1

I was reading about total orders on this website: https://medium.com/@WindUpDurb/on-partial-ordering-total-ordering-and-the-topological-sort-9f9c0d0d812f and they say that $(ℤ, <)$ is a total order, but I think that it's incorrect since for it to be a total order it must first be a partial order, which means the relation $<$ must be reflexive. $<$ is not reflexive, so that shouldn't be a a total order, correct?

cekami7844
  • 686
  • 1
    https://math.stackexchange.com/questions/367583/example-of-partial-order-thats-not-a-total-order-and-why – amWhy Apr 02 '20 at 17:31
  • 1
    Tomayto, tomahto. For some people partial orders are reflexive, for others they are irreflexive. If you like your orders to be reflexive, just change $\lt$ to $\le$. What's the problem? – bof Apr 02 '20 at 17:36
  • @bof I dont get it. My book says: Let $R$ be a relation on a set $S$. $R$ is a partial order if and only if the following three statements are true for every $a$, $b$, and $c$ in $S$: Reflexive Property: $aRa$ Transitive Property: If $aRb$ and $bRc$, then $aRc$. Antisymmetric Property: If $aRb$ and $bRa$, then $a=b$. – cekami7844 Apr 02 '20 at 17:38
  • @cekami7844 Yes, that is the definition of partial order. If you look at the definition of total order on Wikipedia you'll see that both $(\mathbb{Z},<)$ and $(\mathbb{Z},\leq)$ are total orders under that definition. – saulspatz Apr 02 '20 at 17:42
  • @saulspatz, Wikipedia says: "The connex property also implies reflexivity" "a total order is also a (special case of a) partial order, as, for a partial order, the connex property is replaced by the weaker reflexivity property" so how is $(Z,<)$ a total order, if $<$ is not reflexive? that's what I don't understand – cekami7844 Apr 02 '20 at 17:52
  • @amWhy I dont understand the most upvoted answer.. that guy said: "$A⊂A$" how is that correct? – cekami7844 Apr 02 '20 at 17:54
  • @cekami7844 So it does. My mistake. Sorry. Still as bof says, some people define total orders so that $\leq$ is a total order and some so that $<$ is a total order. I'm used to thinking of both as total orders. You'll find this is quite common in math. There are often minor differences in the ways people use the same term, and sometimes the differences aren't so minor. – saulspatz Apr 02 '20 at 17:57
  • Per @bof, there are two types of partial orders, plain ole' partial order (weak) and strict partial order (strong). So your post is not clear, cekami7844. Every partial order is not necessarily a total order, and vice versa, depending on which sort of partial order you have in mind. – amWhy Apr 02 '20 at 17:58
  • cekami7844: See the entire Wiki entry on partial orders. And scroll down, or use this link to take you to the part of that entry that describes non-strict vs. strict partial orders. You'll see that your set defines a strict partial order, meaning it is irreflexive, transitive, and antisymmetric. – amWhy Apr 02 '20 at 18:06
  • @amWhy how is my post not clear, and every total order is a partial order, as I said before, Wikipedia says: "a total order is also a (special case of a) partial order, as, for a partial order, the connex property is replaced by the weaker reflexivity property", and I didn't ask if $(Z,<)$ is a strict partial order, I asked how can $(Z,<)$ be a total order, if a total order requires the relation to be reflexive – cekami7844 Apr 02 '20 at 18:10
  • Also, cekami7844, see the entry discussing Total orders. There is no requirement of reflexivity. "Formally, a binary relation $ \leq$ is a total order on a set $X$ if the following statements hold for all $ a,b and c\in X$ :

    Antisymmetry: If $a \leq b$ and $b\leq a$, then $ a = b$. Transitivity: If $a \leq b$ and $b\leq c$, then $a\leq c and b ≤ c.$ Connexity(Comparabilty) $a\leq b$ or $b\leq a}$ for all $a, b$ in $X$."

     – amWhy Apr 02 '20 at 18:12
  • No, what you want to say is that every total order is a non-strict partial order. Have you read my comments and linked to the sites I suggested? Your relation is not a non-strict partial order, nor is it a total order. It is a strict partial order, which is by definition, NOT reflexive. – amWhy Apr 02 '20 at 18:14
  • @amWhy I understand the difference between strict partial and nonpartial order, and when I say total order I mean total order that I not strict. A strict total order Is a different thing https://mathworld.wolfram.com/StrictOrder.html I think there's a problem with the terminology in general.. – cekami7844 Apr 02 '20 at 18:25
  • Formally, a binary relation $ \leq$ is a total order on a set $ X$ if the following statements hold for all $ a,b$ and $c \in X$:

    Antisymmetry I $a ≤ b$ and $b ≤ a$ , $ a=b;$ Transitivity : If $a\leq b$ and $b\leq c$ then $ a\leq c$; Connexity: If $a ≤ b$ or $b ≤ a$ for all $a, b \in X$

     – amWhy Apr 02 '20 at 18:25
  • Well you have a strict total order on your set, which also defines a strict partial order. It is not reflexive, as the criteria for a strict total order does NOT require reflexivity. It requires that if $a\lt b$, then $a\not\geq b$. – amWhy Apr 02 '20 at 18:34
  • you said that one of the properties that a total order has to satisfy is connexity, Wikipedia states: "The connex property also implies reflexivity" https://en.wikipedia.org/wiki/Total_order so I dont get how it does not require reflexivity.. – cekami7844 Apr 02 '20 at 18:38
  • I quoted from the section in the entry on total order that described non-strict total orders. Please read when you scroll slightly downward on the total order entry, to strict total order. – amWhy Apr 02 '20 at 18:46
  • Note also it does not explicitly forbid saying that $a=a$; we just can't say $a/lt a$. The major point of a total order is that all elements are comparable to all other elements. For any elements $#$ and $%,$ in a set $\text{Foo}$ under the strict total order relation we'll call $\oplus$, that either $# \oplus %$ or $%\oplus #$. – amWhy Apr 02 '20 at 18:51
  • cekami7844 I know this question, and the links I've sent you, and the complication that there are non-strict and strict partial orders, and similarly non-strict and strict total orders has probably been confusing to you. Let me link you to a math.se post about strict total orders: https://math.stackexchange.com/questions/1695465/why-are-is-equal-to-and-total-strict-ordering-mutually-exclusive?noredirect=1&lq=1. Fire off any questions when you've had a chance to look over all the links and comment, and sit on it a while. I'll be around! – amWhy Apr 02 '20 at 18:58
  • My last linked question has no official answer, but you can see in the comments, strict total orders are irreflexive order relations, like you have in $(\mathbb Z, <)$. – amWhy Apr 02 '20 at 19:14
  • Bottom line is this. $(\mathbb Z, <)$ is neither a non-strict partial order, nor a non-strict total order. However, it is both a strict partial order AND a strict total order. So, as an answer to your title question. No it is not a non-strict total order (but to correct your post, neither is it a non-strict partial order). But, it is a strict total order, which also happens to qualify as a strict partial order. – amWhy Apr 02 '20 at 21:23

0 Answers0