Lower Closure of Subset is Subset of Lower Closure

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $X, Y$ be subsets of $S$.


Then

$X \subseteq Y \implies X^\preceq \subseteq Y^\preceq$

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


Proof

Let $X \subseteq Y$.

Let $x \in X^\preceq$.

By definition of lower closure of subset:

$\exists y \in X: x \preceq y$

By definition of subset:

$y \in Y$

Thus by definition of lower closure of subset:

$x \in Y^\preceq$

$\blacksquare$


Sources