2
$\begingroup$

I know that some of these symbols are used in set theory like $A \cup B$, but that's not what I'm talking about. I have seen those symbols used in a way similar to $\Sigma$ summation and $\Pi$ product. I hope this makes sense, but I'm just wondering what it means.

$\endgroup$
5
  • $\begingroup$ They are just a notation , eg, $\bigcup$ is specifying for union of n sets $\endgroup$
    – Max Payne
    Jan 27, 2016 at 14:15
  • $\begingroup$ The $\bigwedge$ and $\bigvee$ are used in lattices and the like, particularly complete lattices. $\endgroup$ Jan 27, 2016 at 14:17
  • $\begingroup$ @Tim That would imply $\bigcup$ is only used for finite intersections, though. $\endgroup$ Jan 27, 2016 at 14:17
  • $\begingroup$ @ThomasAndrews sorry any arbitrary n. and *union not intersection $\endgroup$
    – Max Payne
    Jan 27, 2016 at 14:19
  • $\begingroup$ Can this really be answered without context? $\endgroup$
    – Future
    Jan 27, 2016 at 15:04

3 Answers 3

3
$\begingroup$

Given any monoid (that is a set $A$ equipped with an associative operation $\cdot$ and an identity $1$), we can define "finite products" roughly by:

$$\prod_{i=1}^n a_i = a_1\cdot a_2\cdot \dots \cdot a_n$$

where $\prod_{i=1}^0 = 1$.

Possible monoids are for example $(\mathbb{R},\cdot, 1)$ yielding "$\prod$", $(\mathbb{R},+,0)$ yielding "$\sum$" or $(P(S), \cup, \emptyset)$ yielding "$\bigcup$" and so on and so forth.

So, we can also view a monoid as a set $A$ together with a map $A^* \to A, (a_n) \mapsto \prod_{i=1}^n a_i$ taking lists (words, tuples) of elements of $A$ to elements of $A$.

Occassionally however, we may find "maps" (broadly speaking) which not only accept finite lists, but also infinite lists or even bigger families of elements as objects.

For example, a complete lattice is a set $A$ equipped with maps $\bigvee$ and $\bigwedge$ taking abitrary families of elements of $A$ to elements of $A$.

Intuitively, if you take the set of all "small sets" (this is usually realized as a proper class) as the set $A$, then you get a complete lattice with operations $\bigcup$ and $\bigcap$ called union and intersection, which take families of elements of $A$ (that is sets of sets) to sets.

$\endgroup$
2
  • $\begingroup$ +1 It should be noted that if the monoid is abelian, then you can also write $\prod_{i\in I} a_i$ without requiring an order on $I$, as long as $I$ is finite. In $\vee$ and $\wedge$ and $\cap$ and $\cup$ this extends to infinite cases, too. $\endgroup$ Jan 27, 2016 at 14:42
  • $\begingroup$ @ThomasAndrews Yes, that is indeed noteworthy as well. Thank you. $\endgroup$ Jan 27, 2016 at 17:06
2
$\begingroup$

$$\bigcup_{n=0}^kA_n=A_0\cup A_1 \cup \cdots \cup A_k$$ $$\bigcup_{n=0}^\infty A_n=A_0\cup A_1 \cup A_2 \cup \cdots$$ The same rule applies to others. You can also write $$\bigcup_{n\in \mathcal{I}} A_n$$

even if $\mathcal{I}$ is not countable.

$\endgroup$
0
1
$\begingroup$

In a lattice $L$, you have the meet and join operators $\wedge$ and $\vee$. Since these are associative, we will often write:

$$\bigvee_{i=1}^n x_i = x_1\vee x_2\vee\cdots \vee x_n\\ \quad \bigwedge_{i=1}^n x_i = x_1\wedge x_2\wedge\cdots \wedge x_n$$

When the lattice is "complete," for any $X\subseteq L$, you can write $$\bigwedge X\quad \bigvee X$$

In set theory, if $A$ is a set of sets (and in set theory, every set is a set of sets) then:

$$\bigcup A\quad\text{and}\quad \bigcap A$$

are the union and intersections of the memebers of $A$.

This is often written as indexed sets - give sets $A_i$ for $i\in I$, we write:

$$\bigcup_{i\in I} A_i\quad \text{and}\quad\bigcap_{i\in I} A_i$$

Unlike sums, however, $\cap,\cup,\vee,\wedge$ all can be arranged freely. You have to be careful about infinite sums:

$$\sum_{i\in I} x_i$$

because the value can depend on an order which you give $I$.)

A simple example of a complete lattice that is not boolean is $\mathbb R\cup\{\pm \infty\}$ with $a\vee b= \max(a,b)$ and $a\wedge b = \min(a,b)$ and $\bigvee S=\sup S$ and $\bigwedge S=\inf S$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .