Antisymmetry & Partial Orderings - H&J Ch.2 Section 5 | Peter's Help

[Math Amateur]

I am reading "Introduction to Set Theory" (Third Edition, Revised and Expanded) by Karel Hrbacek and Thomas Jech (H&J) ... ...

I am currently focused on Chapter 2: Relations, Functions and Orderings; and, in particular on Section 5: Orderings

I need some help with H&J's depiction of antisymmetric relations and partial orderings ...The introduction to H&J's section on antisymmetric relations and partial orderings reads as follows:View attachment 7599In the above text from H&J we read the following:

" ... ... A binary relation \(\displaystyle R\) in \(\displaystyle A\) is antisymmetric if for all \(\displaystyle a, b \in A\), \(\displaystyle aRb\) and \(\displaystyle bRa\) imply \(\displaystyle a = b\) ... "Now since \(\displaystyle a = b\) in the above instance \(\displaystyle aRb\) and \(\displaystyle bRa\) can be expressed as \(\displaystyle aRa\) (or \(\displaystyle bRb\) ...) ... so isn't antisymmetry essentially reflexivity ...

Can someone explain to me exactly what the essential difference between antisymmetry and reflexivity ... and, indeed why we don't define partial orderings simply as relations that are reflexive and transitive ... Help will be appreciated ...

Peter

 

[Evgeny.Makarov]

In the above text from H&J we read the following:

" ... ... A binary relation \(\displaystyle R\) in \(\displaystyle A\) is antisymmetric if for all \(\displaystyle a, b \in A\), \(\displaystyle aRb\) and \(\displaystyle bRa\) imply \(\displaystyle a = b\) ... "Now since \(\displaystyle a = b\) in the above instance \(\displaystyle aRb\) and \(\displaystyle bRa\) can be expressed as \(\displaystyle aRa\) (or \(\displaystyle bRb\) ...) ... so isn't antisymmetry essentially reflexivity

I think you should think about this more rigorously. The definition of antisymmetry does not say $a=b$ (and for which $a$ and $b$?). If you think there is an equivalent, simpler statement of antisymmetry, try to write it precisely and then prove that it is equivalent to the original definition. Considering examples of relations that are and are not antisymmetric also helps.

 

[Math Amateur]

I think you should think about this more rigorously. The definition of antisymmetry does not say $a=b$ (and for which $a$ and $b$?). If you think there is an equivalent, simpler statement of antisymmetry, try to write it precisely and then prove that it is equivalent to the original definition. Considering examples of relations that are and are not antisymmetric also helps.

Hi Evgeny,

Yes I was being somewhat informal ...

I should have said ... since aRb and bRa imply a = b ... then we can essentially write aRa or bRb ... that is the condition aRb and bRa gives us reflexivity ... hmm ... yes ... see your point ... what exactly am i saying ...

Have to rethink my question ...

By the way ... can you think of a good example of an antisymmetric relation ...?

Peter

 

[Evgeny.Makarov]

can you think of a good example of an antisymmetric relation ...?

Both non-strict and strict inequalities on numbers are antisymmetric, as are the divisibility relation on integers and the empty relation. Another one is \(\displaystyle \begin{cases}x<y,&x\ge0,y\ge0\\x\le y,&\text{otherwise}\end{cases}\) on real numbers.