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?
Please help me understand this. Thank you.
$W$ 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 $|y| \in W$.