Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question about Equivalence relation.

There is a set A, and for this set I have a Equivalence relation C.

Then there is a subset B ⊂ A

My question is about this expression I don't understand: C ∩ (B x B )

How can be the Relation C intersected with the pair (BxB)? Is C a set? I don't understand how can a relation been intersected with a set?

Thank you

share|cite|improve this question
$C$ is a subset of $A\times A$ – ՃՃՃ Jan 17 '13 at 13:57
Thank you Psx. Ok, I understand, the subset are the pairs of elements tha – Pedro Jan 17 '13 at 13:59
Thank you Psx. Ok, I understand, the subset are the pairs of elements that meet the relation. :-) – Pedro Jan 17 '13 at 14:00
That's right. You're welcome. :-) – ՃՃՃ Jan 17 '13 at 14:02

$C$ is a relation on $A$, which means that $C\subseteq A\times A$, i.e., it is a set of ordered pairs.

For example, if $A=\{0,1,2,3,4,5\}$, then $C=\{(0,0),(1,1),(2,2),(3,3),(4,4),(5,5),(0,1),(1,0),(2,3),(3,2)\}$ is an equivalence relation on $A$. (It corresponds to the partition $\{\{0,1\},\{2,3\},\{4\},\{5\}\}$.)

Now if $B\subseteq A$ then $B\times B$ is a set of ordered pairs of elements from $A$. (Namely, it contains all pairs where both coordinates are from $B$.) So it makes sense to intersect $C$ and $B\times B$.

For example, take $B=\{0,1,2\}$. Then $B\times B=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}$ and
$C\cap (B\times B) = \{(0,0),(1,1),(2,2),(0,1),(1,0)\}$. Notice that $C\cap (B\times B)$ is an equivalence relation on the set $B$.

share|cite|improve this answer
Wonderfull, thank you. – Pedro Jan 17 '13 at 14:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.