Assume S is a finite poset. A lattice is a poset S such that
- S is bounded.
- ∀ x, y ∈ S there exists x ∧ y (existence of Meet/Infimum).
- ∀ x, y ∈ S there exists x ∨ y (existence of Join/Supremum).
Exercise: Check that any two of the three conditions above implies the third one.
I'm particularly interested in the proof of:
S is bounded and for every two elements x, y ∈ S x ∧ y exists $\Rightarrow$ x ∨ y exists.
I know that the proof for this probably involves some kind of induction, but my last use of induction is a long time ago and I'm really desperate to get a proof for this implication.
Thank you for your help!
Suppose that $S$ is bounded, finite, and $x \wedge y$ exists for every $x,y\in S$.
Let $a,b \in S$ and let us see that the least upper bound of $a$ and $b$, denoted $a \vee b$ exists in $S$.
Let $U$ denote the set of upper bounds of $a$ and $b$, that is, $$U = \{ s \in S : a \leq s \;\&\; b \leq s \}.$$ Since $S$ is bounded, $U \neq \varnothing$, and since $U$ is finite, you can use an induction argument to show that $\bigwedge U$, the greatest lower bound of $U$, exists in $S$: it's a matter of picking all the elements $u_1, \ldots, u_k$ of $U$ and compute $u_1 \wedge \cdots \wedge u_k$.
Now you can verify that $a \vee b = \bigwedge U$, that is, the least upper bound of $a$ and $b$ is the greatest lower bounds of their common upper bounds.