Non-transitive relations defined on sets of cardinality 2

Question

When thinking about transitive relations, I came up with this "theorem":

All relations $\mathrm{R}: S\to S$ defined on $S = \left\{1, 2\right\}$ are transitive.

But, I couldn't come up with a proof. So, can somebody help me and either prove the above or provide a counter-example for the same?

score 3 · Accepted Answer · answered Sep 20 '18 at 15:02

3

No. Consider $R=\{(1,2), \, (2,1)\}$.
Then we have $1\,R\,2\,R\,1$ but not $1\,R\,1$.

answered Sep 20 '18 at 15:02

Berci

what about non-symmetric, non-transitive relations? are there any of those? – Truth-seek Sep 20 '18 at 16:46
Since besides this only $R\cup{(1,1)}$ and $R\cup{(2,2)}$ are the non-transitive relations which are symmetric, we can state that every relation on $S$ is either symmetric or transitive. – Berci Sep 20 '18 at 17:21
Apart from the ‘empty’ relation $\mathrm{R}=\emptyset$ (I don’t think we count that as symmetric but I’m not 100% sure actually) – aidangallagher4 Sep 20 '18 at 17:51
The empty relation vacouosly satisfies $aRb\implies bRa$. – Berci Sep 20 '18 at 21:04
@Berci Thnx for all the help. – Truth-seek Sep 21 '18 at 10:49

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