What is the difference between disjoint union and union?
If $S = A \cup B$, then $S$ is the collection of all points in $A$ and $B$
What about $S = A \sqcup B$?, I think disjoint union is the same as union, only $A, B$ are disjoint. So the notation is a bit misleading. Because it is not a new operation, but operation where the pair $A,B$ satisfies $A \cap B = \varnothing$.
So given $A \cap B = \varnothing$, $S = A \sqcup B = A \cup B$.
Is my interpretation correct?
$\endgroup$ 11 Answer
$\begingroup$The notation $A\sqcup B$ (and phrase "disjoint union") has (at least) two different meanings. The first is the meaning you suggest: a union that happens to be disjoint. That is, $A\sqcup B$ is identical to $A\cup B$, but you're only allowed to write $A\sqcup B$ if $A$ and $B$ are disjoint.
The second meaning is that $A\sqcup B$ is a union of sets that look like $A$ and $B$ but have been forced to be disjoint. There are many ways of defining this precisely; for instance, you could define $A\sqcup B= A\times\{0\}\cup B\times \{1\}$. This construction can also be described as the coproduct of $A$ and $B$ in the category of sets.
(This ambiguity is similar to the ambiguity between "internal" and "external" direct sums; see for instance my answer here.)
$\endgroup$ 2
