Explanation of the following notation

Question

I am having a hard time understanding the meaning of the union operation in this equation.

$$C(A)=\bigcup_{x \in A}C(x)$$

For context, here is the sentence:

The candidate set for $x$ is $S \cap C(x)$. The candidate region for a set of points $A$ is $C(A)=\bigcup_{x \in A}C(x)$, with the candidate set $S \cap C(A)$. [1]

Is this an indexed union? Also, I haven't seen a union operation except for when it appears between two sets.

Any advice would be greatly appreciated. Thanks.

[1] Clarkson, 1988. "A randomized algorithm for Closest-Point Queries".

Answer 1

The symbol means that $\displaystyle\bigcup_{x \in A} C(x)$ means the following:

Let $A=\{x_1,x_2,x_3,\cdots\}$. Then, $$\bigcup_{x \in A}C(x)=C(x_1) \cup C(x_2) \cup\cdots\cup C(x_i)\cup \cdots$$

Note that, to illustrate the point, I have assumed that $A$ is countable, but that need not be true.

In words, it means that, to each $x$ in $A$, there exists $C(x)$ and that is contained in the union. Or, better will be: For a point to belong to $C(A)$, it must be belong to one of the $C(x)$ for some $x \in A$. Further, a point that is in $C(x)$ for some $x \in A$ also belongs to $C(A)$.

Actually, this symbol is a short hand notation. For instance, if $A=\mathbb R$, it becomes hard to write out the union as explicitly as I wrote down. There is no notion of next number (say to $0$) that makes sense.

Answer 2

We say $y$ is in $C(A)$ if and only if $y$ is in $C(x)$ for some $x\in A$. In other words, $\bigcup_{x \in A} C(x)$ is the set of points which are in at least one of the $C(x)$. In general, the union of sets $\bigcup A_i$ is the set of elements contained in at least one of the $A_i$, and the subscript under the union sign tells you which sets you are taking the union of.

Answer 3

Yes, this can be called an "indexed union". The set $C(A)$ is defined to consist of all points $p$ which lie in at least one set of the form $C(x)$ with $x\in A$.

For example, if $A = \{ a,b,c\}$, then $C(A)$ is defined to be $C(a)\cup C(b)\cup C(c)$.