I am struggling to understand what is different from 'Godel incompleteness' (GI) and call it 'ordinary completeness' (OC). Godel showed FOL to be complete. Yet, Godel's theorems establish that the language Q (an axiomatic extension of FOL) is incomplete. What distinguishes GI from OC in terms of FOL?
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Through browsing questions asked in the past about Godel's completeness and imcompleteness theorems, I've come to see that the sense of completeness in both of the theorems are different. However, I can't see how they are different!   Does this distinction boil down to that of truth and proof? Is it the case that the completeness theorem talks about the relationship between truth and proof, while the incompleteness theorems only relate to provability?  Could someone please elaborate on the sense in which both notions of completeness are different?
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I'm reading the proof of lemma 15.43 of Jech's Set Theory:     Let $G$ be generic on a complete Boolean algebra $B$. If $M$ is a model of $\mathsf{ZFC}$ such that $V\subset M\subset V[G]$, then there exists a complete subalgebra $D\subset B$ such that $M=V[D\cap G]$.   In the proof of this theorem the author states the following fact: "First we note that since $M$ satisfies the Axiom of Choice, there is for every $X\in M$ a set of ordinals $A_X\in M$ such that $X\in V[A_X].$"  My question is, why is this fact true?  Thanks
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In David Hilbert's 1899 Grundlagen der Geometrie, Hilbert gives a rigorous axiomatization of Euclidean geometry. As I understand it, some of Hilbert's axioms must be expressed in second order logic (for example his "Euclid's axiom") and as his system contains some notion of arithmetic (this is perhaps not correct?), by Gödel's first incompleteness theorem this means that Hilbert's axiomatic system is incomplete (meaning that there exist statements expressed by the vocabulary of the system that cannot be proven to be true) and by Gödel's second incompleteness theorem, his system cannot demonstrate its own consistency.  Later, Tarski gave his own axiomatization of Euclidean geometry that is entirely in first order logic so by Gödel's completeness theorem, it can demonstrate its consistency, it is decidable and complete.   Is the above accurate and if so, is Tarski's axiomatic system still inferior in some ways to Hilbert's (with a lack of expressive power or doesn't cover the entire Euclidean geometry as we understand it)?
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Below, the  of second-order logic is used.    My question is about a second-order analogue of $ZFC$ other than the usual "second-order $ZFC$." Rather than define the latter, I'll just state that its (set-sized) models are exactly the $V_\kappa$s with $\kappa$ strongly inaccessible (or if you prefer, the uncountable Grothendieck universes).  Specifically, I'm interested in the second-order theory consisting of the usual (first-order) axioms of Pairing, Extensionality, Union, Choice, Infinity, and Powerset, and the Separation and Replacement schemes ranging over second-order formulas. I'll call this "$ZFC_2^{scheme}$."    Specifically, for each formula $\varphi(x, y_1,...,y_k)$ of second-order logic with only the displayed variables, each of which is first-order, we have a corresponding Separation and Replacement instance. We do not allow second-order parameters here, so formulas with second-order free variables don't yield Separation/Replacement instances.   The passage from $ZFC$ to $ZFC_2^{scheme}$ has a nice uniformity: it's an example of a more general operation $\mathcal{ZFC}$ which spits out a $ZFC$-analogue given a logic at least as strong as first-order logic.    This is the reason for the parameter restriction above: I want $ZFC_2^{scheme}$ to look just like $\mathcal{ZFC}(\mathcal{L})$ for any other logic $\mathcal{L}$, and not take into account the particular feature that we now have a new kind of variable at our disposal - and in particular, take as "syntax-free" an approach to logics as we can get away with. (This is a choice of arguable value, but it's the one I'm making for now.)   It's easy to show that every model of $ZFC_2^{scheme}$ is well-founded, so to understand its models we only need to look at transitive sets. A natural hope at this point is that $ZFC_2^{scheme}$ is just second-order $ZFC$ in disguise, that is, that for a transitive set $M$ we have $M\models ZFC_2^{scheme}$ iff $M=V_\kappa$ for some strongly inaccessible $\kappa$. However, this isn't at all obvious to me (although it is easy to show the right-to-left direction).     Question. What are the (set-sized) models of $ZFC_2^{scheme}$?   (Note by contrast that the "arithmetic analogues" $PA_2^{scheme}$ and second-order $PA$ are equivalent since each pins down $\mathbb{N}$ up to isomorphism - although that leads to .)  (I'm happy to drop Choice if that would help.)  EDIT: As Hanul Jeon pointed out below,  shows that consistently $ZFC_2^{scheme}$ (there called "$ZFC_2^{def}$") has countable models, so we have a partial answer. However, I don't see at the moment a way to get an outright $ZFC$ proof that $ZFC_2^{scheme}$ is strictly weaker than second-order $ZFC$.
