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Studying an introduction to set theory, we were presented with several axioms as the Axiom Schema of Comprehension and the Axiom of Infinity. This last aximom allows the definition of the inductive set.

The Schema of Comprehension states that every property defines a subset. If Pᵢ(x) is a property of x which can be true or false and Aᵢ is any defined set:

Provided Q(x) ≡ Pᵢ(x) ∧ x∈Aᵢ then, ∃Bᵢ { x ∈Bᵢ | Q(x) }

I have no complete comprehension of what can a property be. Can we define a property for each natural number (#(ϕ)) depending on whether the Gödel coded logical sentence (ϕ) it represents is provable or true? Awareness of the undecidability of this property implies there will be some elements within the subset of natural numbers this property defines that we are just not allowed to know if they belong or not to this new subset. Is this correct and/or allowed?

The main issue I'm trying to settle with this question is to achieve some clarification of which are the rules of logic appliying for the definition of this properties in the Axiom Schema of Comprehension. Can this properties be undecidable? Are there any types of definitions depending on the order of logic? Can someone recomend some reading on this?

I hope I can make the question more clear this time.

UPDATE: Interpretation that Confuses

I’ll write here the main conflict I think is produced between Axiom of Infinity and Axiom of Comprehension. I think this reasoning must not be correct, but I’m having a hard time understanding which are the allowances of the underlying logic that defines properties. Please excuse the untidiness as I’m no mathematician, I’m trying the best I can:

With Axioms of Infinity and the Schema of Comprehension, if we use some type of Gödel encoding for all logical symbols and also include a free variable, g. I will use these expressions:

  • The #’(Ф) is the Gödel coding of logical symbols plus a free pointer variable g
  • True(#’(Ф)) is a logical sentence that holds if Ф is true (we may not be able to decide if it does)
  • Prov(#’(Ф)) is a logical sentence that holds if there exists a logical poof for Ф (we may not be able to decide if it does)

Every natural number encodes to some enumerable phrase in this logical language (that includes g). Some are undecidable, some are ill posed, some have g as a free variable. There is one that states “¬Prov(g)”.

Please consider some group of properties Pᵢ(x), as those which are true only in the case that: While g=i, if True(x) holds, then Pᵢ(x). For example: Pᵢ(x) ↔ g=i → True(x). We have a property for every natural number i. These properties can be undecidable.

For the very specific property in this group, in which i=#’(¬Prov(g)) there will be a member of the inductive set which can’t either belong, neither not belong, to the subset of naturals defined by Pᵢ(x). That member is #(¬Prov(g)).

If ¬True(g), then Prov(g). But if we can prove g, and also, we know that for this particular set i=g, we have that Prov(¬Prov(g)). This means either our logic is faulty or ¬Prov(g) holds.

If True(g), ¬Prov(g), but we also know that Prov(g) because we just prooved ¬True(g) is inconsistent.

So, granted any set Aᵢ, and Q(x) ≡ Pᵢ(x) ∧ x∈Aᵢ then {∃ i∈N | ∄Bᵢ { x∈Bᵢ | Q(x) }}

I think this reasoning is not allowed, or at least not in FOL, it mixes levels. Please if you could point out the flaws.

Original Question Titled: Presentation of the Inductive Set and the Allowed Logical Properties of the Schema of Comprehension

<p>I was studying an introduction to set theory where the axioms are
presented, I believe, according to some systemic expression need. So I
reached a point where we need to define the group of natural numbers,
and with the axioms so far introduced, we cannot postulate the sure
existence of an infinite set. The Axiom of Infinity is then introduced
for the production of the inductive set.</p>

<p>Previously we studied that if we have a set, a property always defines
a subset within it (due to the Axiom Schema of Comprehension). So
following this, it would seem possible to state a property that is not
able to define a subset of naturals. Would this contradict the Axiom
Schema of Comprehension?</p>

<p>Using some Gödel coding, enumerating properties that discriminate upon
provability, and holding every time a different static pointer, until
it matches the index of a Gödel sentence, is it possible to produce
this property? </p>

<p>Please if someone could bring some light into this confusion pointing
out how and why it is or isn’t well conceived, or under what order of
logic could that property be defined. I’ve been very confused after we
reached this axiom because I’m not familiar with the logic rules
underlying the definition of a property, and just know some of the
conventions as presented in the introduction to sets.</p>

