How to show $c-b\lt b-a$

Question

The question:

Let $G$ be an Arf semigroup and $a\lt b\lt c$ be three consecutive elements in $G$. How to show that $c-b\lt b-a$ and how to show that this is not necessarily the case for every semigroup.

Note

the definition of arf semigroup

let $W=\{i_0 =0, i_1,...\}$ be a semigroups of nonnegative integers, where $i_1\lt i_2\lt ...$. Then $W$ is an Arf semigroup if the collection $i_h-i_h=0$, $i_{h+1}-i_h$, $i_{h+2}-i_h$,... Is a semigroup for all choices of $h\ge 0$

What I understand from the question:

$a\lt b\lt c$ be three consecutive elements in $G$. That's, the only element of $G$ in the open real interval $(a,c)$ is $b$

And in order to show $c-a\lt b-a$, I need to show the elements of $G$ get closer.

But, I cannot do these. Please help me solving this question. Thank you:)

Answer 1

This is false and it is irrelevant that $G$ has any structure at all (except that it's infinite).

$G = \{ i_0, i_1, \dots \} \subseteq {\mathbb N}$ with $i_0 < i_1 < i_2 < \dots$ and the questions asks to show that $i_1 - i_0 > i_2 - i_1 > i_3 - i_2 > \dots$. But this is impossible, as these are all natural numbers.