The document summarizes a session where the concepts of sets, relations, and properties like reflexivity, symmetry, and transitivity were discussed. Tables were used to systematically list the ordered pairs for a set to determine which pairs would satisfy each property. For example, to have a reflexive relation, the ordered pairs would need to include (a,a) for all elements a in the set. The document provides an example of building a relation on the set {1,2,3} that is not reflexive, symmetric, or transitive by selecting pairs that do not complete a row in the property tables. It concludes by suggesting building similar tables for the set {1,3,5,8}.

1. Session Notes: Sunday, 23 Feb 2014
Tyler Murphy
February 24, 2014
This evening we investigated some properties of sets and relations. We looked in depth
at the problems from the worksheet given out in class on Friday, Feb 21. We also looked
a little at some of the problems from the homework about transitivity, reflexivity,and
symmetry.
First, let us begin with an exposition of problem 2.2.5.a.
Let A = {1, 2, 3}. List the ordered pairs in a relation on A that is not reflexive, not
symmetric, and not transitive.
Notice that the answer in the back of the book is the set R = {(1, 1), (1, 2), (1, 3)}.
Ask yourself why this is. Surely, (1, 1) is a reflexive pair. Why then is R not reflexive?
Consider the definition of ref lexive. A relation R on set A is reflexive IF AND ONLY IF,
FOR ALL a ∈ A, (a, a) ∈ R.
The keywords here are the ”For all” and ”if and only if.” For all a ∈ A means we would
need (1,1), (2,2), AND (3,3) to be in R.
Now ask yourself what does it mean to be symmetric? It means that if some ordered pair
(x, y) is in R, then (y, x) must be in R.
Finally, consider what it means to have transitivity. For any two pairs (a, b) and (b, c) in
R, you also have (a, c) in R.
So now that we know what this means, what do we do?
First, make a list of all the ordered pairs that can possibly be generated from A x A.
(1,1)
(2,1)
(3,1)
(1,2)
(2,2)
(3,2)
(1,3)
(2,3)
(3,3)
Now we can use this table to generate all the options necessary for what we want.
Make a table of the ordered pairs that, if included in R, will give you the property.
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2. (1,1)
Reflexivity
(2,2)
(3,3)
Note: If we have ALL of these pairs, we have reflexivity.
Now look for transivity.
If You
(1,2)
(1,3)
(2,1)
(2,3)
(3,1)
(3,2)
Transitivity
Have These
(2,1)
(3,2)
(1,3)
(3,1)
(1,2)
(2,1)
Then this
(1,3)
(1,2)
(2,3)
(2,1)
(3,2)
(3,1)
Note that each line will give you transitivity in R. You don’t need EVERY line to have
it. Just that if the first two are in R, the third must be as well.
Finally, consider what is necessary for symmetry.
If
(1,2)
(1,3)
(2,1)
(2,3)
(3,1)
(3,2)
Symmetry
Then
(2,1)
(3,1)
(1,2)
(3,2)
(1,3)
(3,3)
Note that each line will give you symmetry in R. You don’t need EVERY line to have
it. Just that if the first one is in R, then the second must be as well.
That was a lot of prep work. But we are now ready to begin the problem (and all
subsequent ones).
If you want an R that does not have symmetry, reflexivity, or transitivity, we simply need
to pick ordered pairs from each table that does not complete a row in any of our tables.
So we can make R = {(1, 1), (1, 2), (3, 2), (1, 3), (2, 2)} or any other number of combinations
that satisfy what we want, just as long as the pairs we choose do not complete a row in
any of our tables.
With this in mind, it should be easy to build an R that has (or not) any of these properties.
This is a handy method that can really help you wrap your mind around what each of the
properties means.
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3. If you’re up for it and find the time, try the following:
Let A = {1, 3, 5, 8}. Build some tables like we have here to show what pairs would give
you each property.
I will continue this discussion regarding anti-symmetry on another document.
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