Infimum of infima of all subsets the following characterization seems trivial intuitively but it takes some time to prove it. 
Informal version of the problem
Given a set $A$ and a family of subsets of $A$ whose union is $A$, is the infimum of $A$ equal to the infimum of all the infima of those subsets?
Formal version of the problem
Let $A$ be a set and let $A_1, A_2,\dots, A_n$ be subsets of $A$ such that $\bigcup_{i=1}^n A_i = A$.
It seems quite sure that that:
$$
\inf A = \inf \{\inf A_i \mid 1 \leq i \leq n\}
$$
But it seems the proof takes some effort as I struggled for some time on it (or perhaps I have little experience with infima and suprema and this slows me down?).
If you replace "infima" with "minima" then it becomes very easy, but in this case it's less easy because the elements of $A$ may be very different from the elements of $I = \{\inf\ A_i \mid 1 \leq i \leq n \}$, as $I$ contains infima and infima may not be in the subsets $A_1,\dots, A_n$, that is they may not be in $A$.

Let's work in a partially ordered set $S$, with $A\subseteq S$. Also we know that $A=\bigcup_{i\in I} A_i$ and


*

*$a=\inf A$ exists;

*for each $i\in I$, $a_i=\inf A_i$ exists.


Then we claim that $a=\inf\{a_i:i\in I\}$.
Part 1 $a$ is a lower bound for $A_i$, hence $a\le a_i$ for every $i$, because $a_i$ is the greatest lower bound of $A_i$. Therefore $a$ is a lower bound for $\{a_i:i\in I\}$.
Part 2 We claim that $a$ is the greatest lower bound for $\{a_i:i\in I\}$. Indeed, if $b$ is a lower bound, then $b\le a_i$ for every $i$, so $b\le x$, for every $x\in A_i$. If $x\in A$, then $x\in A_i$, for some $i$ and therefore $b\le x$. Hence $b$ is a lower bound for $A$ and therefore $b\le a$.
