Lattice and Ring of Real-Valued Functions forms Ordered Ring

From ProofWiki
Jump to navigation Jump to search


This article needs proofreading.
Please check it for mathematical errors.
If you believe there are none, please remove {{Proofread}} from the code.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Proofread}} from the code.


Theorem

Let $\struct {S, \tau_{_S} }$ be a topological space.

Let $\R$ denote the real number line.


Let $\struct{\R^S, +, \times}$ be the ring of real-valued functions from $S$ to $\R$.

Let $\struct{\R^S, \vee, \wedge, \le}$ be the lattice of real-valued functions from $S$ to $\R$.


Then:

$\struct{\R^S, +, \times, \le}$ is an ordered ring


Proof

From Structure Induced by Group Operation is Group:

the zero of $\struct{\R^S, +, \times}$ is the constant mapping $0_{R^S} : S \to R$ defined by:
$\forall s \in S : \map {0_{R^S}} s = 0_R$


It needs to be shown that the order $\le$ on the lattice of real-valued functions satisies the ring compatible ordering axioms:

\((\text {OR} 1)\)   $:$   $\le$ is compatible with $+$:      \(\ds \forall f, g, h \in \R^S:\)    \(\ds f \le g \)   \(\ds \implies \)   \(\ds \paren {f + h} \le \paren {g + h} \)      
\((\text {OR} 2)\)   $:$   Product of Positive Elements is Positive      \(\ds \forall f, g \in \R^S:\)    \(\ds 0_{\R^S} \le f, 0_{\R^S} \le g \)   \(\ds \implies \)   \(\ds 0_{\R^S} \le f g \)      


$\paren{\text {OR} 1}$

Let $f, g, h \in \R^S$.

Let $f \le g$.

Hence:

$\forall s \in S: \map f s \le \map g s$


We have:

\(\ds \forall s \in S: \, \) \(\ds \map {\paren{f + h} } s\) \(=\) \(\ds \map f s + \map h s\) pointwise addition
\(\ds \) \(\le\) \(\ds \map g s + \map h s\) Real Numbers form Ordered Field
\(\ds \) \(=\) \(\ds \map {\paren{g + h} } s\) pointwise addition


By definition of lattice of real-valued functions:

$\paren {f + h} \le \paren {g + h}$


Since $f, g$ and $h$ were arbitrary:

$\forall f, g, h \in \R^S : f \le g \implies \paren {f + h} \le \paren {g + h}$

$\Box$

$\paren{\text {OR} 2}$

Let $f, g \in \R^S$.

Let $0_{\R^S} \le f, 0_{\R^S} \le g$.

Hence:

$\forall s \in S: 0 \le \map f s, 0 \le \map g s$


We have:

\(\ds \forall s \in S: \, \) \(\ds 0\) \(\le\) \(\ds \map f s \map g s\) Real Numbers form Ordered Field
\(\ds \) \(=\) \(\ds \map {\paren{fg} } s\) pointwise multiplication


By definition of lattice of real-valued functions:

$0_{\R^S} \le f g$


Since $f$ and $g$ were arbitrary:

$\forall f, g \in \R^S : 0_{\R^S} \le f, 0_{\R^S} \le g \implies 0_{\R^S} \le f g$

$\Box$


Thus the ring compatible ordering axioms are seen to hold for the order $\le$ on the lattice of real-valued functions.

$\blacksquare$


Sources

1960: Leonard Gillman and Meyer Jerison: Rings of Continuous Functions: Chapter $1$: Functions on a Topological Space, $\S 1.2$

Retrieved from "https://proofwiki.org/w/index.php?title=Lattice_and_Ring_of_Real-Valued_Functions_forms_Ordered_Ring&oldid=662717"