Join and Meet in Inclusion Ordered Set of Topology
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Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $L = \left({\tau, \preceq}\right)$ be an inclusion ordered set of $\tau$.
Let $X, Y \in \tau$.
Then $X \vee Y = X \cup Y$ and $X \wedge Y = X \cap Y$
Proof
By definition of topological space:
- $X \cup Y, X \cap Y \in \tau$
Thus by Join in Inclusion Ordered Set ans Meet in Inclusion Ordered Set:
- the result holds.
$\blacksquare$
Sources
- Mizar article WAYBEL14:18