Understanding the union operation Suppose we have:
$A = \{(x,v,w):x+v=w\}$
$B = \{(x,v):x=v\}$
$C = \{(w,u):\exists x 2x=w\}$
Can we say that $C = A \cup B$?
 A: I don't really understand what you are asking here.  
The usual way to define a set with this kind of "set builder notation" is to specify a universe $U$, and then use the following form:
$$
\{x : \phi(x)\}
$$
where $\phi$ is a formula with one free variable that is either true or false at each $x\in U$. 
(If you are using higher order logic, or you just want to make the universe explicit, you can also write $\{x\in U :  \phi(x)\}$.)
So, the thing about the formula is that it has to be a well formed formula (wff).  This means that it is either an atomic formula (such as an equation or inequality), or is built up recursively from atomic formulae using appropriate logical operators ($\wedge,\vee,\neg,\to,\forall,\exists$, for instance.)
In your example, it is not clear what the universe is, and the expressions on the right of the "such that" delimiter are not wffs.  
Now, I guess that for the first set $A$, you intend something like:
$$A=\{\langle x,v,w\rangle : x+v-w=0\}.$$
Note that this has a logical formula in it: $\phi(x,v,w)\equiv x+v-w=0$ is either true or false depending on the values of $x$, $v$, and $w$ substituted into it: $\phi(0,0,0)$ is true and $\phi(1,1,1)$ is false.  In this case, $A$ is a plane in $\mathbb{R}^{3}$. 
So here's where it gets hard to follow.  If we assume that $B=\{\langle x,v\rangle:x-v=0\}$, then $B$ is a line in $\mathbb{R}^{2}$.  And $C$ is even harder to interpret, since you've left out the free variable to the left of the "such that" delimiter. The logical formula has only one free variable, so I'd guess that it is a subset of some one dimensional universe ($\mathbb{R}$ perhaps?), but that isn't necessarily the case. 
So, what is the union of a plane and a line? Usually, before you can consider the union of two sets, you must make sure that they are in the same universe.  In this case, the universes are probably $\mathbb{R}^{3}$ and $\mathbb{R}^{2}$. That doesn't mean that we can't take the union (we can always define a bigger universe that contains both $\mathbb{R}^{3}$ and $\mathbb{R}^{2}$), but it does make it hard to see why you would want to.
A: The question has been modified a bit from the previous one. The replacement of $-$ by $=$ makes sense. However, in the definition of $C$, you had
$C=\{w: \exists x(2x=w)\}$, and that should have been kept.  
Presumably the set that our variables range over is the set of natural numbers, or the set of integers, or the set of reals. Which one it is does not matter much.
Whatever we choose, it is not true that $A\cap B=C$. The reason is simple. The set $A$ is a set of ordered triples, the set $B$ is a set of ordered pairs, and $C$ is a set of numbers. 
The intersection of a set of ordered triples of numbers and a set of ordered pairs of numbers is empty. A set of ordered triples of numbers is a different kind of creature than a set of ordered pairs of numbers. 
Added: Define $A$ as the set of triples $(x,v,w)$ such that $x+v=w$. Define $B$ as the set of $(x,v,w)$ such that $x=v$.  Finally, let $C$ be the set of all $(x,v,w)$ such that $\exists t(w=2t)$. 
These definitions are related to yours, but definitely different. 
Then any element of $A\cap B$ is an element of $C$. But definitely not necessarily vice-versa, since $x$ and $v$ have no conditions on them.  
A: Oh, I see what you're trying to do:
\begin{eqnarray}
A&=&\{\langle x,v,w\rangle :x+v=w\}\\
B&=&\{\langle x,v,w\rangle :x=v\}\\
C&=&\{\langle x,v,w\rangle :2x=w\} 
\end{eqnarray}
These are three planes in $\mathbb{R}^{3}$. Planes $A$ and $B$ intersect in a line. Since every solution to the conditions $x+v=w$ and $x=v$ is also a solution to $2x=w$, it follows that the plane $C$ contains the intersection of $A$ and $B$.  Symbolically:
$$
A\cap B \subseteq C.$$
However, $C$ is not equal to this intersection (for a start it's a plane, and not a line). Consider $\langle 1,0,2 \rangle$, for instance. 