<p>Strictly, what can and what cannot be a property? Under which
conditions is this? Maybe you can refer me to some related reading for
this. I found <a href="https://math.stackexchange.com/questions/268044/what-properties-are-allowed-in-comprehension-axiom-of-zfc">this thread</a> that addresses self membership, but
does not consider the fact that naturals can code enumerable logical
sentences. Please help.</p>
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C.Dhio
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  • It might be helpful if you could try to state the comprehension scheme in your own words (as this has the potential to highlight the specific issues/missunderstandings/... you are having with it). Besides that: I have no idea what you're asking in your second paragraph. – Stefan Mesken Mar 30 '16 at 01:32
  • Hi @Stefan, thanks for your answer. What we know so far is that the Schema of Comprehension says that if Pᵢ(x) is a property of x which can be true or false, and Aᵢ any defined set, then: ∃Bᵢ { x ∈Bᵢ | Pᵢ(x) ⋀ x∈Aᵢ } Please correct me if im wrong in this. If we have the Axiom of Infinity and Schema of Comprehension, it seems we can map enumerable logical truths to the naturals in the set of natural numbers. Is it possible to define a property that decides truth for every natural, upon the provability or truth of the logic sentence coded by it? Something like Pᵢ(#(ϕ))↔ϕ – C.Dhio Mar 30 '16 at 13:08
  • These questions are somewhat unrelated. First of all: What you name a "property" is a first-order formula in the language of set theory and "$x \in A$" is such a formula - so there is no need to distinguish between your $P(x)$ and $x \in A$ - they can be combined to the formula $Q(x) \equiv P(x) \wedge x \in A$. Regarding the second questions: There is no formula $P(x)$ such that for every sentence $\phi$, we have $P(#\phi) \leftrightarrow \phi$. This is Tarski's undefinability theorem. – Stefan Mesken Mar 30 '16 at 13:23
  • @Stefan, you are most clarifying. Please comment if the following conclusions are correct: 1. Properties must be defined in FOL using the theory's language. 2. Tarski's Theorem implies that you can't define a property that assigns truth to an element based on the theory's semantics. Regarding point 2, please if could you help me clarify: Is it correct that defining is not the same as deciding in Tarski's Theorem? Could you define a property based on provability? Can you please point some reading to understand Undefinability Theo. in this context? I read wiki but need more introd. Thanks!! – C.Dhio Mar 30 '16 at 13:47
  • Technically, properties don't have to be FOL formulae, but in the given context they are. You can allow more complicated properties, but this isn't what you want to worry about right now. The second point is a bit more complicated. Tarski's Theorem says that there is no single formula that decides truth for all sentences. However, in $ZF$, for every finite number of sentences (and in fact for any level of the Lévy hierarchy such a formula can be explicitly written down - this isn't entirely obvious and requires some work. – Stefan Mesken Mar 30 '16 at 14:24
  • Tarski's undefinability theorem should be covered in any decent introduction to first order logic and personally, I think that it is very important to anyone interested in set theory to have a very solid understanding of these concepts (as they play a role in many technical arguments). Unfortunately, first order logic was amongst those topics that I still learned from German literature and thus I'm not familiar with the English textbooks. – Stefan Mesken Mar 30 '16 at 14:30
  • However, there is a thread here at MSE recommending Good books on mathematical logic. – Stefan Mesken Mar 30 '16 at 14:31
  • Thanks @Stefan for taking the time to answer. Thanks for the recommendations. I'm still abit puzzled in regard to just stating a property that could be undecidable. Why isn't this allowed? I think I should update/reformulate the question in virtue of you clarifications, as it is still On Hold. I believe you point out the main issue is a complete uderstanding of underlying FOL to understand Set Theory. Maybe you can help me clarify this last point. Thanks!! – C.Dhio Mar 30 '16 at 15:07
  • I have no idea what you mean by that and so remain of little help. – Stefan Mesken Mar 30 '16 at 15:56
  • Comprehension , in words states that if $ A$ is a set and $P$ is any property then there is a set $ B$ of all those and only those $x\in A$ for which $P(x)$ is true. If $P(x)$ is is not decidable for any particular $x\in A$. then it is undecidable whether $x\in B$. But the Comp. axiom asserts that the set $B$ exists anyway. For example,in ZFC, let $A$ be the set of injections from $\omega_1$ into $\mathcal R$, and let $ B={f\in A: f$ is a bijection $}$. If ZFC is consistent, that it cannot decide whether $B=\phi,$ but ZFC proves that $B$ exists. – DanielWainfleet Mar 30 '16 at 18:11
