Verifying Proof: If $L$ is a poset with a bottom element, and $\exists \sup(S)$ for every subset $S \subset L$, then $L$ is a complete lattice

Question

I am currently working through Kaplansky for self study and was hoping to get some feedback on this proof. I would appreciate comments on clarity and legability as well.

Claim: If $L$ is a poset with a bottom element, and $\exists \sup(S)$ for every subset $S \subset L$, then $L$ is a complete lattice.

Proof: Let $0 \in L$ be the bottom element and consider some subset $S \subset L$. We construct the set of lower bounds: $$ S' = \{l \in L \mid s \in S \implies l \leq s \}. $$

Clearly, $0 \in S'$ since $\forall l \in L$ we have $0 \leq l \implies \inf(S') = 0$. Additionally, notice that $\exists \sup(S')$ by virtue of the fact that every subset of $L$ has a suprememum. Because $S'$ consists of all lower bounds of $S$, it must be the case that $\sup(S') = \inf(S)$, since any lower bound of $S$ must be in $S'$, and $s' \leq \sup(S')$ $\forall s' \in S'$. Thus, $S$ has an infimum.

Since every subset of $L$ has an $\inf$ and $\sup$, it must be the case that $\forall a,b \in L$, $\exists \sup(\{a,b\})$ and $\exists \inf(\{a,b\})$. Thus, $L$ is a lattice.

$\therefore L$ is a complete lattice.

$\blacksquare$

Answer 1

To show $L$ is a complete lattice, we need to show that every subset $A$ of $L$ has a supremum and an infimum.

If $A = \emptyset$, then $\sup A = \bot$, the bottom element (all elements of $L$ are upper bounds of $\emptyset$, voidly, and the minimum, so $\bot$ (or $0$ if you prefer, more Boolean algebra style), while $\inf \emptyset = \max L = \top$ or $1$ which also exists as $\sup L$ itself.

For all other $A$, $\sup A$ aready exists in $L$ by assumption and for the inf, we need to consider (indeed) the set of lower bounds for $A$. This is defined as

$$L(A)=\{x \in L\mid \forall a \in A: x \le a\}$$

and is non-empty as $0 \in L(A)$ for any $A$. So $s=\sup L(A)$ exists in $L$ and I claim it is $\inf A$: if $a \in A$ then $a$ is an upperbound for $L(A)$ by definition, and as $\sup L(A)$ is the minimal upperbound for $L(A)$, $s \le a$ and as $a \in A$ was arbitrary, $s$ is a lower bound for $A$. Now, if $b$ is any other lower bound for $A$, then $b \in L(A)$ and so $b \le \sup L(A)=s$ and so $s$ is the maximal lowerbound for $A$, i.e. $s=\inf A$.

It follows that all subsets of $L$ have an infimum too. QED