Show that if $F = \emptyset$, then the statement $x\in\bigcap F$ will be true no matter what $x$ is Show that if $F = \emptyset$, then the statement $x\in\bigcap F$ will be true no
matter what $x$ is
I know that $x\in\bigcap F = \forall A \in F, x\in A$ 
But how can $x$ be in any set, much less 'all' sets in $F$ when $F$ it has no sets in it?
I know that  $x\in\bigcup F$ is false because it means that $\exists A \in F$ so that $x \in A$ and that can't be true because no set 'exists' in $F$.
So how can $x\in\bigcap F$ be always true if there is no set in $F$ in which $x$ can be?
 A: The statement
$$x\in\bigcap F$$
means
$$\hbox{for all $S\in F$ we have $x\in S$}.\tag{$*$}$$
Now, what is the truth value of a statement
$$\hbox{for all $S\in F$,}\ldots\langle\hbox{whatever}\rangle$$
when $F$ is the empty set?  The accepted answer is that such a statement is always true, regardless of the "whatever".
To some extent this is a convention, and serves as part of our (mathematical) understanding of the word "all".  It has not always been so: if you look at mathematical and philosophical sources from the 19th century, you can find a good deal of discussion on this.  But this is the way it is taken nowadays, and probably the main reason is the following.
Assume for the time being that $F$ is not empty.  Then hopefully it is clear that the two statements
$$\hbox{for all $S\in F$, statement $X$ is true}\tag1$$
and
$$\hbox{there is no $S\in F$ for which statement $X$ is false}\tag2$$
are logically equivalent: that is, they are really just two ways of saying the same thing.  Now $(2)$ is clearly true when $F$ is the empty set (there is no $S$ in $F$ at all, so there is certainly no $S$ in $F$ for which statement $X$ is false).  So if we want the equivalence of $(1)$ and $(2)$ to also hold for the empty set - that is, we want to treat the empty set as "nothing special" - then we have to agree that $(1)$ is also true when $F$ is empty.
In terms of your question, the conclusion of all this is that if $F$ is empty, then $(*)$ is true, regardless of the value of $x$.
Hope this helps!
A: What you have encountered is the vacuous truthiness of an empty universe.
A universal statement is falsified if there exists a counterexample within the domain of discourse.   However, there are no counterexamples within an empty set, so whatever is asserted about all its members is true.
$$\forall y\in \emptyset ( P(y) ) \iff \neg \exists x\in \emptyset (P(y))$$
Or in your case $\forall A \in \emptyset (x\in A) \iff \neg\exists A\in \emptyset (x\notin A)$
There are no sets in the empty set of which $x$ is not a member, so $x$ is a member of all sets in the emptyset.        
