Prove that the greatest lower bound of $F$ (in the subset partial order) is $\cap F$.

Question

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 ?

Answer 1

HINT: Show it directly from the definition of greatest lower bound. Suppose that $L$ is a lower bound for $F$. Then $L\subseteq X$ for each $X\in F$. What can you say about the relationship between $L$ and $\bigcap F$?

Answer 2

∩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.