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I'm reading Comprehensive Mathematics for Computer Scientists 1. On the second chapter: Axiomatic Set Theory.

He first states the axiom of the empty set, the axiom of equality and then he proceeds to the axiom of union:

Axiom 3 (Axiom of Union) If $a$ is a set, then there is a set:

$\{$$x$ | there exists an element $b\in a$ such that $x\in b$$\}$.

This set is denoted by $\bigcup a$ and is called the union of $a$.

Notation 2 If a = {b,c}. or a = {b,c,d}, respectively, one also writes b $\cup$ c, or b $\cup$ c $\cup$ d, respectively, instead of $\cup$a

I've learned the definition of Union while I was in school, but it wasn't with axioms, they just gave an intuitive example:

$a=\{1,2,3\}$

$b=\{4,5\}$

$a\bigcup b=\{1,2,3,4,5\}$

I can't see how the notion of this intuitive example happens on the axiom of union. In my example, it's easy to understand because there's a mention to another set, where's the mention in this axiom?

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In ZFC, any element of a set is itself a set, so the interpretation of $\cup a$ is the union of all the sets in $a$. –  Kris Aug 24 '12 at 20:54

5 Answers 5

up vote 7 down vote accepted

The connection between your example and the more general definition is that $\bigcup\{a,b\}=a\cup b$. Written out in all its gory details, this is

$$\bigcup\Big\{\{1,2,3\},\{4,5\}\Big\}=\{1,2,3\}\cup\{4,5\}=\{1,2,3,4,5\}\;.$$

Let’s check that against the definition:

$$\begin{align*} &\bigcup\Big\{\{1,2,3\},\{4,5\}\Big\}\\ &\qquad=\left\{x:\text{there exists an element }y\in\Big\{\{1,2,3\},\{4,5\}\Big\}\text{ such that }x\in y\right\}\\ &\qquad=\Big\{x:x\in\{1,2,3\}\text{ or }x\in\{4,5\}\Big\}\\ &\qquad=\{1,2,3\}\cup\{4,5\}\\ &\qquad=\{1,2,3,4,5\}\;. \end{align*}$$

Take a slightly bigger example. Let $a,b$, and $c$ be any sets; then

$$\begin{align*} \bigcup\{a,b,c\}&=\Big\{x:\text{there exists an element }y\in\{a,b,c\}\text{ such that }x\in y\Big\}\\ &=\{x:x\in a\text{ or }x\in b\text{ or }x\in c\}\\ &=a\cup b\cup c\;. \end{align*}$$

One more, even bigger: for $n\in\Bbb N$ let $A_n$ be a set, and let $\mathscr{A}=\{A_n:n\in\Bbb N\}$. Then

$$\begin{align*} \bigcup\mathscr{A}&=\Big\{x:\text{there exists an }n\in\Bbb N\text{ such that }x\in A_n\Big\}\\ &=\{x:x\in A_0\text{ or }x\in A_1\text{ or }x\in A_2\text{ or }\dots\}\\ &=A_0\cup A_1\cup A_2\cup\dots\\ &=\bigcup_{n\in\Bbb N}A_n\;. \end{align*}$$

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If there exists an element $y\in\Big\{\{1,2,3\},\{4,5\}\Big\}$, then shouldn't we have a $\Big\{\{1,2,3\},\{4,5\},y\Big\}$? –  Vladimir Putin Aug 24 '12 at 18:15
2  
@Gustavo: No: $y$ is a dummy name used here to stand for any member of the set $\Big\{\{1,2,3\},\{4,5\}\Big\}$. Here the possible values of $y$ are $\{1,2,3\}$ and $\{4,5\}$. –  Brian M. Scott Aug 24 '12 at 18:17
    
@Gustavo: Yes, $x$ in expressions like $\{x:\text{something}\}$ is also a dummy variable; it can stand for anything that satisfies the condition $\text{something}$. –  Brian M. Scott Aug 24 '12 at 18:34
    
Then I have to choose one of the sets on: $\{\{1,2,3 \},\{4,5 \}\}$ and then choose an element inside the chosen one? –  Vladimir Putin Aug 24 '12 at 18:37
    
Yep. Your last comment seems to tell me that I should consider all possible options, isn't it? –  Vladimir Putin Aug 24 '12 at 18:38

Let $A=\{a,b\}$ (the set whose only elements are $a$ and $b$). Then the union of $a$ and $b$ that you described is what the Axiom of Union produces from $A$.

Remark: Informally, let $A$ be a set whose elements are a bunch of plastic bags with stuff in them (so $A$ is a set of sets). Then the set produced by the Axiom of Union from $A$ dumps the stuff contained in the bags into a single bag. (Duplicates are thrown away.)

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Oh, then $\bigcup a$ acts like a variable. When he says: "There is a set..." he refers to a set that have no name, it's only $\{$x | such that... $\}$ and not $z=\{$x | such that... $\}$, Then I thought that this nameless set were implicit here $\bigcup a$, if it had a name $z$, it would be written like: $z\bigcup a$. Is this right? –  Vladimir Putin Aug 24 '12 at 18:02
1  
@GustavoBandeira: I am having trouble understanding what you mean. Here $\cup$ acts as a unary operator (function). If you apply it to the set $\{a,b,c,d\}$ of sets it produces $a\cup b\cup c\cup d$. It operates similarly on an infinite collection $\{a_1,a_2,a_3,\dots\}$ of sets. –  André Nicolas Aug 24 '12 at 18:11
    
I wasn't aware that I could write it like LISP. –  Vladimir Putin Aug 24 '12 at 18:59
    
@GustavoBandeira: It is a common notation. In principle, all we use is $\in$ and logical symbols, but in doing doing set theory it is then useful (indeed almost necessary!) to introduce abbreviations for important constructions. –  André Nicolas Aug 24 '12 at 19:56

When we write $a\cup b$ we actually mean $\bigcup\{a,b\}$. This is a shorthand instead of writing long formulas every time we want to talk about the union of two sets.

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Yep. Same as I commented here. –  Vladimir Putin Aug 24 '12 at 18:04
    
I've read your comment with attention now. This reminds me of LISP where you can write (+ 2 3). –  Vladimir Putin Aug 24 '12 at 19:41
    
@Gustavo: Think of $\bigcup$ as a LISP function "union": $$(\textrm{union }a\ b\ \ldots)$$ It takes a list of sets and returns their union. The $a\cup b$ notation is a bit like C syntax. –  Asaf Karagila Aug 24 '12 at 20:35

Think of $a$ as a set (or collection, if you like) of other sets. Then $\bigcup a$ is the union of all these sets. So, for instance, in your example:

$$\bigcup \lbrace\lbrace 1,2,3\rbrace,\lbrace 4,5\rbrace\rbrace = \lbrace 1,2,3,4,5\rbrace$$

You may think of $A\cup B$ as shorthand for $\bigcup \lbrace A,B\rbrace$.

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Yep, it's the same as I pointed here. –  Vladimir Putin Aug 24 '12 at 18:03

This axiom talks about a set of sets.

This is because the axiom states $b\in a$ and $x\in b$: $x$ in $b$ tells you that $b$ is a set (and is an element of $a$).

For example: $a=\{\{1\},\{2,3\}\}$ then the axiom states that $\{1\}\cup\{2,3\}=\{1,2,3\}$ exists.

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