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Without using Axiom of Choice, prove that $$\forall\alpha\geq\omega,~~|L_\alpha|=|\alpha|,$$  in which $\alpha$ is an ordinals, $\omega$ is the set of natural numbers, $L_\alpha$ is the $\alpha$-th constructiable set, $|x|$ is the cardinals of $x$.  I am learning a set theory, and I have read a proof by using AC. The AC is used only when $\alpha$ is a limit ordinals during the transfinite inductive method for $\alpha$ . I wonder if there is a proof without AC. 　THANKS!
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Set theory (Jech) $\text{p.}\;27:$     It is an open problem whether one can prove without the axiom of choice that there exists a regular uncountable $\aleph_{\alpha}\;($the informed guess is that one cannot$)$.   My question is on the remark in brackets, in particular the use of the word informed: could someone briefly expand on the reasons for believing that this is likely to be impossible?
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I've been reading about mathematical logic and computability theory, but I'm somewhat confused on one note: what exactly is a language? What does it mean when I am told "let $\langle 0, +, \leq \rangle$ be a language"? I looked this up everywhere I could, but haven't found a satisfactory/clear answer.
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Recently I finished my first pure mathematics course but with some intrigue about some proofs of definitions by contradiction and contrapositive but not direct proofs (the existence of infinite primes for example), I think most of them because the direct proof extends away from a first mathematics course or the proofs by contradiction/contrapositive are more didactic. The one that most bothers me in particular is the demonstration that the empty set is a subset of every set, and it is unique. I understand the uniqueness and understand the proof by contradiction:  "Suppose $\emptyset \subsetneq A$ where $A$ is a set. So it exists an element $x \in \emptyset$ such that $x \notin A$ wich is absurd because $\emptyset$ does not have any elements by definition."  but I would like to know if there exists a direct proof of this and if indeed extends from a first course. Thanks beforehand.
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I'm working with the universal and the existential quantifiers and I was wondering is there an order of operations that applies to them? What is the difference between   $(\forall x)P(x)(\exists x)Q(x)$  And  $(\exists x)Q(x)(\forall x)P(x)$  Lets say we have this relationship  (a) $[(\forall x)P(x) \vee (\exists x)Q(x)] \implies (\forall x)[P(x) \vee Q(x) ]$
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I have a doubt. I need help.  Can the basic axioms of equality be presented as "open axioms"?  I) (reflexivity) $\qquad x = x$              II)  (Substitutivity) $\qquad (x = y) \to \big(F (x, x) \to F (x, y)\big)$
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I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?  Also, more specific questions about the different cases:   Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple? Essential singularity case: Why exactly is $f(\{|z| &gt; r \} \cap f(\{|z|&lt;r\}) \neq \emptyset$? $f(\{|z| &gt; r \})$ is dense, but how does $f(\{|z|&lt;r\}$ being open guarantee that their union is non-empty?
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Is it possible for there to be a model of ZFC with the property that, for every set $S$ in the model, there is a unary predicate in the language of ZFC such that $S$ is the is the only set satisfying the predicate?  I'm pretty sure I've been told the answer is "yes", but I am never able to find a reference when I want one. So this question (assuming it is answered in the affirmative) is to be that reference!
