Prove that for every nonempty family of F of sets,$\cap_{A \in F}\subseteq \cup_{A \in F}$

Question

Prove that for every nonempty family of F of sets,$\cap_{A \in F}\subseteq \cup_{A \in F}$.

I am very confused with this notation and cannot figure out how to prove this. Any help is much appreciated.

Answer 1

There are two sets involved in this problem.

The first one is $$ \cap_{A \in F}$$ and the second on is $$\cup_{A \in F}$$

What you want to show is that the first set is a subset of the second set.

In order to prove that we need to show that every element of $$ \cap_{A \in F}$$ is also an element of $$\cup_{A \in F}$$

Let $$x\in \cap_{A \in F}$$ Then for every $ A \in F$, $x\in A $

Pick an arbitrary $A \in F$.

We know that $$ A\subseteq \cup_{A \in F}$$

Since $x\in A$ and $ A\subseteq \cup_{A \in F}$ we have $$x\in \cup_{A \in F}$$

Which proves $$\cap_{A \in F}\subseteq \cup_{A \in F}$$