I have some difficulties in understanding the [uni]verse platonic view. How are we to understand the existence of a model of a false theory? what is the relationship of this model to the platonic world of sets?

My personal try is the following, let's assume the existence of a platonic universe $P^{sets}$ of sets, and also another universe $P^\in$ that serves as a relational universe, i.e., that can stand as a representative of the relation epsilon $\in$ in the real world. For example $P^\in$ might be a universe of some object representation of "ordered pairs" of sets, these ordered pairs are primitive ones, they stand for really existing objects exemplifying ordered pairs in a platonic realm. So $P^\in$ would be a really existing platonic membership relation between objects of $P^{sets}$.

Now when we say for example the rules of $ZFC$ are true, in a platonic sense, would that be taken to mean that the sets in $P^{sets}$ would be related to each other by the platonic membership relation in a manner that abides $ZFC$ rules.

So now when we say that $ZFC$ can interpret $ZF\neg C$ then this mean that the later is consistent (should $ZFC$ be consistent) so there would exist a model of $ZF\neg C$, but how I'm to understand the existence of this model in the platonic world? is it the case that the interpretation would be a pair of subsets of $P^{sets}$ one representing a domain and the other is a set of set-ordered pairs (non primitive ones, i.e. interpreted ordered pairs as special kinds of sets), and so the rerlation object interpreting the membership relation is not a real relational one, i.e. not composed of primitive membership ordered pairs??? as it is the case with ZFC? so this model is deemed as not real?

Is that a good analogy? I mean I want to conceal two matters: that there EXISTs a model of a false but consistent theory, if we are to hold platonistic views, then this existence must be in some reality? otherwise the assertion "there exists" wold have no meaning, now this reality must be taking part in the true platonic world of sets? where else it would be? On the other hand we must differentiate this kind of existence of a model of a false theory from the existence of a model of a true theory! that's why I used the "primitive ordered pairs" to represent the relations in the theory, a true theory would have its relation sets being real in the platonic sense, i.e. composed of primitive ordered pairs abiding the rules of that theory, while a false one would not have them real, i.e. the relation on its domain is not composed of primitive ordered pairs.

I always had the following impression:

for every effectively generated theory $T$ we have: $$ Con(T) \to \exists T^* (True(T^*) \wedge T^* \ inteprets \ T)$$

the idea can be seen in the above anaology, if a set theory $T$ is consistent but false, still the model of $T$ must be a part of $P^{sets}$ and so there must be a true theory $T^*$ (i.e. $T^*$ satisfied in $P^{sets}$) that can itnerpret $T$, otherwise how can we explain that a model of $T$ must exist? exist where?

Is that impression correct?

I visualize the whole platonic world of mathematics $P^{math}$ as the union of the platonic worlds of the individual disciplines i.e. the union of $P^{arith}$ , $P^{Geom}$ , etc..., I think standard models of those disciplines are the real ones represented by the Platonic realms of those disciplines, i.e. they have real relation sets, as opposed to the fake or false theories which don't have real grounded relations in the platonic world.

I know that this question is in some sense philosophical, but "platonism" especially the [uni]-versed one, is an almost ongoing working assumption of most innovative work in mathematics and set theory, so it is important to at least shed some light on that aspect.

  • 3
    $\begingroup$ What is a "false theory"? I am familiar with the phrase "false statement" but I have never heard of a theory being false or true. By "false theory" do you specifically mean "not ZFC"? $\endgroup$ Mar 7 '18 at 20:24
  • 4
    $\begingroup$ @Zach Hi. For a theory of arithmetic, being true means "being true in the standard model of arithmetic" (that is, the actual natural numbers). For a set theory the notion is harder to define, as it depends on "the true model of set theory". But you can still say a few things, regardless of whether you believe that there is such an object. For instance, surely a theory whose arithmetic consequences are false is a false theory. Most set theorists would say that if your theory proves that, say, $0^\sharp$ does not exist, then it is false. Plus, the notion makes sense within any model of ... $\endgroup$ Mar 8 '18 at 2:14
  • 4
    $\begingroup$ @Zach ... set theory. Say your model is $M $ and, within $M $, $T $ is a theory of sets. If $T $ contradicts (from the point of view of $M $) what $M $ "thinks" is true, then (from $M $'s perspective) $T $ is false. Note that here this is all relative to $M $, regardless of whether or not $M $'s theory is true, or even whether one believes that it makes no sense no ask whether $M $'s theory is true. $\endgroup$ Mar 8 '18 at 2:18
  • $\begingroup$ @ZachTeitler the context of this posting is Platonism with a single universe, so it is relevant to speak of a Platonistic criterion of truth. So a true theory means a theory whose sentences are satisfied in the fragment of the Platonic world that it speaks of (i.e. its symbols are taken to refer to, or range over), so if suppose the sets and the membership relation in our Platonic world obeyed rules of ZFC, then a theory speaking about the same referents of ZFC, i.e. about sets and the membership relation, but had a rule that contradicts one of the rules of ZFC, would be deemed a false theory. $\endgroup$ Mar 8 '18 at 12:11
  • $\begingroup$ @ZachTeitler .....[continuation], now that theory is either inconsistent, or it might be consistent! but here in latter case, it would have a model, and that model is a fragment of the Platonic world itself but it is not the same fragment of the Platonic world that ZFC is referring to! so, in that case, that false theory would turn into a true theory if its referents are changed to the proper ones. $\endgroup$ Mar 8 '18 at 12:17

