1
$\begingroup$

Let $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$. Give an example of a relation $T$ on $A$ with at least three elements that is not reflexive, not symmetric, but transitive. Explain clearly why your solution has/doesn’t have the given properties.

I am stuck on this question as to be transitive i thought you needed to have $(x,y), (y,x)$ and $(x,x)$.

So i have worked out the relation as $T = \{(1,1), (1,2), (2,1)\}$.

So for example $(1,2), (2,1)$ and $(1,1)$ makes it transitive.

and, from what I understand, it is reflexive as $(x,x)$ are elements of $T$ if all $x$ is a elements of $A$.

It is symmetric as $x$ and $y$ are elements of $A$ and $(x,y)$ is a element of T and $(y,x)$ is a element of $T$ which means you cannot find such a set?

$\endgroup$
  • $\begingroup$ Hint: revisit the definition of what it means for a relation to be transitive (which involves three elements of the underlying set, not necessarily distinct) $\endgroup$ – Mark Bennet May 5 '15 at 15:16
  • $\begingroup$ yes sorry i see my mistake straight away as transitive is (x,y) (y,z) and (x,z)??? is this correct? @MarkBennet $\endgroup$ – John May 5 '15 at 15:21
  • $\begingroup$ @thomas Yes, that is correct. $\endgroup$ – Lord_Farin May 5 '15 at 15:31
  • $\begingroup$ How about $<$ on $\{1,2,3\}$? $\endgroup$ – Hagen von Eitzen May 5 '15 at 15:39
1
$\begingroup$

Hint: Since we need to break symmetry, there has to be some $(a,b) \in T$ with $a \ne b$. Let us pick $(1,2) \in T$.

Now we need to add some extra things to $T$ to satisfy the requirement that it have three or more elements. (Recall that $T$ is transitive iff $(x,y), (y,z) \in T \implies (x,z) \in T$.)

Suppose that we add $(2,3)$ to $T$. What is missing from $T$ to make it transitive? Is the result reflexive? Symmetric? Transitive?

$\endgroup$
0
$\begingroup$

xTy iff x > y + 1 , is clearly non symmetric and not reflexive. Suppose y > z + 1; we deduced x > y + 1 > z + 2 > z + 1, hence T is transitive.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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