# Set is Subset of Lower Closure

## Theorem

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

Let $X$ be a subset of $S$.

Then $X \subseteq X^\preceq$

where $X^\preceq$ denotes the lower closure of $X$.

## Proof

Let $x \in X$.

By definition of reflexivity:

$x \preceq x$

Thus by definition of lower closure:

$x \in X^\preceq$

$\blacksquare$