Bottom in Ordered Set of Topology

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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

By Empty Set is Element of Topology:

$\varnothing \in \tau$

By Empty Set is Subset of All Sets:

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

Hence $P$ is bounded below.

Thus by definition of the smallest element:

$\bot_P = \varnothing$

$\blacksquare$


Sources