When does downward closure commute with supremum?

Question

Let $A$ be a suplattices, and suppose we have a family $\{a_i\}_{i\in I}\subseteq A.$

Is $\bigcup_{i\in I}(\operatorname{\downarrow}a_i) = \operatorname{\downarrow} \sup_{i\in I}(a_i)$ in general?

Here $\downarrow a$ means the principal downward closed sets (order ideal) generated by $a.$

I believe this assertion is wrong in general, and my counter example is $I=\{1, 2\}$ and $A=\{0\lt a_1\lt 1, 0\lt a_2\lt a\lt 1\}.$ Then $\operatorname{\downarrow} a_1\cup\operatorname{\downarrow} a_2=\{0, a_1, a_2\}$ while $\operatorname{\downarrow}(a_1\lor a_2)=A.$
Am I correct?

Are there any obvious conditions on $A$ or $\{a_i\}_{i\in I}$ that would guaranteed the property in the above question?

I am looking for a condition like $A$ being distributive. This sounds like a very vague question. But, hopefully more experienced people can tell me something useful.

Answer 1

This is virtually never true. Since $\sup_{i\in I}(a_i)\in \operatorname{\downarrow} \sup_{i\in I}(a_i)$, the only way your equality can hold is if $\sup_{i\in I}(a_i)\in \operatorname{\downarrow}a_i$ for some $i$. That is, you must have $\sup_{i\in I}(a_i)\leq a_i$ for some $i$, which means that $a_i$ is greater than equal to every $a_j$, so your family $(a_i)$ must have a greatest element. (And of course, if the family does have a greatest element $a_i$, then your equality does hold since both sides are just equal to $\operatorname{\downarrow}a_i$.)

Answer 2

As the comment by Eric Wofsey below points out, this condition always fails for $I=\varnothing$. We can however give a characterisation of the algebras where the statement you wrote is true for all nonempty indices $I$. This is the case if and only if $A$ is a linear order with no infinite ascending chain.

First, notice that your condition is equivalent to claim that every nonzero element in $A$ is completely join prime: $x\leq \bigvee_{i\in I} a_i$ entails $x\leq a_i$ for some $i\in I.$ We can show that every element of $A$ is completely join prime if and only if $A$ is a linear order with no infinite ascending chain.

If $A$ is linear and has no infinite ascending chain, then $x\leq\bigvee_{i\in I} a_i=a_j$ for some $j\in I$, hence $x$ is join prime.

Conversely, (i) if $A$ is not linear, then there is some $x=y\lor z$ with $y,z$ incomparable. Hence $x\leq y \lor z$ but $x\nleq y$ and $x\nleq z$, meaning that $x$ is not join prime. (ii) If $A$ is linear but has an ascending chain $\{a_i\}_{i<\omega}$, we obtain $\bigvee_{i<\omega}\{a_i\}\leq \bigvee_{i<\omega}\{a_i\}$ but $\bigvee_{i<\omega}\{a_i\}\nleq a_n$ for all $n\in\omega$, hence $\bigvee_{i<\omega}\{a_i\}$ is not join prime.