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I have been given a question that states,

$A= \{ a|a \in \mathbb{R} \}$
$B= \{1,2,3 \}$
$A \times B=?$

Now I am stuck with how do I represent real numbers? Even if I show an infinite series (by adding ... at both ends of the Cartesian product) where should I take a start to represent the numbers?

Thank you in advance.

Edit:

By represent I mean that how many sets do I make in the Cartesian Product? How can I show all the real numbers in the Cartesian Product?

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I would be thankful if someone could create a Cartesian-Product tag for the community. –  Fahad Uddin Jul 30 '11 at 17:52
4  
I don't understand what you mean by "represent." Do you mean something like $\{ (a, b) : a \text{ real}, b \in \{ 1, 2, 3 \} \}$? –  Qiaochu Yuan Jul 30 '11 at 17:53
    
$A\times B$ is composed of three disjoint copies of the real line. Is that what you mean? –  lhf Jul 30 '11 at 18:10
    
@Qiaochu and lhf: What I mean is that there are infinite terms in a set of all real numbers.Moreover as it includes all the floating point values so it more difficult to represent all of the elements of that set. –  Fahad Uddin Jul 30 '11 at 18:17
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@fahad I do not understand what "represent" means either. But whatever issue you face in representing this set seems to arise in the case of $\mathbb R$ itself. What is special about cartesian products here? –  Srivatsan Jul 30 '11 at 18:22

4 Answers 4

up vote 3 down vote accepted

When a set has only finitely many elements, you can display it by listing its members: the set of digits base ten, for instance, is $\{0,1,2,3,4,5,6,7,8,9\}$. When a set has infinitely many members, you can’t list them all. If they can be arranged in a sequence that has an obvious pattern, you can suggest the set by listing enough of its members to make the pattern clear to the reader: $\mathbb{Z} = \{\dots,-2,-1,0,1,2,\dots\}$. In many cases just about every reader will make the same interpretation. For example, I’d expect practically everyone to interpret $\{2,4,6,8,\dots \}$ as the set of positive even integers. But even in simple cases like this you can’t be sure that everyone’s pattern matching sense will agree with yours, so if you really want to specify this set precisely, you have to write it in a way that includes a precise specification of what qualifies something to be a member of it: $\{n \in \mathbb{Z^+}: 2|n \}$, $\{2n:n \in \mathbb{Z^+}\}$, $\{n:n \mbox{ is a positive even integer}\}$, etc.

Moreover, the ‘suggestive listing’ technique is possible only with countably infinite sets, and not even all of those: it has to be a set that can be arranged in a ‘nice’ sequential order, one that a reader can reliably identify with the set in question. The set $\mathbb{Q}$ of rational numbers would be very difficult to indicate in this way with any real assurance that virtually all readers would correctly identify it.

In your problem there is no hope of such a listing, because the set of real numbers is uncountable, and therefore your set $A\times B$ is also uncountable. This guarantees that there is no way to suggest its membership by listing a few members and expecting the reader to spot a pattern that says ‘These are the real numbers’. You have no choice but to specify the set precisely. There are many slightly different ways to do so; Qiaochu gave one in the comments. A slightly different one is $A\times B =$ $\{\langle x,n \rangle:x \in \mathbb{R} \mbox{ and }n \in \{1,2,3\}\}$. A very short one: $\{\langle a,b \rangle:a\in A \& b\in B\}$. Yet another, much wordier: $\{\langle a,b \rangle:a\mbox{ is a real number and }b\mbox{ is }1,2,\mbox{ or }3\}$. (The short one probably isn’t a good choice if you really want to give the reader an immediate clear indication of what the elements of this Cartesian product actually are.)

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Thanks. Can you please help me out with the notations you used? In {a,b:a is a real number and n is 1,2, or 3} what is 'n' and why is there a need to define it?Moreover how would the person know that 'b' is also a real number? –  Fahad Uddin Jul 31 '11 at 4:17
    
@fahad: The $n$ was a typo; it should have been $b$. I’ll correct it immediately. –  Brian M. Scott Jul 31 '11 at 4:36

Here are a couple of suggestions.

$1$. Take any three distinct lines. For definiteness let them be the lines $y=1$, $y=2$, and $y=3$, though this does not really matter. Then we can think of any element of $A\times B$ as a point on one of these three lines.

$2$. Any real number can be represented uniquely as a decimal, if we specify that if the real number has a terminating decimal expansion, that's the one we choose. So for example the number $1.229999\dots$ would be represented as $1.23$.

Now look at the set $Y$ of real numbers whose first digit after the decimal point is one of $1$, $2$, or $3$. We can identify such real numbers with elements of $A\times B$ as follows. Take such a number $y \in Y$. Let $d$ be the first digit of $y$ after the decimal point. Let $x$ be the number obtained by removing the first digit of $y$ after the decimal point, and moving all the other digits after the decimal point one to the left.

Then $y$ can be thought of as a concrete representation of the ordered pair $(x,d)\in A\times B$, and $Y$ as a concrete representation of $A\times B$.

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I would like to suggest you read the definition of Cartesian product first.

For your question "how many sets do I make in the Cartesian Product", I think you are talking about the cardinality of $A\times B$, which is equal to the cardinality of $A$ in your case.


P.S. Since you mentioned "floating point", I think you are talking about representing the Cartesian product in computer, which is the subject you may want to ask in "stackoverflow".

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The notion of taking "three distinct lines" can be made slightly more "formal" by the fact that

$$\{ 1, 2, \dotsc, n \} \times S \cong \underbrace{S \uplus S \uplus \dotsb \uplus S}_{n \, \text{times}}$$

where $\uplus$ is the disjoint union. So depending on what you mean by "representing" your set, you could represent it as

$$\mathbf R \uplus \mathbf R \uplus \mathbf R$$

Formally, for most ways to construct the cartesian product and disjoint union, the sets will not actually be equal in the strong sence. There is, though, a strong formal sense in which they are equal, namely in the sense that the functors $\{1, 2, 3\} \times -$ and $- \uplus - \uplus -$ are naturally isomorphic.

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