Let E be a non-empty set. Let's consider the inclusion relation over P(E) : (∀x, y ∈ P(E))(X ≤ Y ⇔ X ⊂ Y ). Show that it is an order relation

Question

Let E be a non-empty set. Let's consider the inclusion relation over P(E): (∀x, y ∈ P(E))(X ≤ Y ⇔ X ⊂ Y )

(a) Show that it is an order relation. (b) Show that it is total if, and only if, E = {a}

I know that I have to prove that it is reflexive, antisymmetric and transitive, but how do I do that? And how do I show that it is total when E = {a}?

Answer 1

If $x \leq y$ and $y \leq z$ then $x \subseteq y$ and $y \subseteq z$. That is, every member in $x$ is a member of $y$. But every member of $y$ is also a member of $z$, hence every member of $x$ is a member of $z$, and so $x \leq z$ means this relation is transitive. Of course, it is reflexive since every set is a subset of itself. It is antisymmetric by the Axiom of Extensionality.

Now, suppose the relation is total. That is, for any members $x$ and $y$ of $E$, either $x \leq y$ or $y \leq x$. If $E$ is not of the form $\{a\}$ then there exists distinct elements $a, b \in E$ since $E$ is non-empty and then neither $\{a\} \leq \{b\}$ nor $\{b\} \leq \{a\}$, a contradiction. So $E$ is a singleton. Likewise, if $E$ is a singleton then trivially $P(E)$ is linearly ordered as $\varnothing \leq \{a\}$.