Bottom in Ordered Set of Topology
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $P = \struct {\tau, \subseteq}$ be an inclusion ordered set of $\tau$.
Then $P$ is bounded below and $\bot_P = \O$
Proof
By Empty Set is Element of Topology:
- $\O \in \tau$
By Empty Set is Subset of All Sets:
- $\forall A \in \tau: \O \subseteq A$
Hence $P$ is bounded below.
Thus by definition of the smallest element:
- $\bot_P = \O$
$\blacksquare$
Sources
- Mizar article YELLOR_1:23