Show that $\cup \emptyset =\emptyset$ We know these for the question:
A1. Axiom of empty set
A2. Axiom of extensionality
A3. Axiom of union of sets
Definition. $Sx=\{x\}\cup {x}$.
(Sx: successor)
My question is: How can I prove $\cup \emptyset =\emptyset$?
 A: $$\bigcup\varnothing=\{x\;|\;\exists y(y\in\varnothing\wedge x\in y)\}$$
Now, think about the property that defines the previous class:
$$P(x;y)=\exists y(y\in\varnothing\wedge x\in y)$$
It is clearly false, because there exists no set $y$ such that $y\in\varnothing$. Thus the previous class is defined by a property that is always false, for any set $x$, so it must be equal to the empty set, $\varnothing$.
A: (1) We know that : there is no x such that x belongs to the EmptySet. 
(2) Now suppose some object a belongs to the EmptySet. This supposition contradicts proposition  (1). Since we have a contradiction, anything follows. In particular it follows that : the object a belongs to the set U(EmptySet).  Generalize this : 
      Forall x, if x belongs to EmptySet, then x belongs to U(EmptySet) 

which means that : EmptySet is included in U(EmptySet). 
(4) Suppose some object a belongs to U(EmptySet). It means there is some set S such that : (S belongs to Emptyset) & ( a belongs to S). Use " &-elimination" to get : " S belongs to EmptySet". This conjunct contradicts proposition (1). Once again, we have a contradiction, and anything follows from it. In particular, it follows that : the object a belongs to EmptySet. Generalize this : 
   for all x , if x belongs to U(EmptySet), then x belongs to EmptySet, 

which means that : U(EmptySet) is included in EmptySet. 
(5) We have shown reciprocal inclusion, and therefore set identity : 
          U(EmptySet) = EmptySet 

