1
$\begingroup$

This question already has an answer here:

If A is non-empty set then how may " transitive relation " can be made by A×A ?

$\endgroup$

marked as duplicate by Anurag A, Theo Bendit, Thomas Shelby, Mees de Vries, Shailesh Aug 14 at 9:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

There is no simple way to get a solution but you can interpretate your problem in this way:

For each relation $\sim$ of $A$ we can define the map

$\psi_\sim: A\to \mathcal{P}(A)$

such that

$\psi_\sim(a):=\{b\in A: a\sim b\}$

In this case you have that if $\sim$ is transitive then for each $b\in \psi_\sim(a)$

$\psi_\sim(b)\subseteq \psi_{\sim}(a)$

So

$\{\sim : \sim transitive \}\cong \Lambda$

where $\Lambda:= \{\psi:A\to \mathcal{P}(A): \forall a,b \ if \ b\in \psi(a) \ then \ \psi(b)\subseteq \psi(a)\}$

Now we want prove to determine the cardinality of $\Lambda$. For simplicity $A=\{1,\dots , n\}$

We suppose that $\psi(i)=A$ for each $i< n$ then the only choice of $\psi(n)$ to get $\psi$ transitive can be $\psi(n)=\{n\}$ or $\psi(n)=A$. So in this case we have

$|\{\psi\in \Lambda : \psi(i)=A \forall i< n\}|=2$

We suppose that $\psi(n-1)\neq A$ while $\psi(i)=A$ for each $i<n-1$ . Then $\psi(n-1)$ does not contain $1,\dots ,n-2$ because otherwise $A=\psi(i)\subseteq \psi(n-1)$ and it is not possible. So we can have only

$\psi(n-1)=\{n-1\}$ or $\psi(n-1)=\{n-1, n\}$ or $\psi(n-1)=\{n\}$

In the first case we can choice $\psi(n)=\{n\}$ or $\psi(n)=\{n, n-1\}$ or $\psi(n)=A$ while in the second case we can have

$\psi(n)=\{n\}$ or $ \psi(n)=\{n,n-1\}$

and in the third case we have

$\psi(n)=\{n\}$

So we have 6 cases.

$\endgroup$

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