A question in maximal element in an ordering in which I am confused.

Question

This is an example on page $14$ of Algebra by Thomas Hungerford.

$0, 3,4,5,6,...,-1, -2, -3..., 1,2.$

$2$ is a maximal element in this ordering.

Definition of maximal element: If ($A$, $\leq$) is a poset. An element $a\in A$ is maximal in $A$ if for every $c \in A$ which is comparable to $a$ , $c\leq a$.

So, How is $2$ maximal in this ordering as $3,4$ are greater than $2$ ?

Can anyone please explain?

Edit:

Answer 1

The ordering in this example is not the usual order on $\mathbb{Z}$. The author defined a completely different order. Let's call it $<^*$.

$(1)$ $0$ is the first element.

$(2)$ The next elements are the elements of $\{n\in\mathbb{N}: n\ge 3\}$, they are ordered in the usual $<$ ordering.

$(3)$ The next elements are the negative integers, ordered by the rule $n<^*m$ iff $|n|<|m|$.

$(4)$ Finally, $1$ and $2$ are the last elements with $1<^*2$.

Answer 2

He defines an order on $\Bbb Z$ such that $2$ is the maximal element, simply by naming it last and having nothing following it. The elements on the $\ldots$ can be filled in by the reader, e.g. it's clear that in this new order $0$ is the smallest , followed directly by $3$ and then all other positive integers $\ge 4$ in order, but any negative number is larger then all of them, but among the negative numbers $-1$ is smallest, then $-2$ etc. (in order of absolute value). $1 < 2$ are mentioned as last and so are the largest. It's just an informal way of defining a well-order of type $\omega+ \omega+2$ (if you do set set theory you'll learn about ordinal numbers..).