A Boolean algebra $\mathcal{B}:=(B,\leq,\lor,\land,^c,0,1)$ is said to be complete if every non-empty subset of $B$ has a greatest lower bound (g.l.b). Show that for $\mathcal{B}$ to be complete, it is necessary and sufficient that every non-empty subset have a least upper bound (l.u.b).
Here's my work, albeit incomplete:
Consider $\phi \neq X \subseteq B$, and let $\phi \neq Y = \{x\in B:x^c \in X\}$. We start by assuming that $\mathcal{B}$ is complete, so $Y$ has a g.l.b, say $b$.
Claim: If $a = b^c$, then $a$ is the l.u.b of $X$.
Proof of Claim: $\forall x \in X$, $x^c \in Y$ so $b\leq x^c$ which means $x\leq b^c = a$. So $a$ is an upper bound for $X$. If $m$ is also an upper bound for $x$, then $\forall x \in X$, $x\leq m$ which means $m^c\leq x^c$. By definition of $Y$, if $x\in X$ then $x^c \in Y$, so $m^c$ is a lower bound for $Y$. Since $b$ = g.l.b($Y$), $m^c\leq b$ and $a = b^c\leq m$ which implies that $a$ is the l.u.b of $X$.
In conclusion, what I have shown is that for every subset $X \subseteq B$, we can find a subset $Y\subseteq B$ which has a g.l.b (follows from completeness). Using the claim proved above, we can find a l.u.b for $X$, and this works for all $X\subseteq B$. Hence, all non-empty subsets $X$ of $B$ have a least upper bound.
I think this only proves the necessary part of the argument. How do I show that this is also sufficient?
You proved that if $\bigwedge Y$ exists, for all $\varnothing \neq Y \subseteq B$, then $\bigvee Y$ exists too.
Suppose now that $\bigvee Y$ exists for all $\varnothing \neq Y \subseteq B$.
Define $X = \{y^c : y \in Y\}$ and prove that $\bigwedge Y = \bigvee X$.
(It's exactly the same reasoning you used above.)