I am just starting to learn about lattices. I am trying to see what examples I can come up with. It is helpful if I can have some outside confirmation about my thinking.
Let $X = \{1,3,4,12\}$
Define poset $\{X,\mid\}$ where $\mid$ refers to divisibility. (So, $4\mid1$ but $1\nmid4.)$
I am under the impression that $X$ is a lattice. Divisibility makes $\wedge = gcd$ and $\vee = lcm$
In particular,
$$(1 \wedge 3) = 1 \in X$$
$$(1 \vee 3) = 3 \in X$$
$$(1 \wedge 4) = 1 \in X$$
$$(1 \vee 4) = 4 \in X$$
$$(1 \wedge 12) = 1 \in X$$
$$(1 \vee 12) = 12 \in X$$
$$(3 \wedge 4) = 1 \in X$$
$$(3 \vee 4) = 12 \in X$$
$$(3 \wedge 12) = 3 \in X$$
$$(3 \vee 12) = 12 \in X$$
$$(4 \wedge 12) = 4 \in X$$
$$(4 \vee 12) = 12 \in X$$
Sorry to just write out all the cases. A simple "Yes" or "No" with a brief explanation if No would be very helpful.
That's exactly right... for the opposite relation, $$x \preceq y \iff x \text{ divides } y \text{.}$$ Note: standard usage is that "$x\mid y$ " means this relation "$x$ divides $y$", not "$x$ is divisible by $y$".
For $\preceq$ as above, $\wedge$ and $\vee$ are gcd and lcm respectively. Of course all your cases are correct.
However, the relation "is divisible by" is the reverse of $\preceq$: $$ x \sqsubseteq y \iff y \preceq x $$ so with respect to $\sqsubseteq$, meet and join are interchanged: $\wedge_{\sqsubseteq}$ is lcm and $\vee_{\sqsubseteq}$ is gcd.
If an ordering is a lattice ordering then so is its reverse. If you confirm that an ordering is a lattice ordering (sup and inf exist for all pairs of elements), and then use $\wedge$ and $\vee$ to denote the sup and inf that you proved to exist, then you don't have to prove that the lattice axioms hold for $\wedge$ and $\vee$: they do, by definition. That's the case here.