I am given the example that the reflexive closure of the relation less on integers r(less) is lessOrEqual. Furthermore, closures can yield all possible pairs; e.g. r(notEqual) = $\mathbb{Z} * \mathbb{Z}$.
I want to know from there, how to describe the following closures on the integers:
a) s(less)
b) t(less)
c) s(notEqual)
d) t(notEqual)
But I am confused by the example provided what exactly is being described?
Ah I see.
If $foobar = \{(a,b)\in \mathbb Z\times \mathbb Z| a \ foobar \ b\}$ then
the kumquatative closure of $foobar$ is $k(foobar)=\{(a,b)|a \ foobar\ b\}$ but $foobar$ is extended so that that set is now kumquatative.
So
A) $foobar$ = $less$ so
$foobar = \{(a,b)| a< b\}\subset \mathbb Z\times \mathbb Z$.
1) $kumquat$ive means reflexive.
So $r(less) = \{(a,b)|a < b\} \cup \{(c,c)\} = $
$\{(a,b)|a < b\} \cup \{(a,b)| a= b\} =$
$\{(a,b)|a< b$ or $a=b\} = \{(a,b)|a\le b\} = lessOrEqual$.
2) symmetric
$s(less) = \{(a,b)|a < b\} \cup \{(b,a)| a < b\}=$
$\{(a,b)|a< b$ or $b > a\} = \{(a,b)|a\ne b\} = notEqual$
3) transitive
$\{(a,b)|a< b\} \cup \{(a,c)|\exists b, a<b; b < c\}$
okay.... $less$ was transitive in the first place so the closure is itself.
Or other words. If $a < b$ and $b < c$ then $a < c$ so $\{(a,c)|\exists b, a<b; b < c\} \subset \{(a,b)|a < b\}=lessThan$ so
$\{(a,b)|a< b\} \cup \{(a,c)|\exists b, a<b; b < c\}= \{(a,b)|a < b\}=lessThan$
B) $notEqual$.
1) reflexive:
$r(notEqual)= \{(a,b)|a \ne b\} \cup \{(c,c)\}=$
$\{(a,b)|a \ne b\}\cup \{(a,b)|a=b\}=$
$\{(a,b)|a \ne b$ OR $a= b\} =$
$\{(a,b)|a$ and $b$ do whatever they freakin' want$\} = $
$\{(a,b)|a\in \mathbb Z, b\in \mathbb Z\}= \mathbb Z\times \mathbb Z$.
2) symmetric
$s(notEqual) = \{(a,b)|a\ne b\} \cup \{(b,a)| a\ne b\}=$
$\{(a,b)|a\ne b$ or $b\ne a\} = \{(a,b)|a \ne b\}=$
$notEqual$
Which ... as $notEqual$ is symmetric, its closure must be itself.
3) transitive
$t(notEqual) =$
$\{(a,b)| a \ne b\} \cup \{(a,c)|\exists b; a\ne b, b\ne c\}$.
Okay... reality check. If $a$ is $a$ and $c$ is $c$ we can always find a $b$ so that $a \ne b$ and $b \ne c$. For instance $b = \max (a,c)+1$. Then $b > \max (a,c) \le a,c$ so $b\ne a,c$.
So $\{a,c)|\exists b; a\ne b, b\ne c\} = \{(a,c)\} = \mathbb Z\times \mathbb Z$.