\begin{array}{c}ϵ\in \mathbb{R}\\ ϵ>0\end{array}}{\exists }_{s\in S}x\le s+ϵ.$$

That is, the given upper bound $x$ is the least upper bound of $S$ if and only if, for all real numbers $ϵ>0$, there exists some $s\in S$ for which $x\le s+ϵ.$

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In metric spaces §

Bounded sets and diameters §

Definition: Bounded sets in metric spaces

Let $(X,d)$ be a metric space with a subset $A\subset X$. Then $A$ is bounded if there exists a point $x\in X$ and some $r>0$ such that $A$ is contained in the $r$-ball about $x$ given by

$$B(x,r)=\{y\in X|d(x,y)<r\}.$$

Definition: Diameter of a bounded set in a metric space

Let $(X,d)$ be a metric space and $A\subseteq X$ be bounded. Then the diameter of $A$ is defined by

$$\text{diam}(A)=\text{sup}\{d(x,y)|x,y\in A\}.$$

Note that $A$ is bounded if and only if the diameter is finite.

Proposition

Let $A\subseteq X$ be bounded. Then $\text{diam}(A)=\text{diam}(\overline{A})$, where $\overline{A}$ is the closure of $A$.

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Bounded sequences, limit superior and inferior §

Definition: Bounded sequences in metric spaces

Let $(X,d)$ be a metric space. We say a sequence $(x_n{)}_{n\in \mathbb{N}}$ in $X$ is bounded if there exist $x\in X$ and $r>0$ such that for all $n\in \mathbb{N}$, we have $d(x_n,x)<r$.

Definition: Sequence escaping to infinity in \mathbb R

We say a sequence $(x_n{)}_{n\in \mathbb{N}}$ in $\mathbb{R}$ escapes to infinity if for every $C\in \mathbb{N}$, there exists $N\in \mathbb{N}$ such that $x_n\ge C$ for all $n\ge N$.

Definition: Limit superior of a sequence in \mathbb R

Let $(x_n{)}_{n\in \mathbb{N}}$ be a sequence $\mathbb{R}$, and let $A$ be the set of all sub-sequential limits of $(x_n{)}_{n\in \mathbb{N}}$. Then the limit superior, denoted $\lim \sup x_n$, is the value given by:

• $\sup A$, if $A$ is bounded and not empty;
• $\infty$, if there exists a sub-sequence of $(x_n{)}_{n\in \mathbb{N}}$ that escapes to infinity;
• $-\infty$, if the entire sequence $(x_n{)}_{n\in \mathbb{N}}$ escapes to $-\infty$.

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Review §

• Explain why a set that has a least upper bound must be non-empty. ⭐
• Consider the set $X=\mathbb{Q}$ with subset $S=\{x\in \mathbb{Q}\mid x^2<2\}$. Prove that $S$ would have a least upper bound would have $\text{lub}(S{)}^2=2$. Since $\sqrt{2}\notin S$, the subset $S$ does not have a least upper bound in $\mathbb{Q}$. ⭐

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