What does |sa−x|∈W|s_a-x|\in W mean where WW is a nbd of 0 ?0\ ?

Theorem 3.73.7 of this paper by Das & Savas.

Theorem 3.73.7. Let (L,د„)(L, د„) be a locally solid Riesz space. Let {sخ±:خ±âˆˆD},{tخ±:خ±âˆˆD},{vخ±:خ±âˆˆD}\{s_خ± : خ± ∈ D\} , \{t_خ± : خ± ∈ D\} , \{v_خ± : خ± ∈ D\} be three nets such
that s_خ± ≤ t_خ± ≤ v_خ±s_خ± ≤ t_خ± ≤ v_خ± for each خ± ∈ Dخ± ∈ D. If I_د„-\lim s_خ± = I_د„-\lim v_خ± = x_0I_د„-\lim s_خ± = I_د„-\lim v_خ± = x_0, then I_د„-\lim t_خ± = x_0I_د„-\lim t_خ± = x_0.

Proof: Let UU be an arbitrary د„د„-neighborhood of zero. Choose V, W ∈ N_{sol}V, W ∈ N_{sol} such that W + W âٹ‚ V âٹ‚ UW + W âٹ‚ V âٹ‚ U.
Now by our assumption
A = {خ± ∈ D : s_خ± − x_0 ∈ W} ∈ F(I)A = {خ± ∈ D : s_خ± − x_0 ∈ W} ∈ F(I)
and
B = {خ± ∈ D : v_خ± − x_0 ∈ W} ∈ F(I).B = {خ± ∈ D : v_خ± − x_0 ∈ W} ∈ F(I).
Then A ∩ B ∈ F(I)A ∩ B ∈ F(I) and for each خ± ∈ A ∩ Bخ± ∈ A ∩ B
s_خ± − x_0 ≤ t_خ± − x_0 ≤ v_خ± − x_0s_خ± − x_0 ≤ t_خ± − x_0 ≤ v_خ± − x_0
and so \color{blue}{|t_خ± − x_0| ≤ |s_خ± − x_0| + |v_خ± − x_0| ∈ W + W âٹ‚ V}.\color{blue}{|t_خ± − x_0| ≤ |s_خ± − x_0| + |v_خ± − x_0| ∈ W + W âٹ‚ V}. Since VV is \color{blue}{solid}\color{blue}{solid} so
\color{blue}{t_خ± − x_0 ∈ V âٹ‚ U.}\color{blue}{t_خ± − x_0 ∈ V âٹ‚ U.}
Hence A ∩ B âٹ‚ \{خ± ∈ D : t_خ± − x_0 ∈ U\}A ∩ B âٹ‚ \{خ± ∈ D : t_خ± − x_0 ∈ U\} which implies that \{خ± ∈ D : t_خ± − x_0 ∈ U\} ∈ F(I)\{خ± ∈ D : t_خ± − x_0 ∈ U\} ∈ F(I) and this completes the
proof of the theorem.

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

Please help me understand this. Thank you.

=================

  

 

What does it mean for a neighbourhood of 00 to be solid? (What is the definition?)
– Daniel Fischer♦
Oct 20 at 20:33

  

 

Oh I know. It’s because they are solid neighbourhoods.Right? @DanielFischer
– user118494
Oct 20 at 20:33

  

 

Yes. By the solidity of WW, we have s_{\alpha} – x_0 \in W \implies \lvert s_{\alpha} – x_0\rvert \in Ws_{\alpha} – x_0 \in W \implies \lvert s_{\alpha} – x_0\rvert \in W, and ditto for v_{\alpha}v_{\alpha}. Then \lvert t_{\alpha} – x_0\rvert \leqslant \lvert s_{\alpha} – x_0\rvert + \lvert v_{\alpha} – x_0\rvert\lvert t_{\alpha} – x_0\rvert \leqslant \lvert s_{\alpha} – x_0\rvert + \lvert v_{\alpha} – x_0\rvert and the solidity of VV yield t_{\alpha} – x_0 \in Vt_{\alpha} – x_0 \in V.
– Daniel Fischer♦
Oct 20 at 20:43

=================

2 Answers
2

=================

WW and VV are defined to be solid sets. The definition (SS is solid if y \in S, |x| \le |y| \Rightarrow x \in Sy \in S, |x| \le |y| \Rightarrow x \in S) implies that if y \in Wy \in W then |y| \in W|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||x| as \max (x, -x)\max (x, -x). The rest follows easily.