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I have two questions, and I address them in two seperate paragraphs below.

I read that if we do not accept the Completeness Axiom for the reals and fields but accept all the other axioms for fields and the reals, then given any two-element field, whose elements are denoted by $0$ and $1$, we can show that the field does not have a supremum. I'm not sure how to prove this, but I believe it may be a consequence of the fact this two-element field cannot be totally ordered but I don't know how to then show that this field does not have a supremum; for, it seems to me that the supremum is simply $1$ since it is the smallest element such that all the other elements ($0$) in the field are smaller than it.

Also, I have another question, in my textbook, where it tries to verify that this set of reals will be field it assumes that $0+0=1+1=0$. Now, I can't simply understand how $1+1=0$ but I do believe that the author is trying to show, somehow, the inverse of $1$, but surely that would be $-1$? Now, if it were group theory, this might have made sense since $1+1=0$ for the set of integers in modulo $2$. After some speculation, I noticed that the author said, for fields, we define two operations to have the properties of inverses, identities, associative law Et Cetera. So, does it mean that for this field $1+1=0$ is defined this way?

(Note: By the other "axioms" I mean the distributive, associative, and commutative laws, the identity property and the inverse property. Also, the reason I am asking this question is to clarify a statement the author makes to show that without the completeness axioms some fields may not have a supremum: "we will see in Example 1.1.2 (the field wih elements $0$ and $1$) that properties A-H does not gurantee that every nonempty set that is bounded above has a supremum". And, the text I am quoting from is http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF)

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    $\begingroup$ Where did you read what you say you read? Makes very little sense to me. The elements of the two-element field are not real numbers. (Or at least thinking of them as reals is a very bad idea.) Saying the two-element field has no supremum makes no sense because there's no order on it, as you note. Yes, when they define that field they define the operations, and $1+1=0$ is part of that definition. $\endgroup$
    – David C. Ullrich
    Aug 8, 2015 at 20:39
  • $\begingroup$ You probably read instead that a field that is not complete (but has all the other axioms of the real field) has not the least upper bound property. $\endgroup$
    – Tryss
    Aug 8, 2015 at 20:42
  • $\begingroup$ This question is quite incomplete. It would be better if you could be more specific, including the language of the source, if possible $\endgroup$
    – Thomas Andrews
    Aug 8, 2015 at 20:43
  • $\begingroup$ I may be misinterpreting (having read it again I believe I am, in the sense, as you noted, that they are not a field on the reals), but it's from ramanujan.math.trinity.edu/wtrench/texts/… on the first couple of pages. $\endgroup$
    – Reinhild Van Rosenú
    Aug 8, 2015 at 20:43
  • $\begingroup$ @DavidC.Ullrich Sorry for the terrible question and my misrepresentation of the text and the words "we will see" in the text as they go on to prove, as you also said, that the field will not be totally ordered so it having a supremum does not make sense. Although this question should be deleted for its vague character now that I have re-read it, I believe it useful to someone who is wondering the same thing and looks for some guidance. $\endgroup$
    – Reinhild Van Rosenú
    Aug 8, 2015 at 21:08

1 Answer 1

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In a finite set that is linearly ordered every non-empty subset has a supremum that is actually a maximum.

"all the other axioms for fields and the reals" is highly non-specific, and precision in matters of this sort may be just the thing needed to clear up confusions.

A field with two elements can have its elements linearly ordered so that $0<1$, but that is not what is called an "ordered field" because it fails to satisfy the rule that if $a<b$ then $a+c<b+c$. Thus \begin{align} 0 & < 1 \\[10pt] \text{but }0+1 & \not < 1+1. \end{align}

Your question isn't clear enough to me for me to be confident as to whether this answers it or not.

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  • $\begingroup$ By the other axioms, or properties, I mean the inverse property, the identity property, the associative property, the distributive property, and the commutative property. $\endgroup$
    – Reinhild Van Rosenú
    Aug 8, 2015 at 20:47
  • $\begingroup$ I edited the question with additional detail if that helps. $\endgroup$
    – Reinhild Van Rosenú
    Aug 8, 2015 at 20:56

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