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I am developing a ZFC axiomatic system where together with the empty set, there is a singular (and huge) set of constants that are themselves sets and form a complete ordered field (cof) these constants could be called Rcof, +cof, .cof, <cof, 0cof, 1cof, etc.

For sure, in this ZFC axiomatic system, the Real numbers R could be constructed in the standard Dedekin's cuts way. By construction ( 0 = {}, 1 = {0} = {{}}, 2 = {0,1} = {{},{{}}}, etc ), 1 ∈ 2 is obviously true for Real numbers.

My question is: 1cof ∈ 2cof can be proved, disproved or neither? (That is using the cof definition only). In standard ZFC, (∗) there is, upon isomorphism, just one cof . This mean that must be an isomorphism between R (build in my ZFC system) and Rcof but it seem useless to prove 1cof ∈ 2cof.

May be (∗) is false my ZFC axiomatic system and I have to live there with two cof. Is this a serious problem in my system?

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    $\begingroup$ Your title was my thought exactly. $\endgroup$
    – Todd Trimble
    Commented Nov 14, 2014 at 20:23
  • $\begingroup$ @ToddTrimble sorry for "destroying" your comment; I had not seen it before I started my edit. $\endgroup$
    – user9072
    Commented Nov 14, 2014 at 20:27
  • $\begingroup$ Your answer ("your system cannot settle the question of whether 1cof∈2cof") is enough for me, Thank you. $\endgroup$
    – Carlos Freites
    Commented Nov 14, 2014 at 20:37
  • $\begingroup$ I deleted my comment and simply posted the same idea as an answer. $\endgroup$
    – Joel David Hamkins
    Commented Nov 14, 2014 at 20:43

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Assuming that ZFC is consistent, then your system will not settle the question of whether 1cof is an element of 2cof or vice versa. One can see this by observing that from any model of ZFC we can make various models of your system, by interpreting your new constant symbols 1cof, 2cof etc. in any of the various ways that we know how to construct a complete ordered field. Some of these constructions will give rise to models of your system where 1cof $\in$ 2cof, and some will not have this. We could also make 2cof $\in$ 1cof, or any other crazy pattern, simply by choosing sets to satisfy the desired type, and then imposing the complete ordered field structure upon that set of choices.

You didn't say how many constants you had, but your remark that it was "huge" suggests that you possibly intended to have more than continuum many such constants. That is no problem, because if ZFC is consistent, then by the Löwenheim-Skolem theorem there is a model of ZFC whose reals are at least that large, and we can use this model as the starting point for the rearranging process.

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  • $\begingroup$ I mean that in the same way ZFC can be introduced the empty set ∅ as For all y, Not(y ∈ ∅) $\endgroup$
    – Carlos Freites
    Commented Nov 14, 2014 at 21:26
  • $\begingroup$ Ah, so you want the constants to be atoms? (see en.wikipedia.org/wiki/Urelement). This contradicts ZFC, of course, and if this is what you mean you should be talking about ZFA or some other set-theory accommodating urelements. But in that case, if you wanted the constants to all be atoms, then of course 1cof $\notin$ 2cof would be an immediate consequence of this, and the theory would decide it. (Which I suppose goes to show that if one has an axiomatic system in mind, it helps to be clear about exactly what it is.) $\endgroup$
    – Joel David Hamkins
    Commented Nov 14, 2014 at 21:35
  • $\begingroup$ I mean that in the same way in ZFC can be (there are others ways) introduced the empty set ∅ as For all y, Not(y ∈ ∅) I state that R, +, ., <, 0, 1 form a complete ordered field. I was sure should not be problem with this, until I ask myself the relation between this constants and the Real numbers that can be constructed in the usual way inside my system. This is the origin of my question. I interpreted your original comment "your system cannot settle the question of whether 1cof∈2cof" as "nether 1cof∈2cof or Not(1cof∈2cof) can be proved" and that was the confirmation I was expecting. $\endgroup$
    – Carlos Freites
    Commented Nov 14, 2014 at 21:59
  • $\begingroup$ No. I intent all my constants as being sets. This very important because I do not want stay away from ZFC. $\endgroup$
    – Carlos Freites
    Commented Nov 14, 2014 at 22:02
  • $\begingroup$ So you don't want urelements, and you seem to want to introduce them as defined objects in ZFC. This will require fixing a particular representation of the complete ordered field in ZFC (which is a silly thing to do for all the usual reasons that are commonly given). If you define them as equivalence classes of cauchy sequences or Dedekind cuts in $\mathbb{Q}$, etc., then you will have 1cof $\notin$ 2cof, since no such equivalence class is an element of another (although this depends on how you define $\mathbb{Q}$ and how you define sequences, etc., an essentially similar issue).... $\endgroup$
    – Joel David Hamkins
    Commented Nov 14, 2014 at 22:05

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