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I think the distinction of the 0 and the 1 of a field is something that should be deduced out of the field axioms and this distinction is not an additional axiom of the field definition. In different texts we might encouter different appearing definitions for fields but remember that all these definitions are equivelant. The axioms that Dan has checked are ...

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In response to comments about numbers defined as sets: Numbers can be sets but not all sets are numbers. 0 = $\emptyset$ 1=$\{\emptyset\} = \{0\}$ 2=$\{0,1\}=\{\emptyset,\{\emptyset\}\}$ n = {0,1,2,3,.....,n-1} etc. Note, all numbers are finite. $\mathbb N =\{0,1,2,3,4,....\}$ is not finite and is not a number. We can define n + 1 as $n \cup \{... 3 I somewhat disagree with Ittay. This is not a question of size. It's a question of purpose. Numbers, and here I'm referring specifically to real numbers,1 have a particular purpose. They model the notion of length. When we think about something being long, we think about an interval of time, or distance or otherwise a notion of size, which is a real number. ... 2 It's a question of size. The real numbers are of finite magnitude. There is no reason to expect a bunch of finite things to sum up to a finite thing. The correct analogy would be to compare addition of numbers with union of finite sets. Then you also would not expect the union of a bunch of finite sets to be finite. It requires more care. Similarly, you ... 0 "If our set theory does not have an axiom of infinity which allows us to define N, then are we disallowed the use of induction?" Sometimes, the inductive step of an inductive proof can get defined something like the following: For any n, if P(n) holds, then P(S(n)) holds, where S(n) indicates the successor of n in the sequence. Suppose that you can do ... 4 Suppose that you work with the following naive theory: Extensionality, so you can say when two sets are equal; Comprehension, so you can say that if you can describe a collection, then it is a set; and Choice, if you fancy doing mathematics with the axiom of choice. Comprehension does not restrict what it means to be a property. So it includes things ... 1 I think he just means that simply replacing Full Comprehension with Restricted Comprehension is not enough. Given sets$X$and$Y$, you could infer the existence their union$X\cup Y$using Full Comprehension alone. You cannot do that with Restricted Comprehension, so you would also need an axiom of unions (as in ZFC). 0 To answer your question "Do formal models justify intuitive mathematics?" I would answer that it is just the opposite: intuitive mathematics justify the formal models. In physics it is rather clear that the bottom line is the physical phenomena and the theories physicists develop are ultimately justified only to the extent they seem to fit with the former. ... 0 "Intuitive maths" isn't a model in and of its own, it is a convenient abstraction over formal models which are essentially meaningless outside said field. 1) Not "intuition", which isn't well-defined, but intuitive maths is often simply thought of as an abstraction, which is valid. 2) Because it allows justifying some intuition. Essentially, it is used to "... -1 Induction still works in the absence of the axiom of infinity. Without AxInf, you still have$0 = \emptyset, 1 = \{0\}$,$2 = \{0, 1\}$, etc. For a set$X$, if we define the the successor of$X$to be$X \cup \{X\}$, then every set has a successor. In short, we have a collection whose members are exactly$0, 1, 2, 3, \dots$This collection is not a set, ... 0 There is no way to "prove" or "justify" one's intuition. What you can do, is view properties that you would expect something to have, and check to see that the "set theoretic" construction agrees with them (with proof.) Similarly, I think that you can take a construction that you find "intuitive" and prove that it is equivalent to a more "formal," or "... 0 Hint 1)$(∀x)(∀y) \ ((Ax → Rxy) → ¬Ay)$--- premise 2)$(Ax → Rxx) → ¬Ax$--- from Ax:$\forall x \alpha \to \alpha[t/x]$and 1) by modus ponens twice 3)$Rxx$--- assumed [a] 4)$Rxx \to (Ax \to Rxx)$--- Ax.1 5)$Ax \to Rxx$--- from 3) and 4) by mp 6)$¬Ax$--- from 5) and 2) by mp 7)$Rxx \to ¬Ax$--- from 3) and 6) by Deduction Th 8)$\forall ...

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Don't know which syntax/basic rules you need to use, but generally it'll be something like this: $$(\forall x)(\forall y)((A_x \rightarrow R_{xy}) \rightarrow \neg A_y)$$ $$(\forall x)((A_x \rightarrow R_{xx}) \rightarrow \neg A_x)$$ ($\forall$ elimination) $$(\forall x)(\neg (A_x \rightarrow R_{xx}) \vee \neg A_x)$$ (definition of $\rightarrow$ elimination,...

