Why $\mathcal A$ is not bounded below?

Question

This is an example of Amann analysis book, what I dont understand clearly:

Let $X$ be a set with at least two elements and $\mathcal X:=\mathcal P (X)\setminus\{\emptyset\}$ with the inclusion order. Suppose further that $A$ and $B$ are nonempty disjoint subsets of $X$ and $\mathcal A:=\{A,B\}$.

Then $\mathcal A\subseteq\mathcal X$ and $\sup(\mathcal A)=A\cup B$, but $\mathcal A$ has no maximum, and $\mathcal A$ is not bounded below. In particular $\inf(\mathcal A)$ does not exist.

What I cant see is why the empty set is not the infimum of $\mathcal A$, because we have that $\emptyset \subseteq A$ and $\emptyset\subseteq B$, i.e.

$$\emptyset\subseteq x,\forall x\in \mathcal A$$

Then, what is wrong in my interpretation? Thank you in advance.

Answer 1

By fiat (first sentence of your quote), the empty set $\emptyset$ is assumed to not be an element of $\mathcal{X}$, your partial order.

Answer 2

The empty set would serve as the infimum of $\mathcal{A}$, and the union $A \cup B$ serves as an upper bound of both $A$ and $B$. But since $A \nleq B$ and $B \nleq A$, no maximum exists in $\mathcal{A}$ (though it can be found in $\mathcal{X}$). And since $\emptyset \not\in \mathcal{X}$, the infimum cannot be found in $\mathcal{X}$ at all.

Remember that $\mathcal{X}$ is not an algebra, so it isn't bound to be closed under complements, etc., and that's partially the point of the example.