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:



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

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?

  • $\begingroup$ 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 $a$. $\endgroup$ – Kris Aug 24 '12 at 20:54
  • $\begingroup$ The exposition in that book is rather terse, especially for a book addressed to non-mathematicians (something noted in its Amazon reviews.) There are better examples of the axiom of union at work in Hrbacek and Jech, p. 10, although unlike everything in the answers shown below, the examples of Hrbacek and Jech use "pure" set theory i.e. without urelements like $a$ or $b$. (In layman's terms, this means only the empty set and braces are used to build sets.) It's somewhat insightful to see how the axiom works out in that context as well. $\endgroup$ – Fizz Apr 12 '15 at 3:16
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    $\begingroup$ Also if you do have urelements (aka atoms) other than the empty set, then the axiom of union loses the "top level" ones. E.g. $\bigcup \{a,\{b\}\}$ is just $\{b\}$. This is one of the troubles with urelements and why the axiom looks strange with urelements. Hat tip to Tourlakis' book for mentioning this. $\endgroup$ – Fizz Apr 12 '15 at 4:06

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


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

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

  • $\begingroup$ If there exists an element $y\in\Big\{\{1,2,3\},\{4,5\}\Big\}$, then shouldn't we have a $\Big\{\{1,2,3\},\{4,5\},y\Big\}$? $\endgroup$ – Billy Rubina Aug 24 '12 at 18:15
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    $\begingroup$ @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 $\{4,5\}$. $\endgroup$ – Brian M. Scott Aug 24 '12 at 18:17
  • $\begingroup$ @Gustavo: Yes, $x$ in expressions like $\{x:\text{something}\}$ is also a dummy variable; it can stand for anything that satisfies the condition $\text{something}$. $\endgroup$ – Brian M. Scott Aug 24 '12 at 18:34
  • $\begingroup$ Then I have to choose one of the sets on: $\{\{1,2,3 \},\{4,5 \}\}$ and then choose an element inside the chosen one? $\endgroup$ – Billy Rubina Aug 24 '12 at 18:37
  • $\begingroup$ Yep. Your last comment seems to tell me that I should consider all possible options, isn't it? $\endgroup$ – Billy Rubina 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.)

  • $\begingroup$ 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? $\endgroup$ – Billy Rubina Aug 24 '12 at 18:02
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    $\begingroup$ @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. $\endgroup$ – André Nicolas Aug 24 '12 at 18:11
  • $\begingroup$ I wasn't aware that I could write it like LISP. $\endgroup$ – Billy Rubina Aug 24 '12 at 18:59
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ Yep. Same as I commented here. $\endgroup$ – Billy Rubina Aug 24 '12 at 18:04
  • $\begingroup$ I've read your comment with attention now. This reminds me of LISP where you can write (+ 2 3). $\endgroup$ – Billy Rubina Aug 24 '12 at 19:41
  • $\begingroup$ @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. $\endgroup$ – 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 $\bigcup \lbrace A,B\rbrace$.

  • $\begingroup$ Yep, it's the same as I pointed here. $\endgroup$ – Billy Rubina 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.

  • $\begingroup$ In ZFC, everything is a set of sets (except the empty set, which is a set of nothing) $\endgroup$ – étale-cohomology Jun 19 '17 at 21:21

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