Need help understanding the formal definitions of $\cap\mathcal{F}$ and $\cup\mathcal{F}$ Some context: I'm working through Velleman's $\textit{How to Prove it}$ and am currently at section 2.3.5.
I'm having some trouble grasping the definitions of $\cap\mathcal{F}$ and $\cup\mathcal{F}$. On an intuitive level, it's fairly obvious what each one means:


*

*The intersection of a family of sets are the elements which are common to all the sets in $\mathcal{F}$. 

*The union of a family of sets are all the elements which make up the sets in $\mathcal{F}$.
This intuitive understanding doesn't seem to match up with the formal definitions provided in the book:


*

*$\cap\mathcal{F}=\{x|\forall x \in \mathcal{F}(x \in A)\}=\{x|\forall A(A \in \mathcal{F} \to x \in A)\}$

*$\cup\mathcal{F}=\{x|\exists A \in \mathcal{F}(x \in A)\}=\{x|\exists A(A \in \mathcal{F} \land x \in A)\}$
Here's how I'm parsing (obviously incorrectly) each statement in english:


*

*$\cap\mathcal{F}=$ "For every set A, if A is in $\mathcal{F}$ then the element x is in A?" This would make more sense to me if it was something along the lines of $(A \in \mathcal{F} \land x \in A)$... i.e "every set A is in $\mathcal{F}$ and a given element x is in each set A."

*$\cup\mathcal{F}=$ "For some set A, A is in $\mathcal{F}$ and the element x is in that set A." Again, this is doesn't seem to match up with my intuitive understanding of the union concept.
I would appreciate some guidance on how to correctly interpret these formal definitions! Thanks in advance.
 A: 
$∩ \mathcal F=$  "For every set $A$, if $A$ is in $\mathcal F$ then the element $x$ is in $A$?"
This would make more sense to me if it was something along the lines of $(A∈ \mathcal F ∧ x∈A)$... i.e "every set $A$ is in $\mathcal F$ and a given element $x$ is in each set $A$."

NO: it must be evident from your reading in plain English: "every set $A$ is in $\mathcal F$...". But we do not assert that every set must belong to $\mathcal F$.
We need the conditional in order to "select" only those $A$ that belong to $\mathcal F$.

Similar for union: $x \in A \cup B$ iff $x$ belongs to at least one of $A$ and $B$.
In the same way, $x \in \cup \mathcal F$ iff $x$ belongs to at least one of the elements of $\mathcal F$, i.e. iff for some set $A \in \mathcal F$: $x \in A$.
Try to use the definition to answer the question: "(for $x$ whatever) does $x$ belongs to $\cup \mathcal F$ ?"
Consider $x$: if we find an $A \in \mathcal F$ such that $x \in A$, then it is true that "$x \in A$ and $A \in \mathcal F$, for some $A$".
Thus, we can conclude with: $x \in \cup \mathcal F$.
A: 
  
*
  
*The intersection of a family of sets are the elements which are common to all the sets in $\mathcal{F}$.
  

Correct, so $x$ will be an element of this intersection iff for every $A\in\mathcal F$ we have: $x\in A$. 
In mathematical notation iff $\forall A\in\mathcal F[x\in A]$, leading to: $$\cap\mathcal F=\{x\mid \forall A\in\mathcal F[x\in A]\}=\{x\mid \forall A[A\in\mathcal F\implies x\in A]\}$$

  
*
  
*The union of a family of sets are all the elements which make up the sets in $\mathcal{F}$.
  

Correct if "make up" is interpreted as: the union contains all elements of all the sets in $\mathcal F$. This can be reworded as $x\in\cup\mathcal F$ iff for some $A\in\mathcal F$ we have: $x\in A$.
In mathematical notation iff $\exists A\in\mathcal F[x\in A]$, leading to: $$\cup\mathcal F=\{x\mid \exists A\in\mathcal F[x\in A]\}=\{x\mid \exists A[A\in\mathcal F\wedge x\in A]\}$$
A: $\cap \mathcal F~{ = \{x\mid \forall A\in \mathcal F: x\in A\} \\= \{x\mid \forall A: (A \in\mathcal F\to x\in A)\}}$
It is the set built of $x$ such that if $A$ is an element of $\mathcal F$ then $x$ is an element of $A$.
That is: "The set of elements that are in every set that are in $\mathcal F$."

$\cup \mathcal F ~{= \{x\mid \exists A\in \mathcal F: x\in A\}\\=\{x\mid \exists A: (A\in F~\wedge~x\in A)\}}$
It is the set of $x$ such that there is an $A$ in $\mathcal F$ and $x$ is in $A$.
"The set of elements that are in some set that is in $\mathcal F$."
