Intersection of Set A and a set containing empty set Give set $A = \{\emptyset, \{\emptyset\} , \{\{\emptyset\}\}\}$, is $\{\emptyset\} \cap A = \{\emptyset\}$ or just $\emptyset$?  I understand that $\emptyset \cap A = \emptyset$ (and why this is true), but I'm thrown off by the repeated nesting of empty sets within sets.
Given that, I am under the impression that $\emptyset \cup A = A =\{\emptyset\} \cup A$.  Is this also true?
 A: In general,
$$\emptyset \cup S = S \not=\{\emptyset\} \cup S$$
Note that in our case, $A =  \{\emptyset, \{\emptyset\} , \{\{\emptyset\}\}\}$, $A $ is a set of sets and $\emptyset $ is just a set with nothing. $B = \{\emptyset\}$ is also a set of sets, but happens to only contain the empty set. Then we have
$$A = \{\emptyset, B, \{B\}\} $$
And thus $A \cap B = B = \{\emptyset\}$ given that $B \subset A $.
Also, for the $A $ given, $A \cup B = A $ only because $B \subset A$. There are plenty of sets $S $ where $S \cup B \not= S$.
A: "Nothing exists" is one (tongue in cheek) way of thinking about the axiom of the empty set.
The empty set $\emptyset$ exists by fiat; which is to say (either directly as an axiom or derived from other axioms), $\exists \emptyset : \forall x, x \notin \emptyset$.
Once we have any set $x$, we have rules by which we can prove the existence of other sets; such as the set $\{x\}$ whose only element is $x$, and so on. $\emptyset$ is no different in this respect; it is a set, and we have rules that show how to construct other sets from sets whose existence we have already proven.
The union of two sets $A$ and $B$ is a set with the property $\forall x: (x \in A \cup B \iff x \in A \lor x \in B)$. So thinking about $\emptyset \cup A$, we are talking about a set:
$$\forall x: (x \in \emptyset \cup A \iff x \in \emptyset \lor x \in A)$$
But by definition, it can never be true that $x \in \emptyset$; so the above boils down to:
$$\forall x: (x \in \emptyset \cup A \iff x \in A)$$
or in other words (see axiom of extensionality); $\emptyset \cup A = A$. 
On the other hand there does exist some set $x$ s.t. $x \in \{\emptyset\}$; namely, the set $x = \emptyset$.  So when we consider $\{\emptyset\} \cup A$,we are talking about the set:
$$\forall x: (x \in \{\emptyset\} \cup A \iff x \in \{\emptyset\} \lor x \in A)$$
Thus $\{\emptyset\} \cup A$ is not necessarily equal to $A$; because (depending on $A$) we might not have $\emptyset \in A$, but we do have $\emptyset \in \{\emptyset\}$; and therefore $\emptyset \in \{\emptyset\} \cup A$.
A: Note that $\emptyset=${ } which contains no element while the set $X=${$\emptyset$} is non-empty containing an element $\emptyset$.
{$\emptyset$}$\cap A$ represents the intersection of $X$ with $A$ which is clearly the set $X$ itself as $X\subset A$
A: I think you are confused between subsets and members of a set.
$\emptyset$ is a subset of every set but it is not necessarily a member of every set but in your question set A has $\emptyset$ as a member and the set$\{\emptyset\}$ also has $\emptyset$ as a member, as they have only one member in common $A \cap \{\emptyset\}=\{ \emptyset\}$ because $A \cap B$ is the set of all common members of A and B.
Let $A \cap {\emptyset}=B$ then B is the set of all common members in set A and ${\emptyset}$ as they have no common members their intersection B is  the empty set, hence $A \cap {\emptyset}=B={\emptyset}$.
 yes $\emptyset \cup A = A =\{\emptyset\} \cup A$ is a true statement but it's not true for any set A in general.
A: Let $A=\{0,1,2\}$ and $B=\{0\}$. Now your question is (up to renaming symbols) if $A\cap B=B$ or $A\cap B =\emptyset$, which I think you can answer with ease.
