Equivalence of Definitions of Upper Section

Theorem
The definitions of upper set are equivalent. Specifically:

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $U \subseteq S$.

Then the following are equivalent:

where $U^\succeq$ is the upper closure of $U$.

$(1)$ implies $(2)$
Suppose that:


 * $\forall u \in U: \forall s \in S: u \preceq s \implies s \in U$

Let $k \in U^\succeq$.

Then by the definition of upper closure, there is some $u \in U$ such that $u \preceq k$.

Since $k \in U^\succeq \subseteq S$, the premise proves that $k \in U$.

Since this holds for all $k \in U^\succeq$, it follows that:
 * $U^\succeq \subseteq U$

$(2)$ implies $(3)$
Suppose that $U^\succeq \subseteq U$.

Let $u \in U$.

Then since $U \subseteq S$, $u \in S$ by the definition of subset.

Since $\preceq$ is reflexive:
 * $u \preceq u$

Thus by the definition of upper closure:
 * $u \in U^\succeq$.

Since this holds for all $u \in U$:
 * $U \subseteq U^\succeq$

Thus by definition of set equality:
 * $U^\succeq = U$

$(3)$ implies $(1)$
Suppose that $U^\succeq = U$.

Let $u \in U$.

Let $s \in S$.

Let $u \preceq s$.

Then by the definition of upper closure, $s \in U$.

Thus we have shown that:
 * $\forall u \in U: \forall s \in S: u \preceq s \implies s \in U$