Infimum in Ordered Subset

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

Let $R = \left({T, \preceq'}\right)$ be an ordered subset of $L$.

Let $X \subseteq T$ such that
 * $X$ admits an infimum in $L$.

Then $\inf_L X \in T$
 * $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$

Proof
By definition of ordered subset:
 * $T \subseteq S$

and
 * $\forall x, y \in T: x \preceq' y \iff x \preceq y$

Sufficient Condition
Let $\inf_L X \in T$.

By definition of infimum:
 * $\inf_L X$ is lower bound for $X$ in $L$.

By definition of $\preceq'$:
 * $\inf_L X$ is lower bound for $X$ in $R$.

We will prove that
 * $\forall x \in T: x$ is lower bound for $X$ in $R \implies x \preceq' \inf_L X$

Let $x \in T$ such that
 * $x$ is lower bound for $X$ in $R$.

By definition of $\preceq'$:
 * $x$ is lower bound for $X$ in $L$.

By definition of infimum:
 * $x \preceq \inf_L X$

Thus by definition of $\preceq'$:
 * $x \preceq' \inf_L X$

Hence $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$

Necessary Condition
assume that
 * $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$

By definition of infimum:
 * $\inf_R X \in T$

Thus by assumption:
 * $\inf_L X \in T$