Prove that a solid linear subspace of a Riesz space is a Riesz subspace

Question

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?

Answer 1

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.