I've looked at a few Q/A and not found a satisfactory answer to even the basic question. The extended question is:
Given a set of sets, or a set of tuples:
- Is there a standard way of denoting whether an object is an element of one of the inner sets/tuples?
- Is there a standard way of identifying these objects?
- Is there a standard way of denoting the multiplicity across inner sets/tuples?
By way of example, we'll take the set of primitive Pythagorean triples as tuples:
$$S := \{(a,b,c) \in \mathbb{N}^3 \mid [a^2 + b^2 = c^2] \land [\gcd (a,b,c) = 1] \land [a < b < c] \}$$
$$S = \{(3,4,5), (5,12,13), (8, 15, 17), \cdots\}$$
Obviously $5 \not \in S$, because every element of $S$ is a tuple. But $5$ appears in two of the tuples, as does every $4k+1$ prime, and other integers appear more than twice. Is it correct to write, for instance, "$5 \in \in S$" and "$6 \not \in \in S$"? If not, is there a standard correct method?
I think that if we have indexed/ordered sets and tuples, we can identify individual elements with typical notation. For instance, if we use $c$ to order the tuples, we could say that $s_{1,3} = s_{2,1} = 5$. But how does one specify the ordering of the tuples mathematically? Or is a well-ordered set of tuples a mathematical impossibility?
And is there any notation for the multiplicity of an object across the inner sets/tuples of the set? One can certainly imagine uses for it, but I don't think I've ever seen it notated. Obviously, one can calculate it using a computer if need be, but is there something like $\text{mult}(5) = 2$ that is commonly used, or even correctly used, for this?
(I considered adding "soft-question" as a tag but uncertain whether it'd be correct here.)
Best Answer
In set theory, given a set $S$ and a set $x$ which belongs to an element of $S$, you can use the following way to represent the relation of $x$ and $S$:
\begin{equation*} x\in\bigcup S=\bigcup_{X\in S} X, \end{equation*} where $\bigcup S$ is a collection of all the elements of all the elements of $S$. Formally, $\bigcup$ is a generalized union which is defined as
$$\bigcup S:=\{x\mid x\in X\text{ for some } X\in S\}.$$
For example, if $x=0$ and $S=\{\{0,1\},\{1,2,3\}\}$, then $\bigcup S=\{0,1\}\cup\{1,2,3\}=\{0,1,2,3\}$, and hence $x=0\in \{0,1\}$ and $x=0\in\bigcup S$.
But for your question, note that it's a little complicated because the definition of tuples. In set theory, we define 2-tuples as
$$(a,b):=\{\{a\},\{a,b\}\},$$
and 3-tuples as
\begin{align*} (a,b,c):=&((a,b),c)\\ =&\{\{(a,b)\},\{(a,b),c\}\}\\ =&\{\{\{\{a\},\{a,b\}\}\},\{\{\{a\},\{a,b\}\},c\}\}. \end{align*}
and $n$-tuples ($n\geq 2$) as
\begin{align*} (a_0,\cdots,a_{n-1}):=((a_0,\cdots,a_{n-2}),a_{n-1}). \end{align*}
So when using $\bigcup$ notations to show the belonging relations you talked about, be careful of the numbers of $\bigcup$ to be used. For example, if $S=\{(a_0,a_1)\mid \cdots\}$ then
\begin{align*} x=a_0\in &\bigcup(\bigcup S)=\bigcup\textstyle^2~S,\\ x=a_1\in &\bigcup(\bigcup S)=\bigcup\textstyle^2~S, \end{align*}
and so
\begin{align*} x=a_i\in &\bigcup(\bigcup S)=\bigcup\textstyle^2~S; \end{align*}
if $S=\{(a_0,a_1,a_2)\mid\cdots\}$, then
\begin{align*} x=a_2\in &\bigcup(\bigcup S)=\bigcup\textstyle^2~S,\\ x=a_1\in &\bigcup(\bigcup(\bigcup S))=\bigcup\textstyle^3~S,\\ x=a_0\in &\bigcup(\bigcup(\bigcup S))=\bigcup\textstyle^3~S,\\ \end{align*}
and so
\begin{align*} x=a_i\in\bigcup_{2\leq j\leq 3}(\bigcup\textstyle^j S). \end{align*}
And more generally, for $S$ whose elements are $n$-tuples such as $(a_0,\cdots,a_{n-1})$ you can show that
\begin{align*} x=a_i\in\bigcup_{2\leq j\leq n}(\bigcup\textstyle^j S). \end{align*}
If you don't want to count the numbers of $\bigcup$, maybe the following way is better: suppose $x$ is related to some element of $S$, saying $x\in Y$ and $\phi(Y)\in S$ (for which given $S$ both the $Y$ and $\phi(Y)$ are usually easy to be found), then
$$x\in \bigcup\{Y\mid \phi(Y)\in S\}=\bigcup_{\phi(Y)\in S}Y.$$
Note that @1Rock 's $5 \in \bigcup_{(a,b,c) \in S} \{a,b,c\}$ is a practice over this idea where $x=5$, $Y=\{a,b,c\}$ and $\phi(Y)=(a,b,c)$, and that in particular when $Y=\phi(Y)$ we have
\begin{align*} x\in \bigcup\{Y\mid \phi(Y)\in S\}&=\bigcup_{\phi(Y)\in S}Y\\ &=\bigcup_{Y\in S}Y=\bigcup S. \end{align*}