# A meet-semilattice with top element that is not a lattice?

I am reading Francis Borceux’s “Handbook of Categorical Algebra I” and on page 135 it says

In particular a finite version of 4.2.5 does not hold: a finitely complete and well-powered category certainly admits finite intersections of subobjects (see 4.2.3), but not in general finite unions of subobjects. Finite unions have been constructed in 4.2.5 using possibly infinite intersections. For a counterexample, just consider a ∧-semi-lattice with a top element which is not a lattice.

I get the argument for not being able to construct a union through finite intersection. But I have trouble proving it with the mentioned counterexample. Specifically, I can’t picture a “ ∧-semi-lattice with a top element which is not a lattice” in my mind. Why can’t the top element be the union for a finite set of elements if there is no other “better” element? Does the construction of this counterexample have anything to do with (in)finiteness?

It is well-known that a finite meet-semi-lattice with a maximum element is a lattice. The reason is that we can define $$a \vee b := \wedge \{c\colon \textrm{c is an upper bound for a,b}\}$$, where this set is non-empty (since we have a maximum) and finite (since the poset is finite), and finite meets exist by supposition that we have a meet-semi-lattice.
But this is not true for infinite posets. Let $$P := (\{(a,b)\colon 0\leq a,b \leq 1\}\setminus \{(1,1)\}) \cup \{(a,a)\colon 1 < a \leq 2\}$$, with the usual partial order $$(a_1,b_1)\leq (a_2,b_2)$$ iff $$a_1 \leq a_2$$ and $$b_1 \leq b_2$$. Then $$P$$ is a meet-semi-lattice (with $$(a_1,b_1)\wedge (a_2,b_2)=(\mathrm{min}(a_1,a_2),\mathrm{min}(b_1,b_2)$$) and it has a maximum element $$(2,2)$$. But $$(1,0)$$ and $$(0,1)$$ lack a join.
• Or the subset of $\mathcal P(\mathbb N)$ consisting of $\emptyset$, $\{1\}$, $\{2\}$, and the sets $\{1,2\}\cup\{n,n+1,n+2,\dots\}$ for $n\ge3$.