Lower Set with no Greatest Element is Open in GO-Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space.

Let $L$ be a lower set in $S$ with no greatest element.


Then $L$ is open in $\left({S, \preceq, \tau}\right)$.


Proof

By Maximal Element in Toset is Unique and Greatest, $L$ has no maximal element.

By Lower Set with no Maximal Element:

$\displaystyle L = \bigcup \left\{{l^\prec: l \in L }\right\}$

where $l^\prec$ is the strict lower closure of $l$.

By Open Ray is Open in GO-Space and the fact that a union of open sets is open, $L$ is open.

$\blacksquare$


Also see