What is the logic-level interpretation of $rConsider the two following statements, where $\epsilon$ is a positive real number and $a,b \in \mathbb R$.

*

*$|a-b| \leq \epsilon$


*$|a-b| \geq \epsilon$
The absolute value function $|*|$ is defined as $$
|x|=
\begin{cases}
-x&\text{if } x < 0\\
x&\text{if } x\geq 0
\end{cases}
$$

Applying the definition of the absolute value function to the first statement (1), we produce two sub-statements:
A1) $a-b \geq 0 \implies a-b \leq \epsilon$
B1) $a-b \lt 0 \implies -(a-b) \leq \epsilon \iff a-b \geq -\epsilon$
In the same manner, we can produce two sub-statements for the second statement (2):
A2) $a-b \geq 0 \implies a-b \geq \epsilon$
B2) $a-b \lt 0 \implies -(a-b) \geq \epsilon \iff a-b \leq -\epsilon$

In the case of A1 and B1, I often see these two sub-statements combined in the following manner:
$-\epsilon \leq a-b \leq \epsilon$
In the case of A2 and B2, I have only seen the phrase: "Either $a-b \geq \epsilon \ $ OR $\ a-b \leq -\epsilon$."
I have always interpreted a structure like $r<s<t$ to mean that $s$ is greater than $r$ AND $s$ is less than $t$. However, should I instead be reading "$-\epsilon \leq a-b \leq \epsilon$" as $a-b$ is less than or equal to $\epsilon$ OR $a-b$ is greater than or equal to $-\epsilon$...such as is done with the second statement?
I can visualize the picture of the solution sets for the first statement (1) being "continuous" (i.e. the two solution sets for $a-b \geq 0$ and $a-b \lt 0$ touch each other )...and I can visualize the picture of the solution sets for the second statement (2) being disjoint (i.e. the two solution sets do not touch each other and are separated by $2\epsilon$)...but I'm not really sure if this is useful.
My question is thus: At the logic level, how should I interpret the inequality: $r<s<t$?
 A: Notation like "$u<v<w$" does indeed mean a conjunction of inequalities. But this doesn't cause any tension with $(2)$: the expression in $(2)$ simply can't be put in that form, because it doesn't describe the right "shape."
I'm going to get rid of the $a-b$ stuff in the interests of simplicity. Let's just consider the statements "$\vert x\vert\le y$" and "$\vert x\vert \ge y$," which I'll call "$(1)$" and "$(2)$" respectively.
We have, assuming $y>0$, that:

*

*$(1)$ is equivalent to "if $x\ge 0$ then $x\le y$ and if $x<0$ then $x\ge -y$."


*$(2)$ is equivalent to "if $x\ge 0$ then $x\ge y$ and if $x<0$ then $x\le -y$."
You've correctly observed this. However, despite their apparent similarity they do not actually behave similarly.
To see this, let's write them in terms of intervals:

*

*$(1)$ is equivalent to "Either $x\in [0,y]$ or $x\in [-y,0)$."


*$(2)$ is equivalent to "Either $x\in [y,\infty)$ or $x\in (-\infty, -y]$."
That is, each of $(1)$ and $(2)$ picks out a "possibility set" for $x$ in terms of $y$. But these two possibility sets are quite different: the former is $[0,y]\cup [-y,0)=[-y,y]$, whereas the latter is $(-\infty, -y]\cup[y,\infty)$ which is not simplifiable to a single interval. Notation like "$u\le v\le w$" says "$v$ is in the interval $[u,w]$," but that notation isn't applicable in $(2)$ since the possibility set isn't an interval.
