Infimum of infima of all subsets

Question

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 $A$.

Answer 1

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

1. $a=\inf A$ exists;
2. 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 $b\le a$.

Answer 2

I write here the proof I made just in case someone else will be struggling with this in the future.

This proof works with the assumption that we work in a totally ordered set, that is $ A \subseteq S $ where $ S $ is totally ordered (for example the proof works with natural or real numbers).

If anyone knows a better, simpler or shorter proof I am happy to accept that as the right answer instead of this one.

Proof

First I define some new symbols to make the proof easily readable. $$ \begin{array}{l} a = inf\ A \\ a_1 = inf\ A_1 \\ \vdots \\ a_n = inf\ A_n \\ I = \{a_1, \dots, a_n\} \\ a' = inf\ I \end{array} $$ Therefore, in order to prove this, we need to prove that $ a = a' $. This is done by proving that $ a \ngtr a' $ and $ a \nless a' $.

First we prove that $ a \ngtr a' $. Because of the properties of infima, since each $ A_i $ is a subset of $ A $, then its infimum $ a_i $ cannot be smaller than $ a $, that is $ a \leq a_i $. Since this is true for all the $ a_i $ then $ a $ is a lower bound for $ I $. Then we can conclude that $ a \ngtr a' $ because otherwise $ a' $ would not be the greatest lower bound of $ I $ (that is its infimum).

Now we prove that $ a \nless a' $. We prove this by contradiction, that is we assume that $ a < a' $. Then there exists at least one $ x \in A $ such that $ x < a' $ because otherwise if all the elements of $ A $ are greater or equal to $ a' $ then $ a' $ would be a lower bound for $ A $ and therefore $ a $ would not be the greatest lower bound of A (that is the infimum). So let $ x $ be such that $ x < a' $. Since $ x \in A $ and $ A $ is equal to the union of all the $ A_i $, then there exists $ A_i $ such that $ x \in A_i $, then since $ a_i $ is the infimum of $ A_i $ it follows that $ a_i \leq x $. But since $ x < a' $ then by transitivity $ a_i < a' $, which is impossible because otherwise $ a' $ would not be a lower bound for $ I $, then $ a' $ would not be the infimum of $ I $ as assumed. $ \blacksquare $

I hope this proof helps whoever is facing mathematical problems which involve proving this property of infima.