6
$\begingroup$

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?

$\endgroup$

1 Answer 1

12
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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 $n\ge3$. $\endgroup$
    – bof
    Aug 31, 2020 at 2:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.