Prove that the greatest lower bound of $F$ (in the subset partial order) is $\cap F$. This is one of the question I'm working on:

Suppose $A$ is a set, $F \subseteq \mathbb{P(A)}$, and $F \neq
 \emptyset$. Then prove that the greatest lower bound of $F$ (in the
   subset partial order) is $\cap F$.

Now this is my attempt at this problem:
We know that $\cap F$ is a lower bound of F since $\forall X \in F (\cap F \subseteq$ X). Now we need to prove that this is the greatest lower bound of $F$. 
Now I'm stuck here. How to show that it is the greatest lower bound ?

∩F is the smallest element in the subset partial order ( as you pointed out  ∀X∈F(∩F⊆ X)). Therefore it is the greatest lower bound.
