# Bottom in Ordered Set of Topology

Jump to navigation Jump to search

## Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $P = \left({\tau, \subseteq}\right)$ be an inclusion ordered set of $\tau$.

Then $P$ is bounded below and $\bot_P = \varnothing$

## Proof

$\varnothing \in \tau$
$\forall A \in \tau: \varnothing \subseteq A$

Hence $P$ is bounded below.

Thus by definition of the smallest element:

$\bot_P = \varnothing$

$\blacksquare$