0

I want to proof the following theorem:

Let R be an equivalence relation on set A. Then {R[a]:a that belongs to A} is a partition of A.

So long I have manage to proof that each a that belongs to A, it belong to the partition (by using reflexive property); and that if S and T belongs to a partition then S=T (by using symmetry and transitive property). The part that I cannot prove is the partition property that says:

If S and T are partitions then S intersection T is equals to the empty set.

How can I prove that?

Lila
  • 479

1 Answers1

0

We can prove the third property by contradiction. Let $P = \{R[a] \mid a \in A\}$.

Let $R[a]$ and $R[b]$ be distinct sets in $P$ and suppose $\exists\; x \in A$ with $x \in R[a]$ and $x \in R[b]$.

Then $aRx$ and $xRb$. By transitivity and symmetry, $aRb$ and $bRa$.

Therefore, for any $y \in R[a],\; yRa$ so by transitivity, $yRb$ and so $y \in R[b]$.

Similarly, for any $z \in R[b],\; zRb$ so by transitivity, $zRa$ and so $z \in R[a]$.

This contradicts that $R[a]$ and $R[b]$ are distinct sets. So there cannot be such an $x$. Therefore, $R[a]$ and $R[b]$ are disjoint, as the third property requires.

For your first two properties, you might want to compare them against the three properties here to make sure you have them right.

Mick A
  • 10,208