Simple question on Union of sets: Proof of associativity 
I need to prove $A\cup B \cup C=(A\cup B)\cup C=A\cup (B\cup C)$.

I already proved that $(A\cup B)\cup C=A\cup(B\cup C)$ by proving they are subsets of one another.
My question is now, how do I prove that they are equal to  $A\cup B\cup C$? And is it necessary to include that part? I felt that it is, as the question asks to prove all 3 parts equal to each other. Please advise me.
Thanks a lot!
 A: You might say that there is actually no such thing as $A \cup B \cup C$, in a way. There are two possible orders of evaluation here, $(A \cup B) \cup C$ and $A \cup (B \cup C)$ which you showed to be equal. $A \cup B \cup C$ is just a notation to signify this equality and order-independence. 
Together with commutativity, assuming you also proved that, so that $A \cup (B \cup C)$ is same as $(B \cup C) \cup A$, you can show that all possible ways of taking the union of three sets gives the same result. Thus we call it just $A \cup B \cup C$.
For four sets, you need to establish the equality of 24 permutations, I guess. But then you can write them all as $A \cup B \cup C \cup D$.
You can see this is getting tedious so an appeal to induction to show that this is true for all n is wise.
A: What you should have asked yourself before asking the question is what does $A\cup B\cup C$ mean? Which order is the operations supposed to be done? Or do you use a more clever definition for that(*)?
Actually in this case the associativity is used to give the expression $A\cup B\cup C$ a meaning: it says that we don't have to specify in which order the operations has to be done since $(A\cup B)\cup C=A\cup(B\cup C)$. As we don't have to do that we can just say that $A\cup B\cup C$ is defined as $(A\cup B)\cup C)  = A\cup(B\cup C)$
(*) One could of course use more advanced machinery to define $A\cup B\cup C$ as the set consisting of elements that are in at least one of the sets $A$, $B$ or $C$. In that case one would have the case that there's something to be proven. One way to do it all is by using truth table:
$$\begin{matrix}
x\in A & x\in B & x\in C & x\in (A\cup B) & x \in (A\cup B)\cup C & x \in (A\cup B\cup C) \\
\hline
f & f & f & f & f & f \\
f & f & T & f & T & T \\
f & T & f & T & T & T \\
f & T & T & T & T & T \\
T & f & f & T & T & T \\
T & f & T & T & T & T \\
T & T & f & T & T & T \\
T & T & T & T & T & T \\
\end{matrix}$$
The $x\in (A\cup B)$ column is true whenever $x$ in one of $A$ and $B$ that is at least one of the first two columns is true. In similar way the next column is formed as true iff at least one of the $x\in C$ and $x\in (A\cup B)$ is true. The last is true whenever at least one of the first three is true.
That the last two columns are identical means that $x\in (A\cup B)\cup C)$ is equivalent to $x\in (A\cup B\cup C)$. 
A: Let $a\in A \implies a \in A\cup B\cup C \ also a\in A\cup (B\cup C) \ and\ a \in (A\cup B)\cup C$.
 Similarly, Let $b\in B \implies b\in A\cup B\cup C \ also b\in A\cup (B\cup C) \ and\ b \in (A\cup B)\cup C$.
 Similarly, Let $c\in C \implies c\in A\cup B\cup C \ also c\in A\cup (B\cup C) \ and\ c \in (A\cup B)\cup C$.

So, $A\cup B\cup C = (A\cup B)\cup C =  A\cup (B\cup C)$

A: I have just loudly championed a different answer in comments, but on further consideration, I do think there's more to it than that.
First, let us write down some definitions


*

*If $A$ and $B$ are sets, then $A\cup B$ means the set whose elements are exactly every $x$ that is an element of at least one of $A$ and $B$.


*If $A$, $B$ and $C$ are sets, then $A\cup B\cup C$ means the set whose elements are exactly every $x$ that is an element of at least one of $A$, $B$, and $C$.

Once we have given $A\cup B\cup C$ a definition it is not immediate that the set described by that definition is the same set as the one we get by doing the union in two separate steps as $(A\cup B)\cup C$. It is true that they are the same (fortunately so, because otherwise the notation $A\cup B\cup C$ would be really misleading), but it does deserve a proof of its own.
Proving this is not really more difficult than proving that $(A\cup B)\cup C = A\cup(B\cup C)$ by showing inclusions in both directions.

To show that this is not just empty pedantry, let's consider another example where it does not go as nicely:



*If $A$ and $B$ are sets, then $A\mathop\triangle B$ means the set whose elements are exactly every $x$ that is an element of exactly one of $A$ and $B$.


*If $A$, $B$ and $C$ are sets, then $\triangle(A,B,C)$ means the set whose elements are exactly every $x$ that is an element of exactly one of $A$, $B$, and $C$.

The binary $\triangle$ operation defined in (3) is quite nice and well-behaved (and useful enough in certain contexts that the notation $\triangle$ is somewhat standard). In particular, it is straightforward to prove that $\triangle$ is commutative and associative.
However, it is not true that $\triangle(A,B,C)$ is the same as $(A\triangle B)\triangle C$. (For a counterexample, consider $A=\{1,2\}$, $B=\{1,3\}$, $C=\{1,4\}$. Then $(A\triangle B)\triangle C=\{1,2,3,4\}$, but $\triangle(A,B,C)=\{2,3,4\}$).
So $\triangle$ is an example of an operator that seems to behave just as nice as $\cup$ does -- in particular, it is associative -- but where applying it to one argument at a time does not reproduce the straightforward generalization of the intuitive definition of the binary operator.
(What actually works is to say that $A_1\triangle A_2\triangle\cdots\triangle A_n$ is the set whose elements are exactly every $x$ that is an element of an odd number of the $A_i$s. Note that when there are just two operands, "an odd number" is the same as "exactly one", but "odd" is the description that generalizes well).
