Examples 4.6

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(a)

To simplify letter typing, denote $A$ as $\mathcal A$ and $\leq$ as $\subseteq$ in the original text, reminding $\mathcal A \subseteq \mathcal{P}(X)​$ is non-empty.

$\sup (A) = \bigcup A​$

• $A$ is bounded above, since $X \in \mathcal P(X)$ is an upper bound of $A$.
• $\bigcup A = \sup (A)​$, since for every upper bound $B​$ of $A​$, every element $a \in A​$, $a \leq B​$. Therefore $\bigcup A \leq B​$ is the least upper bound of $A​$.

$\inf (A) = \bigcap A​$

• $A$ is bounded below, since $\emptyset \in \mathcal P(X)$ is an lower bound of $A$.
• $\bigcap A = \inf(A)​$, since for every lower bound $B​$ of $A​$, every element $a \in A​$, $B \leq a​$. Therefore $B \leq \bigcap A​$, $\bigcap A​$ is the greatest lower bound of $A​$.

[muzimuzhi]

(b)

$X$ has at least two elements, $\mathcal X = \mathcal P(X) \setminus {\emptyset}$. $A$ and $B$ are nonempty disjoint subsets of $X$, $\mathcal A = \{A, B\}$.

$\mathcal A \leq \mathcal X$

• Since $A$ and $B$ are nonempty, $A, B \leq \mathcal X$. Hence for every $a \in \mathcal A$, $a \in \mathcal X$. This is equivalent to $\mathcal A \leq \mathcal X$.

$\sup(\mathcal A) = A \cup B$

• $\mathcal A$ is bounded above since $X$ is an upper bound of $\mathcal A$
• $A \cup B = \sup(A)$ since both $A$ and $B$ are subset of every upper bound of $\mathcal A$.

$\mathcal A$ has no maximum

• Since $A$ and $B$ are disjoint, neither $A \leq B$ nor $B \leq A$ holds.

$\mathcal A$ is not bounded below

• If $x \in \mathcal X$ is an lower bound of $\mathcal A$, then $x \leq A \cap B = \emptyset$. This conflicts with $\emptyset \notin \mathcal X$.

$\inf(\mathcal A)$ does not exist

• This is a simple corollary of the last one.