Let $(L, \sqsubseteq)$ be a poset. In every textbook on lattice theory, you find a property of complete lattices stating that the following three are equivalent:
- $\forall X \subseteq L$, there exists $\bigwedge X$
- $\forall X \subseteq L$, there exists $\bigvee X$
- $L$ is a complete lattice
The usual procedure to prove, for example, $(1) \implies (2)$, is by using the fact that $\bigwedge X^u = \bigvee X$, and that $\bigwedge X^u$ exists by hypothesis. However, apart from a (not-so-strong) intuition, I couldn't find a way to prove $\bigwedge X^u = \bigvee X$ in details. Does anyone have a suggestion that could point me in the right direction? It seems to me that the best way to go would be to show that $\bigwedge X^u$ is itself a upper bound of $X$.
Your approach will indeed work.
Suppose that $x\in X$. Then $x\sqsubseteq y$ for each $y\in X^u$, i.e., $x$ is a lower bound for $X^u$, and hence $x\sqsubseteq\bigwedge X^u$. That is, $x\sqsubseteq\bigwedge X^u$ for each $x\in X$, and therefore $\bigwedge X^u$ is an upper bound for $X$.