Infimum of Infima

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Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $\mathbb T$ be a collection of subsets of $S$.

Suppose all $T \in \mathbb T$ admit an infimum $\inf T$ in $S$.


Then:

$\inf \bigcup \mathbb T = \inf \left\{{\inf T: T \in \mathbb T}\right\}$

as soon as one of these two quantities exists.


Proof

Suppose that $s = \inf \bigcup \mathbb T \in S$.

By Set is Subset of Union, $T \subseteq \bigcup \mathbb T$ for all $T \in \mathbb T$.

Hence by Infimum of Subset:

$\forall T \in \mathbb T: s \preceq \inf T$


Suppose now that $a \in S$ satisfies:

$\forall T \in \mathbb T: a \preceq \inf T$

Then by transitivity of $\preceq$:

$\forall t \in T: a \preceq t$

Since this holds for any $T \in \mathbb T$, also:

$\forall t \in \bigcup \mathbb T: a \preceq t$

Hence $a \preceq s$, by definition of infimum.


That is, $s = \inf \left\{{\inf T: T \in \mathbb T}\right\}$.

$\Box$


Suppose now that $r = \inf \left\{{\inf T: T \in \mathbb T}\right\} \in S$.

By definition of infimum, for all $T \in \mathbb T$ and $t \in T$:

$\inf T \preceq t$

By transitivity of $\preceq$:

$\forall T \in \mathbb T: \forall t \in T: r \preceq t$

Hence for all $t \in \bigcup \mathbb T$:

$r \preceq t$


Suppose that $a \in S$ satisfies:

$\forall t \in \bigcup \mathbb T: a \preceq t$

In particular, for any $T \in \mathbb T$, since $T \subseteq \bigcup \mathbb T$:

$a \preceq \inf T$

and therefore by definition of infimum, also:

$a \preceq r$


That is, $r = \inf \bigcup \mathbb T$.

$\blacksquare$