Definition:Pointwise Supremum

From ProofWiki
Jump to navigation Jump to search


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

Let $\left({f_i}\right)_{i \in I}, f_i: S \to T$ be an $I$-indexed collection of mappings.

Suppose that for all $s \in S$, it holds that:

$\displaystyle \sup_{i \mathop \in I} f_i \left({s}\right) \in T$

where the supremum is taken in $T$.

Then the pointwise supremum of $\left({f_i}\right)_{i \in I}$, denoted $\displaystyle \sup_{i \mathop \in I} f_i: S \to T$, is defined by:

$\displaystyle \left({\sup_{i \mathop \in I} f_i}\right) \left({s}\right) := \sup_{i \mathop \in I} f_i \left({s}\right)$

where the latter supremum is again taken in $T$.

By assumption, this supremum is guaranteed to exist.

Thence it can be seen that pointwise supremum is an instance of a pointwise operation.

Also known as

Because of the way $\displaystyle \sup_{i \mathop \in I} f_i$ is defined, there is usually no need to distinguish between the left- and right-hand side of the definition.

Thus $\displaystyle \sup_{i \mathop \in I} f_i \left({s}\right)$ is commonly used instead of $\displaystyle \left({\sup_{i \mathop \in I} f_i}\right) \left({s}\right)$.

Also defined as

Sometimes the imposition that all suprema exist in $T$ is considered too strong.

In these cases, some suitable extension of $\preceq$ to a suitable ordered set $\overline T$ may be created, in which the suprema do exist.

This $\overline T$ is then set to be the codomain of the pointwise supremum.

For example, this is done on Pointwise Supremum of Real-Valued Functions.