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

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?

• Just curious, what's 'lattice' in your definition? Jul 31, 2020 at 11:24
• 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. Jul 31, 2020 at 11:26
• 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). Jul 31, 2020 at 11:35
• 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. Jul 31, 2020 at 11:36
• 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 $g$. Jul 31, 2020 at 11:46

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.