Ordered Set of Closure Systems is Ordered Set

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

Then $\operatorname{ClSystems}\left({L}\right)$ is an ordered set,

where $\operatorname{ClSystems}\left({L}\right)$ denotes the ordered set of closure systems.

Proof
By definition of ordered set of closure systems:
 * $\operatorname{ClSystems}\left({L}\right) = \left({X, \precsim}\right)$

where
 * $X$ is the set of all closure systems of $L$,
 * dor all closure systems $s_1 = \left({T_1, \preceq_1}\right), s_2 = \left({T_2, \preceq_2}\right)$ of $L$: $s_1 \precsim s_2 \iff T_1 \subseteq T_2$

Reflexivity
Let $s = \left({T, \preceq'}\right)$ be a closure system.

By Set is Subset of Itself:
 * $T \subseteq T$

By definition of $\precsim$:
 * $s \precsim s$

Thus by definition
 * $\precsim$ is reflexive.

Transitivity
Let $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$, $s_3 = \left({T_3, \preceq_3}\right)$ be closure systems such that
 * $s_1 \precsim s_2$ and $s_2 \precsim s_3$

By definition of $\precsim$:
 * $T_1 \subseteq T_2$ and $T_2 \subseteq T_3$

By Subset Relation is Transitive:
 * $T_1 \subseteq T_3$

By definition of $\precsim$:
 * $s_1 \precsim s_3$

Thus by definition:
 * $\precsim$ is transitive.

Antisymmetry
Let $s_1 = \left({T_1, \preceq_1}\right)$, $s_2 = \left({T_2, \preceq_2}\right)$ be closure systems such that
 * $s_1 \precsim s_2$ and $s_2 \precsim s_1$

By definition of $\precsim$:
 * $T_1 \subseteq T_2$ and $T_2 \subseteq T_1$

By definition of set equality:
 * $T_1 = T_2$

By definition of ordered subset:
 * $\mathord\preceq_1 = \mathord\preceq \cap \left({T_1 \times T_1}\right) = \mathord\preceq_2$

Then
 * $s_1 = s_2$

Thus by definition:
 * $\precsim$ is antisymmetric

So, by definition:
 * $\precsim$ is ordering.

Hence $\operatorname{ClSystems}\left({L}\right)$ is an ordered set,