I'll answer this from a Platonic viewpoint. Consider a theory $T$ that is false when its primitive concepts are interpreted in the standard way. For example, $T$ could be the theory ZF$\neg$C that you mentioned in this connection. If we interpret the primitive concepts "set" and "element" in the standard way, so that "set" refers to all the sets in our Platonic world and "element" refers to the actual elementhood relation between these sets, then the axiom of choice is true, so ZF$\neg$C is a false theory. But, because it is consistent, there are other interpretations of the primitive concepts that make ZF$\neg$C true. That is, one can define (within the Platonic universe, using the standard notions of set and element) concepts that I'll (for the purpose of this answer) call "pseudo-set" and "pseudo-element" inn such a way that all of the axioms of ZF$\neg$C become true when one interprets "set" and "element" to mean "pseudo-set" and "pseudo-element". More formally, if you take any axiom of ZF$\neg$C, replace every $\in$ in it with the "pseudo-element" relation, and replace every quantifier "for all $x$" or "there exists $x$" with "for all pseudo-sets $x$" or "there exists a pseudo-set $x$" (respectively), then the resulting statement is true in the Platonic universe (and in fact is provable in ZFC). So the answer to "where do models of false theories exist" is "in the Platonic universe" --- but they depend on using unusual interpretations of the primitive terms of those false theories. To put it another way: A false theory becomes true if its primitive concepts are suitably reinterpreted. (A Platonist might well say "misinterpreted" instead of "reinterpreted".)

  • $\begingroup$ Yes, but what is the thumbprint of truth? I mean from a Platonic perspective. To emphasize what I mean you said $\in$ is "actual elementhood relation between sets" what does "actual" mean here? Should I take the "pseudo-element" relation to be 'not actual relation'? how 'actual' translates in Platonic terms? In my account, 'actual' relation is marked by being composed of primitive ordered pairs existing in that platonic universe, while "not actual" or "re-interpreted" relation is composed of "interpreted ordered pairs" which are not the primitive ones, they are sets working as ordered pairs. $\endgroup$ Mar 7 '18 at 7:55
  • $\begingroup$ So in my account one can discriminate between true relations and false ones by the type of ordered pairs they use, so there is some Platonic mark of truth. In your account I don't see enough information to make that discrimination (I mean in Plastonistic terms), for instance, I can take $ZF \neg C$ to be the true one and consider $ZFC$ to be the re-interpreted one, both have their domains in $P^{sets}$ and there is no clear Platonic feature by which I can discriminate 'true' relations from 'false' ones? $\endgroup$ Mar 7 '18 at 8:06
  • $\begingroup$ as an analogy, the box and sphere optical illusion is not solvable, it only depends on our choice, while the wine glass two faces optical illusion is in some sense solvable towards the glass being the principal image and the two faces being the alternative (re-interpreted image), there is a thumbprint that settles the preference here, since the image fulfills more properties of the glass than of the faces. Your account looks like the first one, so relative, while I try to conceptualize some platonic features that marks true models from false ones. $\endgroup$ Mar 7 '18 at 8:17
  • $\begingroup$ On the other hand, I like your answer "in the platonic universe", this motivates my claim that every consistent effectively generated system is interpretable in some TRUE theory. $\endgroup$ Mar 7 '18 at 8:20
  • 3
    $\begingroup$ Yes. To say that a consistent theory $T$ has a model means exactly that the primitive notions in $T$ can be interpreted in such a way as to make all the axioms of $T$ true. It does not guarantee that this interpretation will agree with the one that you had in mind when formulating $T$, nor with the one that most people would like to use. $\endgroup$ Mar 8 '18 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.