MHB Why doesn't it come from a cartesian product of sets?

[evinda]

Hello! (Wave)

There is the following sentence in my notes:

Let $A$ be a set. We define the set $I_A=\{ <a,a>, a \in A \}$.

$$A \times A=\{ <a_1,a_2>: a_1 \in A \wedge a_2 \in A \}$$

Then $I_A$ is a relation, but does not come from a cartesian product of sets.

Could you explain me the last sentence? (Thinking)

 

[Siron]

To illustrate with an example. Let $A = \{a_1,a_2\}$ then $I_A = \{(a_1,a_1),(a_2,a_2)\}$ and $A \times A = \{(a_1,a_1),(a_1,a_2),(a_2,a_1),(a_2,a_2)\}$. Clearly they're not the same and as you mentioned $I_A$ does not come from a cartesian product. Look at how $I_A$ is defined, it only contains the couples where both elements are the same. The cartesian product contains all the possible couples.

 

[evinda]

I still haven't understood why $I_A$ does not come from a cartesian product of sets.. (Sweating)
Could you explain it further to me? (Thinking)

 

[Evgeny.Makarov]

I still haven't understood why $I_A$ does not come from a cartesian product of sets..

It apparently means that $I_A$ is not equal to a Cartesian product of two sets if $|A|>1$. Indeed, suppose that $a,b\in A$ and $a\ne b$. Then $\langle a,a\rangle\in I_A$ and $\langle b,b\rangle\in I_A$. Suppose now that $I_A=B\times C$. Then $B$ must contain all first elements of pairs from $I_A$; in particular, $a,b\in B$. Similarly, $C$ contains all second elements of pairs from $I_A$; in particular, $a,b\in B$. But then $\langle a,b\rangle\in B\times C$ even though $\langle a,b\rangle\notin I_A$.

 

[evinda]

It apparently means that $I_A$ is not equal to a Cartesian product of two sets if $|A|>1$. Indeed, suppose that $a,b\in A$ and $a\ne b$. Then $\langle a,a\rangle\in I_A$ and $\langle b,b\rangle\in I_A$. Suppose now that $I_A=B\times C$. Then $B$ must contain all first elements of pairs from $I_A$; in particular, $a,b\in B$. Similarly, $C$ contains all second elements of pairs from $I_A$; in particular, $a,b\in B$. But then $\langle a,b\rangle\in B\times C$ even though $\langle a,b\rangle\notin I_A$.

I understand... Thank you very much! (Smile)