Consider a collection of subsets $A_i$ of an ordnered vector space or field.
I am trying to find out, under what (minimal) conditions the following holds.
$\inf \bigcap_{i\in I}A_i \le \sup_{i\in I}\left(\inf A_i\right)$
So far I have one necessary condition:
$\bigcap_{i\in I}A_i =\emptyset \Longleftrightarrow A_i = \emptyset $
Another requirement might be something like convexity.
Thanks for your help
On
The equality seems to hold for sets of the form $\{x|a\le x \}$. As well as intervals $\{x|a\le x \le b\}$.
Also for upper sets, (i.e. sets that contain with any $x$ also every $y\ge x$), the equality is equivalent to the complete distributive law of a lattice.
Let $A=\bigcap_{i\in I} A_i $. For every $i$ we have $A\subset A_i$, hence $\inf A\ge \inf A_i$. It follows that $\inf A\ge \sup_{i\in I} (\inf A_i)$. So, the reverse inequality you seek can only hold as equality.
There are some sufficient conditions for $$\inf A = \sup_{i\in I} (\inf A_i)\tag{1}$$ to hold. For example, if $A_1\subseteq A_2\subseteq \dots$ then $A=A_1$, so both sides of (1) are equal to $\inf A_1$. This does not work in the opposite direction: if $A_1\supseteq A_2\supseteq \dots$ then (1) may be a strict inequality even when the intersection is nonempty. Indeed, let $A_n=\{2\}\cup (0,1/n)$: here $\inf A_n=0$ for all $n$, but $\inf A=2$.
When all $A_n$ are convex subsets of $\mathbb R$ (that is, intervals) and the intersection $A$ is nonempty, (1) holds. Indeed, in this case the intersection is also an interval: say, $A=\langle a,b\rangle$, where angle brackets could be $($ or $[$. It is easy to see that $a=\sup_{i} a_i$ where $a_i$ is the left endpoint of $A_i$.
I don't think there is any minimal condition for (1) to holds. After all, we can always cheat and include $A$ among the family $A_i$, no matter how badly the family behaves otherwise.