# A confused question about a partially ordered set

Let $X$ be a nonempty set and $\mathbb{F}$ be the collection of all extended nonnegative-valued functions $f \colon X \to [0, +\infty]$.

I am wondering that when this set $\mathbb{F}$ is equipped with the usual pointwise ordering $\geq$, then is this $(\mathbb{F}, \geq)$ a partially ordered set?

I know that the collection of all nonnegative real-valued functions $g \colon X \to \mathbb{R}_+$ is a partially ordered set when the ordering is defined pointwise. However, I am not quite sure for the case of extended real-valued functions space.

Could anyone help me out please? Any idea or suggestions are much appreciated! Thank you very much in advance!

• What are the properties we require of a relation in order to call it a partial order? – Arthur Jul 8 '18 at 9:49
• Thanks @Arthur . To be a partial order , we require a binary relation on a set is reflexive, transitive and antisymmetric. Many thanks again:) – Paradiesvogel Jul 8 '18 at 9:55

## 1 Answer

Fact:

If $(Y, \le_Y)$ is a partially ordered set, $X$ is a set and on $\mathbb{F} = \{f: X \to Y\}$ we define $$f \le_F g \iff \forall x \in X: f(x) \le_Y g(x)$$ then $(\mathbb{F}, \le_F)$ is also a partially ordered set.

The proof is just plugging in the definitions, nothing fancy.

And both $[0,+\infty)$ and $[0,+\infty]$ are partially ordered sets in their natural order (of course we define $x \le \infty$ for all $x$ in the second case).

So both function sets are partially ordered, using the fact.

• Thank you @Henno Brandsma. It is a huge help. :-) Thanks – Paradiesvogel Jul 8 '18 at 10:00
• @Paradiesvogel Glad to be able to help. Probably (eastern) Dutch by the nickname? ("paradysfûgel" I would have said in Frisian). – Henno Brandsma Jul 8 '18 at 10:15
• Yeah, you are right! Glad to meet you :-) – Paradiesvogel Jul 8 '18 at 10:21