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I am a set theorist in my orientation, and while I did take a few courses that brushed upon categorical and algebraic constructions, one has always eluded me.  The inverse limit. I tried to ask one of the guys in my office, and despite a very shady explanation he ended up muttering that "you usually take an already known construction."  The  presents two approaches, the algebraic and the categorical. While the categorical is extremely vague for me, the algebraic one is too general and the intuition remains hidden underneath the text in a place I cannot find it.  Since I am not too familiar with categories, the explanation most people would try to give me which is categorical in nature seems to confuse me - as I keep asking this question over and over every now and then.  Could anyone explain to me in non-categorical terms what is the idea behind an inverse limit? (I am roughly familiar with its friend "direct limit", if that helps)  (While editing, I can say that the answers given so far are very interesting, and I have read them thoroughly, although I need to give it quite some thinking before I can comment on all of them right now.)
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I have read somewhere there are some theorems that are shown to be "unprovable". It was a while ago and I don't remember the details, and I suspect that this question might be the result of a total misunderstanding. By the way, I assume that unprovable theorem does exist. Please correct me if I am wrong and skip reading the rest.  As far as I know, the mathematical statements are categorized into: undefined concepts, definitions, axioms, conjectures, lemmas and theorems. There might be some other types that I am not aware of as an amateur math learner. In this categorization, an axiom is something that cannot be built upon other things and it is too obvious to be proved (is it?). So axioms are unprovable. A theorem or lemma is actually a conjecture that has been proved. So "a theorem that cannot be proved" sounds like a paradox.  I know that there are some statements that cannot be proved simply because they are wrong. I am not addressing them because they are not theorems. So what does it mean that a theorem is unprovable? Does it mean that it cannot be proved by current mathematical tools and it may be proved in the future by more advanced tools that are not discovered yet? So why don't we call it a conjecture? If it cannot be proved at all, then it is better to call it an axiom.  Another question is, how can we be sure that a theorem cannot be proved? I am assuming the description might be some high level logic that is way above my understanding. So I would appreciate if you put it into simple words.  Edit- Thanks to a comment by @user21820 I just read two other interesting posts,  and  that are relevant to this question. I recommend everyone to take a look at them as well.
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Godel`s incompleteness theorem states that there may exist true statements which have no proofs in a formal system of particular axioms. Here I have two questions; 1) How can we say that a statement is true without a proof? 2) What has the self reference to do with this? Godel sentence "G" can say that SUB(a,a, no prove) but could be this just arbitrarily judgement about non-provability of "a" because it may simply has a proof which is not yet revealed or discovered?
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I am very confused about how to interpret the following.  If we write $\{0,1\}^{+}$=$\cup_{n \in \mathbb{N}} \{0,1\}^{n}$  then I am told that elements  of $\{0,1\}^{+}$ are called words and subsists of it are called languages,  Then I was asked,   if we consider $$g: P( \{0,1\}^{+}) \to P( \{0,1\}^{+})$$ defined as  $g(X)=X \cup \{w01:w \in X\} \cup \{\tau\}$  where $\tau$ represents the word of zero length, then find the least fixed point of g ordered by inclusion.  I have been trying to make sense of it but cant make progress. I know that a fixed point is an element $X$ such that $g(X)=X$  I know that since we want the least, the fixed point has to be less then any other fixed points ordered by inclusion to.  Can anyone help to make sense of this? Thank you
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Besides the axioms of extensionality and regularity, all of the axioms of ZFC either postulate the existence of a set or give a method for generating new sets from existing sets.  Extensionality then gives an equivalence relation on sets which allow us to confirm our definitions we create are well defined.  Regularity then gives a rule that all sets must satisfy.   From all the other axioms (besides regularity) we can start generating sets and decide which generated sets are equivalent. For all generated sets they either will satisfy the proposition in the axiom of regularity or they will not.  If all sets generated do satisfy the axiom of regularity then what is the point of the axiom of regularity? It adds no additional structure to the sets of ZFC.  If there exists a generated set that does not satisfy the axiom of regularity then aren't the axioms inconsistant?  I believe my misunderstanding stems from a misunderstanding with how logic/formal systems work. Any clarification would be greatly appreciated.