Bounded strictly totally ordered semigroup

Question

Is it possible that a strictly totally ordered ($<$) infinite algebraic structure has both maximum and minimum?

There is an example of a strictly totally ordered infinite magma bounded from two sides:
the interval of real numbers $[0,1]$ with operation $x \cdot y = (x + y)/2$.
The operation is compatible with the natural order of the segment.
However, the operation is not associative.

I am looking for an example of a strictly totally ordered infinite semigroup bounded from two sides. I assume the operation on the semigroup is compatible with the order:
$a < b \implies a \cdot c < b \cdot c$ and $a < b \implies c \cdot a < c \cdot b$ for any elements $a$, $b$, $c$ of the semigroup.

Answer 1

I claim that a strictly totally ordered semigroup with minimum and maximum element can't have more than one element.

Call the minimum element $0$ and the maximum element $1$. From the strict total ordering it follows that $$y\cdot1=0\cdot x\implies x=1,\ y=0.$$ This is because $$x\lt1\implies0\cdot1\le y\cdot1=0\cdot x\lt0\cdot1$$ and $$0\lt y\implies0\cdot1\lt y\cdot1=0\cdot x\le0\cdot1.$$

Now by associativity we have $$(0\cdot0)\cdot1=0\cdot(0\cdot1)\implies0\cdot1=1$$ and $$(0\cdot1)\cdot1=0\cdot(1\cdot1)\implies0\cdot1=0,$$ so $$1=0\cdot1=0.$$

Answer 2

EDIT. This was an answer to the original question, without the added condition on the order. I let it as it is to clarify the discussion below.

$({\Bbb N} \cup \{+\infty\}, +)$ is an example of a strictly totally ordered and bounded infinite semigroup. But it is not idempotent, contrary to your claim.