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In D.H. Fremlin, Topological Riesz Spaces and Measure Theory [14F] it is stated that

Let $E$ be a Riesz space. A Riesz subspace of $E$ is a linear subspace which is also a sublattice. A solid linear subspace is always a Riesz subspace.

Recall that if $E$ is a Riesz subspace (an ordered linear space which is also a lattice), a subset $F$ of $E$ is said to be solid if, for every pair $x,y$ of elements of $E$, the following implication is satisfied: $$ 0 \le x \le y \in F \Rightarrow x \in F. $$

Question:

How can I prove that a solid linear subspace of a Riesz space is a Riesz subspace?

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    $\begingroup$ Just curious, what's 'lattice' in your definition? $\endgroup$ Jul 31, 2020 at 11:24
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    $\begingroup$ Lattice is a non-empty set with endowed with a partial order $\le$ for which for every couple $x,y$ of elements of $E$ the set $\{x,y\}$ has supremum and infimum. $\endgroup$
    – Logos
    Jul 31, 2020 at 11:26
  • $\begingroup$ So Riez space seems to be a vector space given an 'order'. And the supremum and infimum seem not to have to be contained in the set of couples (x,y). $\endgroup$ Jul 31, 2020 at 11:35
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    $\begingroup$ A Riesz space is an ordered linear space (order is compatible with linear structure) for which every pair of elements have a supremum and infimum, which is not necessarily one of them. $\endgroup$
    – Logos
    Jul 31, 2020 at 11:36
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    $\begingroup$ Yes, think of the space $\Bbb{R}^E$ of all the functions defined on $E$ with values ins $\Bbb{R}$. This is a Riesz space and if you consider $E = [0,1]$, the sup of $f(x) = x$ and $g(x) = 1-x$ is neither $f$ nor $g$. $\endgroup$
    – Logos
    Jul 31, 2020 at 11:46

1 Answer 1

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With the definition of solid subspace given, the statement is false, unless it is given also that the linear subspace $F$ is a Riesz space with the order induced by $E$.

With this extra hypothesis, if $x$ and $y$ are two elements of $F$, there are two least upper bounds: one due to the order structure of $E$, viz. $\sup_E \{x,y\}$ and one on $F$ given by the induced order, viz. $\sup_F \{x,y\}$.

Let us notice, also, that $\sup_E\{x,y\} \le \sup_F\{x,y\}$ because $F \subseteq E$. As a result, since $0 \in F$, it follows that $$ 0 \le \sup\nolimits_E \{x,0\} \le \sup\nolimits_F \{x,0\} \in F $$ for all $x \in F$. Thus $x^+ = \sup_E \{x,0\} \in F $ because $F$ is solid.

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