  • Thanks @user254665 for your kind comment. I think you point out with precision the confusion I'm trying to clear up. Does this mean that we can define (not decide) a subset of natural numbers that contain all #(ϕ) for which ϕ? Under which order of underlying logic can this be achieved (FOL or HOL)? Wouldn’t this produce some conflict between the Axiom of Infinity and the Schema of Comprehension? Please if someone can shed more light over this, I'd be most thankful. – C.Dhio Mar 30 '16 at 19:28
  • There is no conflict between Comp. and Infinity. I dk what $#(\phi)$ denotes. In the 19th century it was often supposed that we could declare the existence of a set ${x:P(x)}$ for any property $ P$. Until Russell's Paradox. A limitation is to restrict it to ${x\in A :P(x)}.$ The idea being to extract the set of members of some $A$ (that intuitively we "already have") that satisfy $ P.$ – DanielWainfleet Mar 31 '16 at 01:52
  • @user254665 $#\phi$ is the Gödel number of $\phi$. There is an (in fact, many different) effective way to assign a natural number to each formula and thereby formalize deduction system in PA. – Stefan Mesken Mar 31 '16 at 08:59
  • @C.Dhio What does it mean to you to "decide" a subset of natural numbers? (Note that as a consequence of Tarski's theorem - the set $\mathcal T = { #\phi | \phi \text{ is true} }$ provably doesn't exist in any reasonable and consistent set theory - as it would define a truth predicate via $P(x) \equiv x \in \mathcal T$.) – Stefan Mesken Mar 31 '16 at 09:00
  • Thanks @Stefan. I mean by decide that you can't tell if some members belong or not to the set, yet it exists. You mean that the set doesn't exist at all as a consequence of Undef. Theorem? I'll update the main idea of the conflict I observe. Can you please help me identify why this is not allowed. – C.Dhio Mar 31 '16 at 14:20
  • @C.Dhio Yes, we can prove that this set can not exist. The reason for this is exactly the fixpoint argument that proves Tarski's theorem (so you should look up the proof, if you want to understand why this set can't exist). Regarding your notion of decidability: As stated this doesn't make a lot of sense, because we'd have to either restrict ourselfs to sets that provably exist in all models we are interested in or work in a fixed universe that contains all models we would like to consider to even formalize the question whether a fixed element in in some set or not. – Stefan Mesken Mar 31 '16 at 15:09
  • Consider the following example: We may fix a real number $x$ and now ask ourselfs whether $\phi(x)$ is true for some formula $\phi$. If $x$ provably exists in every model we are interested in, this is reasonable question. But it turns out that there are many real numbers that don't exist in every model of ZFC (or some subset/extension thereof). In fact, for every model $M$ of ZFC we can show that it is consistent to have a model $N$ of ZFC such that $M \subseteq N$ and $N$ contains a real number $y$ that is not an element of $M$. It then doesn't even make sense to ask "$M \models \phi(y)$?" – Stefan Mesken Mar 31 '16 at 15:14
  • Dear @Stefan, you are most enlightening in your replies. I have a couple of comments: 1.- I understand what you say about undecidability. I was just stating what I meant when I said "...we can define (not decide) a subset of naturals..." not decidability in general. 2.- I'm trying to understand Tarski's Undefinability Theorem (not quite there :D) which appears as the main missing piece. I must understand that from it we can prove, not that we cannot prove truth for all #(ϕ), if Pᵢ(#(ϕ))↔ϕ, but instead, just that ∄Pᵢ. Is this correct? Thanks again! – C.Dhio Mar 31 '16 at 15:50
  • Yes, there is no formula $P$ such that for every formula $\phi \colon P(# \phi) \leftrightarrow \phi$. Or in plain English: The truth predicate isn't definable. – Stefan Mesken Mar 31 '16 at 15:59
  • Thanks for your help @Stefan. I'm getting out of topic, but it seems that it has that as a rather epistemological consequence: we are not able to reffer to "all truths" without getting inconsistent (at least in FOL). I'd like to undestand why so I'll study more to undestand Tarski's Theorem. – C.Dhio Mar 31 '16 at 16:20
  • In first-order logic you can't define $ S= {$#$\phi | \phi},$ or, equivalently, $\forall n\in N; n\in S\iff \exists \phi ; (n=$#$\phi}$ because you can't quantify over sentences, so you would have to write an infinitely long sentence incorporating every $\phi$ written out in its full first-order glory. – DanielWainfleet Mar 31 '16 at 16:37
  • Thanks @user254665, you state it very clearly. I agree the reasoning I posted cannot be stated in first order. Does the scope of Tarski's Theorem include HOL? – C.Dhio Mar 31 '16 at 16:54
  • @user254665 If by $S$ you mean the set of Gödel numbers of all formulae, that isn't true. We can define the set $S$ and in fact, we can construct our Gödel numbering such that $S = \omega$ (or $\mathbb N$, if you prefer that). What we can't define is the set of those Gödel numbers for formulas, that are true - at least in ZFC. – Stefan Mesken Mar 31 '16 at 16:55
  • S is not the set of all Godel numbers.It's the set of the Godel numbers of the true sentences. – DanielWainfleet Mar 31 '16 at 17:34

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