Talk:Total order/Archive 1

Archive 1

Errata

Why do we have a section called "Mathematical errata"? Is there something wrong with those statements? —Vivacissamamente 13:08, 4 January 2006 (UTC)

The answer seems to be that an anonymous contributor added this last August, and no-one questioned it at the time. I can't find any justification for this heading, so I'll remove it. Elroch 20:55, 21 February 2006 (UTC)

Examples?

'Natural numbers, integers, rational numbers, and real numbers ordered by the standard less than (<) or greater than (>) relations are all total orders.'

Shouldn't this be with <= and >=? With strict inequalities the relations aren't total. -- C (Sept 18, 2005)

True, the relations are not total. But they are transitive and trichotomous, and therefore fit with the alternative definition given just before the examples section. That is, they are total orders in the irreflexive sense, rather than the reflexive sense. You can change them if you like, but they are not really wrong. --Zundark 19:14, 18 September 2005 (UTC)

Connected?

The unique smallest unbounded connected total-ordered set is the real numbers.

What does "connected" mean here? -- JanHidders (06:21, Aug 16, 2001)

Maybe connected in the order topology? This should be explained, I agree. (06:40, Aug 16, 2001 AxelBoldt)

Rationals the smallest dense total-ordered set?

I think the statement about the rationals being the smallest dense total-ordered set is false. For instance Q[0,1] is dense and smaller. --AxelBoldt (12:54, Aug 16, 2001)

I'm not an expert in this area, but what definition of "smaller" are you using? I would say in this case that A <= B iff there is an order homomorphism from A to B. In that case obviously Q\[0,1] <= Q, but also Q <= Q\[0,1] by, for example, f : Q -> Q\[0,1] such that

f(x) = x if x < 0 and

f(x) = x + 2 if x >= 0.

--JanHidders (05:14, Aug 17, 2001)

I think A <= B is supposed to mean that A is order-isomorphic with a subset of B. However, I don't think it's very clear, and would have no objection if someone decided to remove the whole paragraph. Note that Q\[0,1] is order-isomorphic to Q. While I'm here, I'd like to suggest that we try to use "totally ordered" rather than "total-ordered". This is far more common in my experience, and is consistent with our use of "partially ordered" (rather than "partial-ordered"). Zundark, 2001-08-17

I must be missing something. Zundark, isn't my definition equivalent with yours? My f is an order-isomorphism from Q to Q\[0,2), which is a subset of Q\[0,1]. No!?

Btw., I agree with "totally ordered" replacing "total-ordered". In the computer-science literature I know, this is also the more common term. --JanHidders (06:23, Aug 17, 2001)

Sorry, I wasn't being very clear. I wasn't disagreeing with you, I was just responding to Axel. If by "order homomorphism" you meant x < y implies f(x) < f(y), then your definition is the same as mine. Zundark, 2001-08-17

Q[0,1] (if it means Q∩[0,1], as I guess) is not isomorphic to Q as it has both a greatest and a least element, whereas Q has neither. I guess you actually mean Q(0,1).--Army1987 19:50, 19 September 2005 (UTC)

No, it was Q\[0,1], that is, the set of rationals not in [0,1]. This can be seen from the diff of my first edit of 17 August 2001. Changes to the software during the last four years must have messed it up. I've now restored the comments to their original form (I hope). --Zundark 08:13, 20 September 2005 (UTC)

---

I agree with everything that was said above. I guess one could justify the statement about the rationals after all. I also like totally ordered better. --AxelBoldt

---

What about the set of all rational numbers that in lowest terms take the form a/2^n? This is dense and has neither an upper nor lower bound. Is it order-isomorphic to the rationals?

Yes. Every countably infinite totally ordered set which is dense and has no greatest or least element is order-isomorphic to the rationals. --Zundark, Wednesday, April 17, 2002

Could you give an example of order isomorphism between {a/2ⁿ: a and n are integers} and Q? (Army1987 09:44, Mar 30, 2005)

I can give such an example, but it may take a while to write. Michael Hardy 23:47, 29 Apr 2005 (UTC)

Isn't the question mark function one of these?--Army1987 20:49, 3 September 2005 (UTC)

(8 years later :-) I wonder, why not mention that all countable dense unbounded totally ordered sets are isomorphic? See Back-and-forth method#Application to densely ordered sets. Boris Tsirelson (talk) 17:31, 12 November 2013 (UTC)

Exclusive or?

I may be wrong but shouldn't the totalness requirement be "either aRb or bRa"? Isn't what we want an exclusive or?

I don't think so. If a and b are actually equal, then we want both aRb and bRa. If they are not equal, then we have an exclusive or because of the antisymmetry. -- Jitse Niesen 14:55, 15 Apr 2005 (UTC)

if a ≤ b and b ≤ a then a = b (antisymmetry)
if a ≤ b and b ≤ c then a ≤ c (transitivity)
a ≤ b or b ≤ a (totalness) 

Should "or" here be logical disjunction? Isnt that the emphasis of the comparison with antisymmetry? -SV|t 00:55, 30 Apr 2005 (UTC)

Yes this "or" should be (and is) a logical disjunction. Paul August ☎ 03:15, Apr 30, 2005 (UTC)

http://planetmath.org/encyclopedia/TotalOrder.html http://mathworld.wolfram.com/TotallyOrderedSet.html These pages say it should be either/or. --Charlatino (talk) 05:36, 21 July 2008 (UTC)

Or is generally assumed to be and/or in most contexts. Dcoetzee 05:47, 21 July 2008 (UTC)

Completeness

A totally ordered set is said to be complete if every subset that has an upper bound, has a least upper bound.

Am I right in thinking this statement is self-dual?--Malcohol 14:18, 23 October 2006 (UTC)

Totality implies reflexivity?

“Totality implies reflexivity, that is, a ≤ a.”

In (Partee, ter Meulen & Wall 1987: 51) an order (strict or weak) is said to be total if it's connected. And the definition of connectedness (p. 42) says: “A relation $R$ in $A$ is connected (or connex) if and only if for every two distinct elements $x$ and $y$ in $A$, $\langle x,y\rangle \in R$ or $\langle y,x\rangle \in R$ (or both)”

I don't know how widespread this definition is, but at least it's seems that it's not true universally that totality implies reflexivity.

Partee, B., ter Meulen, A., & Wall, R. (1987) Mathematical Methods in Linguistics. Dordrecht: Kluwer.

129.67.168.192 (talk) 12:53, 23 January 2010 (UTC)

Operator notation

The make the article more general and readable to the non-expert, it may be helpful to define the operator "*" . --69.88.160.1 (talk) 21:52, 18 September 2011 (UTC)