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If $A$ is a linear subspace of the Riesz space $E$, then $A=A^{dd}$.

Definition. Let $E$ be a Riesz space. The elements $f$ and $g$ in $E$ are said to be disjoint, we write $f\perp g$, if $|f| \wedge |g|=0$.

Definition. Let $E$ be a Riesz space and $A \subseteq E$. The disjoint complement of $A$ is defined to be $$A^d = \{f \in E: f \perp g, \forall g \in A\}.$$

I know what I should to do for the proof that $A^{dd} \subseteq A$. What I want to know is how to prove that $A \subseteq A^{dd}$. The proof in the book said that the proof is evident. But, it doesn't look clear to me. The proof is as follows: By definition, we have that $$A^{dd} = \{f \in E: f \perp h, \forall h \in A^d \}.$$ Since $A \perp A^d$, then $A \subseteq A^{dd}$.

Thanks in advanced for better explanation.

lap lapan
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  • Something's odd with the title: the equality $A = A^{dd}$ is not true for general linear subspaces $A$, but if and only if $A$ is a band. – Jochen Glueck May 16 '22 at 11:47
  • What if $A$ is a linear subspace? – lap lapan May 16 '22 at 12:53
  • For linear subspaces - and more generally for all sets $A$ - one has the inclusion $A \subseteq A^{dd}$. As I said, equality is true if and only if $A$ is a band. (I would even use this as the definition of a band - but there are alternative definitions which are equivalent). – Jochen Glueck May 16 '22 at 12:59

1 Answers1

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Just use the definition. Since $A \perp A^d$ and $A^{dd}$ is a set of all $f \in E$ for which $f \perp h$ for all $h \in A^d$, it follows that all elements in $A$ is contained in $A^{dd}$. Thus, $A \subseteq A^{dd}$. To see this in another way, consider the case by look the Venn's diagram approach and of course, use the definition. Since $A^d$ is a set of all elements $f\in E$ for which $f \perp g$ for all $g \in A$, it follows that $A$ is contained in $A^d$. Similarly, $A^d$ is contained in $A^{dd}$. Hence, $A \subseteq A^{dd}$.

Hope it helps.

lap lapan
  • 2,188