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:
 * $X \cup Y, X \cap Y \in \tau$

Thus by Join in Inclusion Ordered Set ans Meet in Inclusion Ordered Set:
 * the result holds.