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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 $\cup \emptyset =\emptyset$?

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    $\begingroup$ Well, state what the axiom of union tells you about your set...conclude from there that it has no elements. Then use the axiom of extensionality to conclude that it's the exact same set you started with. $\endgroup$ – Ian Oct 2 '16 at 19:02
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    $\begingroup$ I believe you mean $Sx := x \cup \{x\}$ $\endgroup$ – GFauxPas Oct 2 '16 at 19:06
  • $\begingroup$ @GFauxPas Yes.. $\endgroup$ – PozcuKushimotoStreet Oct 2 '16 at 19:14
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    $\begingroup$ @Kahler: You already have one inclusion, i.e., $\emptyset \subseteq \bigcup \emptyset$, because $\emptyset$ is subset of every set. Suppose $\bigcup \emptyset \not= \emptyset$ and choose any $x \in \bigcup \emptyset$. Then (by the definition of $\bigcup$) there is a set $Y \in \emptyset$ with $x \in Y$. But this is impossible because $\emptyset$ contains no set. Therefore, the supposition $\bigcup \emptyset \not= \emptyset$ was false, that is, $\bigcup \emptyset = \emptyset$. $\endgroup$ – Moritz Oct 3 '16 at 11:48
  • $\begingroup$ @Moritz Thanks. $\endgroup$ – PozcuKushimotoStreet Oct 3 '16 at 12:28
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$$\bigcup\varnothing=\{x\;|\;\exists y(y\in\varnothing\wedge x\in y)\}$$

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, $\varnothing$.

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(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 
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