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Perhaps it is a cultural thing. ZFC is NOT the basis for "mathematics". Much of modern mathematics can be shown to be a logical result of ZFC axioms, but then there are a variety of OTHER alternative bases. ZFC is a basis for much, but NOT all, of modern mathematics, and has found to be convenient to use by those interested in constructing proofs with some ...

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Your point is very well taken because even when the axioms are used in a proof in an essential way, they are not mentioned explicitly. This is because the lecturers typically rely on naive set theory rather than any formalisation, often because they are no more familiar with such formalisations than you are. For example, in a real analysis or measure theory ...

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Given a set $X$, we write $\mathcal{P}(X)$ for the set of all subsets of $X$. This is called the powerset of $X$. Here's an example of how powersets are used. Let $X$ denote a vector space. Maybe you want to define the concept "linear subspace of $X$" in a formal way. One viewpoint is that this concept "is" the function $$f : \mathcal{P}(X) \rightarrow \{\... 28 "My house is supposedly built on concrete pad foundations, but I've been fixing the pipes upstairs, and I haven't seen any foundation yet." This is analogous - you don't see them because they're so deep beneath the surface of where you're working. If you lifted the floorboards and poked around, you'd find the foundations. Though they might actually be ... 9 You probably use ZFC all the time! By analogy, in university level real analysis, the first order of business is to clearly define the set of real numbers. In many modern texts, the set of reals is defined as the unique, complete, ordered field. This simply means that the real numbers satisfies a particular list of properties - like$$a+b=b+a$$for all ... 3 Consider the question: what is a set? ZFC gives an answer to this question. If you feel that your intuition is good enough to solve the problems you are faced with, without asking this question, then you do not need a good understanding of ZFC. This question is fundamental, since the other mathematical concepts like numbers, functions, relations, groups ... 80 I think this is a very good question. I don't have the time right now to write a complete answer, but let me make a few quick points (and I'll add more when I get the chance): Most mathematics is not done in ZFC. Most mathematics, in fact, isn't done axiomatically at all: rather, we simply use propositions which seem "intuitively obvious" without comment. ... 1 In the article,in the section Metamathematics, 4th paragraph, it states that the axioms cannot disprove x=Sx, which I think should say "cannot disprove \exists x\;(x=Sx)." 0 We have found a proof and formalized it using the Coq proof assistant, the details are here: https://hal.inria.fr/hal-01332044 6 Recall that x\in L if and only if there is some \alpha such that x\in L_\alpha. And that is exactly the content of the axiom V=L: For every x, there exists an ordinal \alpha such that x\in L_\alpha. But wait, what is L_\alpha anyway? Ah, well, luckily, it is the unique set X such that there exists a function f with domain \alpha+1,... 3 If by "what does it mean" you mean "what does it mean" I have to ask what do you mean, what does it mean? Saying that ZF has infinitely many axioms means exactly what it says. There are infinitely many primes. There are only finitely many natural numbers less than 3. There are infinitely many axioms of ZF. Probably you actually understand that, and you're ... 0 The full statement of the axiom schema is as follows: given \varphi(x,y,u),$$\forall u(\forall x(\forall a(a\in x\rightarrow\exists y(\varphi(a,y,u)\land\forall z(\varphi(a,z,u)\rightarrow y=z)))\rightarrow\exists w(\forall y(y\in w\leftrightarrow\exists a(a\in x\land\varphi(a,y,u))))) Namely, if $\varphi$ is a formula which defines a function on $x$, ...

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The idea is that you are defining $f(y)$ to be the unique $x$ such that $P(y,x)$ is true. If it said simply ${\forall}y({\exists}x:P(y,x))$ then you would be right that it needs to say $\exists!$ instead of just $\exists$. But notice that it says not just that there is an $x$ such that $P(y,x)$ but that there is an $x$ such that $P(y,z)\Leftrightarrow z=x$ ...

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No. The axiom already implies that there's only one such $x$. Given $y$, let $x$ and $x'$ be elements with ${\forall}z(P(y,z){\iff}(x=z))$ and ${\forall}z(P(y,z){\iff}(x'=z))$. Using $z=x$ in the first statement yields $P(y,x)\iff(x=x)$, and thus $P(y,x)$; then using $z=x$ in the second statement yields $P(y,x)\iff(x'=x)$, and thus $x'=x$.

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