
MHB Master
#1
December 7th, 2019,
12:36
Hey!!
I am looking at the following:
There are the terms reflexive, symmetric, antisymmetric and transitive.
Give for each combination of the properties (if possible) a set $M$ and a relation $R$ on $M$, such that $R$ satisfies these properties.
What is meant exactly? Every possible combination? So do we have to give a set and a relation that satisfies the below properties?
 reflexive, symmetric
 reflexive, antisymmetric
 reflexive, transitive
 symmetric, antisymmetric
 symmetric, transitive
 antisymmetric, transitive
 reflexive, symmetric, antisymmetric
 reflexive, symmetric, transitive
 reflexive, antisymmetric, transitive
 symmetric, antisymmetric, transitive
 reflexive, symmetric, antisymmetric transitive

December 7th, 2019 12:36
# ADS
Circuit advertisement

#2
December 7th, 2019,
16:31
Well, you have four properties so 4!= 24 possible "combinations" (25 if you include the "empty combination" that none of those properties are true).
I think you should give special consideration to two of those properties. What must be true of a set so that it will be both "symmetric" and "antisymmetric"?

MHB Master
#3
December 7th, 2019,
17:32
Thread Author
Originally Posted by
HallsofIvy
Well, you have four properties so 4!= 24 possible "combinations" (25 if you include the "empty combination" that none of those properties are true).
Since the order of the properties doesn't matter do we not have $\binom{4}{2}+\binom{4}{3}+\binom{4}{4}=11$ combinations?
Originally Posted by
HallsofIvy
I think you should give special consideration to two of those properties. What must be true of a set so that it will be both "symmetric" and "antisymmetric"?
A relation R is symmetric if it holds $aRb \iff bRa$ for $a,b\in M$.
A relation R is antsymmetric if it holds $aRb$ then it doesn't hold that $bRa$ for $a\neq b\in M$.
So this combination is not possible, right?

#4
December 8th, 2019,
16:05
Originally Posted by
HallsofIvy
Well, you have four properties so 4!= 24 possible "combinations" (25 if you include the "empty combination" that none of those properties are true).
I believe there are $2^4=16$ possible combinations of 4 properties.
Originally Posted by
mathmari
A relation R is symmetric if it holds $aRb \iff bRa$ for $a,b\in M$.
A relation R is antsymmetric if it holds $aRb$ then it doesn't hold that $bRa$ for $a\neq b\in M$.
So this combination is not possible, right?
It is possible.

MHB Master
#5
December 8th, 2019,
16:14
Thread Author
Originally Posted by
Evgeny.Makarov
Ah yes! For the empty set, right?
So do we have the following?
We consider the set $M=\{0,1,2\}$.
 reflexive, symmetric : $R=\{(0,0), (1,1), (2,2), (0,1), (1,0)\}$.
 reflexive, antisymmetric : $R=\{(0,0), (1,1), (2,2), (0,1)\}$.
 reflexive, transitive : $R=\{(0,0), (1,1), (2,2), (0,1), (1,2), (0,2)\}$.
 symmetric, antisymmetric : $R=\emptyset$.
 symmetric, transitive : $R=\{(0,0), (0,1), (1,0)\}$.
 antisymmetric, transitive : $R=\{(0,0), (1,1), (0,1)\}$.
 reflexive, symmetric, antisymmetric : $R=\emptyset$.
 reflexive, symmetric, transitive : $R=\{(0,0), (1,1), (2,2), (0,1), (1,0), (1,2),(2,1),(0,2), (2,0)\}$.
 reflexive, antisymmetric, transitive : $R=\{(0,0), (1,1), (2,2), (0,1), (1,2),(0,2), (2,1)\}$.
 symmetric, antisymmetric, transitive : $R=\emptyset$.
 reflexive, symmetric, antisymmetric, transitive : $R=\emptyset$.

#6
December 8th, 2019,
16:22
Originally Posted by
mathmari
For the empty set, right?
No, for any subset of the diagonal relation $\Delta=\{(x,x)\mid x\in M\}$.
I don't have time now to check all your answers, but as I said, there are 16 combinations, and they should be enumerated in a systematic way that makes it easy to see that nothing is missing. In your enumeration the first item "reflexive, symmetric" does not mean "reflexive, symmetric, not antisymmetric and not transitive" because this relation is transitive.

MHB Master
#7
January 25th, 2020,
11:42
Thread Author
Originally Posted by
Evgeny.Makarov
No, for any subset of the diagonal relation $\Delta=\{(x,x)\mid x\in M\}$.
I don't have time now to check all your answers, but as I said, there are 16 combinations, and they should be enumerated in a systematic way that makes it easy to see that nothing is missing. In your enumeration the first item "reflexive, symmetric" does not mean "reflexive, symmetric, not antisymmetric and not transitive" because this relation is transitive.
So a relation that is symmetric, antisymmetric and transitiveis is every relation in the form $\{(x,x)\mid x\in M\}$, right? But the empty set satisfies also these properties, or not?

#8
January 25th, 2020,
16:13
Originally Posted by
mathmari
So a relation that is symmetric, antisymmetric and transitiveis is every relation in the form $\{(x,x)\mid x\in M\}$, right?
$\{(x,x)\mid x\in M\}$ is a fixed set. You would not say, "Any number of the form 5", right? But yes, a relation is both symmetric and antisymmetric iff it is a subset of the diagonal relation $\Delta=\{(x,x)\mid x\in M\}$. The empty relation is also a subset of $\Delta$ and is symmetric and antisymmetric.

MHB Master
#9
January 25th, 2020,
16:24
Thread Author
Originally Posted by
Evgeny.Makarov
$\{(x,x)\mid x\in M\}$ is a fixed set. You would not say, "Any number of the form 5", right? But yes, a relation is both symmetric and antisymmetric iff it is a subset of the diagonal relation $\Delta=\{(x,x)\mid x\in M\}$. The empty relation is also a subset of $\Delta$ and is symmetric and antisymmetric.
Ah ok!
So $\Delta$ is symmetric and antisymmetric and so also transitive, right?

#10
January 25th, 2020,
16:31
Yes, every subset of $\Delta$, including $\Delta$ itself, is transitive.