About the union of disjoint sets and the disjoint union I know that $A\sqcup B\equiv (A\times \{1\})\cup (B\times \{2\})$ for arbitrary sets $A$ and $B$, where the symbol $\equiv$ means that these notations represent isomorphic sets, and generally the above is used as the definition for the disjoint union. 
My question is: it is correct to represent the union of two already disjoint sets, namely $C$ and $D$, with the notation $C\sqcup D$ instead of just $C\cup D$? In other words: it is in some sense $C\cup D$ equivalent to the set $(C\times \{1\})\cup (D\times \{2\})$ when $C$ and $D$ are disjoint?
 A: It depends on context, but I would say that you are unlikely to cause confusion if you write $A \sqcup B$ for the union of $A$ and $B$, where $A$ and $B$ are disjoint subsets of a larger space.  For example, it is a fun fact about the $p$-adic numbers (denoted by $\mathbb{Q}_p$, where $p$ is some fixed prime number) that the unit ball in $\mathbb{Q}_p$ is composed of precisely $p$ congruent balls of radius $\frac{1}{p}$.  Indeed, we could reasonably write
$$ B(0,1) = \bigsqcup_{j=0}^{p-1} B(j, \tfrac{1}{p}).$$
Formally, as is noted in the question, the set $A\sqcup B$ is not the same thing as the set $A \cup B$.  On the one hand, the union of two sets is defined to be
$$A \cup B := \{ x : x\in A \lor x\in B \},$$
where $A$ and $B$ are subsets of some (thus far unspecified) universal set.
On the other hand,
$$A \sqcup B := (A\times\{0\}) \cup (B\times\{1\}), $$
where $X\times Y$ denotes the Cartesian product and $\cup$ is the union in the first sense.  Note that the choice of $\{0\}$ and $\{1\}$ is arbitrary.  Really, we just need two distinct elements:  one to pair with elements of $A$, and another to pair with elements of $B$.
However, if $A, B \subseteq \mathscr{U}$ and $A \cap B = \emptyset$ (that is, if $A$ and $B$ are disjoint), then there is a correspondence between the union ($\cup$) and the disjoint union ($\sqcup$):  define the function $f : A\sqcup B \to A\cup B$ by
$$ f(x,k) := x $$
It is not too hard to see that this function is surjective: if $y \in A \cup B$, then either $y \in A$ and so $(y,0) \in A\sqcup B$ is mapped to $y$ by $f$, or $y \in B$ and so $(y,1) \in A\sqcup B$ is mapped to $y$ by $f$.  Injectivity is similarly simple to show.
From a purely set-theoretic point of view, the two sets are isomorphic, i.e.
$$ A \cup B \equiv A \sqcup B. $$
With a bit more work, we can show that this same map preserves other structures (under the right hypotheses).  For example, if $A$ and $B$ are subsets of a topological space, then there is a natural topology on the set $A \sqcup B$ such that the disjoint union is homeomorphic to the union.
