# Join and Meet in Inclusion Ordered Set of Topology

## 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:

$X \cup Y, X \cap Y \in \tau$
the result holds.

$\blacksquare$