# What does $|s_a-x|\in W$ mean where $W$ is a nbd of $0\ ?$

Theorem $3.7$ of this paper by Das & Savas.

Theorem $3.7$. Let $(L, τ)$ be a locally solid Riesz space. Let $\{s_α : α ∈ D\} , \{t_α : α ∈ D\} , \{v_α : α ∈ D\}$ be three nets such that $s_α ≤ t_α ≤ v_α$ for each $α ∈ D$. If $I_τ-\lim s_α = I_τ-\lim v_α = x_0$, then $I_τ-\lim t_α = x_0$.

Proof: Let $U$ be an arbitrary $τ$-neighborhood of zero. Choose $V, W ∈ N_{sol}$ such that $W + W ⊂ V ⊂ U$. Now by our assumption $$A = {α ∈ D : s_α − x_0 ∈ W} ∈ F(I)$$ and $$B = {α ∈ D : v_α − x_0 ∈ W} ∈ F(I).$$ Then $A ∩ B ∈ F(I)$ and for each $α ∈ A ∩ B$ $$s_α − x_0 ≤ t_α − x_0 ≤ v_α − x_0$$ and so $$\color{blue}{|t_α − x_0| ≤ |s_α − x_0| + |v_α − x_0| ∈ W + W ⊂ V}.$$ Since $V$ is $\color{blue}{solid}$ so $$\color{blue}{t_α − x_0 ∈ V ⊂ U.}$$ Hence $A ∩ B ⊂ \{α ∈ D : t_α − x_0 ∈ U\}$ which implies that $\{α ∈ D : t_α − x_0 ∈ U\} ∈ F(I)$ and this completes the proof of the theorem.

I have problem lies in the $\color{blue}{blue\ part}.$ From the definitions it is known that for the said $\alpha$,$s_\alpha - x_0\in W$ but how does that imply that $|s_\alpha - x_0|\in W.\ ?$ and what does it even mean?

• What does it mean for a neighbourhood of $0$ to be solid? (What is the definition?) – Daniel Fischer Oct 20 '16 at 20:33
• Yes. By the solidity of $W$, we have $s_{\alpha} - x_0 \in W \implies \lvert s_{\alpha} - x_0\rvert \in W$, and ditto for $v_{\alpha}$. Then $\lvert t_{\alpha} - x_0\rvert \leqslant \lvert s_{\alpha} - x_0\rvert + \lvert v_{\alpha} - x_0\rvert$ and the solidity of $V$ yield $t_{\alpha} - x_0 \in V$. – Daniel Fischer Oct 20 '16 at 20:43
$W$ and $V$ are defined to be solid sets. The definition ($S$ is solid if $y \in S, |x| \le |y| \Rightarrow x \in S$) implies that if $y \in W$ then $|y| \in W$.
If it is the "modulus" notation that you are asking about, look at definition 2.1 of that paper: one defines $|x|$ as $\max (x, -x)$. The rest follows easily.