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11

You can't. It is consistent that the axiom of foundation fails, and there are sets of the form $x=\{x\}$. More generally there are "Anti-Foundation Axioms" which assert the existence of ill-founded sets. The more famous ones are by Boffa and by Azcel.


0

I'm not certain why this isn't in the previous answer, but it seems worth mentioning that in first-order PA you cannot restrict the models to the specific one you want (the natural numbers 0, 1, 2, ...). Put another way, the same axioms of PA are satisfied by, for one example, a model where you have 0, 1, 2, ..., a, b, c, d, ..., where the latter set of ...


3

In $\sf ZFC$, as said, every object in the universe is a set. Including the objects we use to represent the natural numbers with. Note, by the way, that this is not a very peculiar idea. We use Dedekind cuts to represent real numbers. In that case $\{1_\Bbb R,2_\Bbb R,3_\Bbb R\}$ is a set of subsets of the rational numbers, and their union is actually ...


2

See Axiom of union : Given any set $A$, there is a set $B$ such that, for any element $c$, $c$ is a member of $B$ if and only if there is a set $D$ such that $c$ is a member of $D$ and $D$ is a member of $A$. In symbols : $\forall A \exists B \forall c(c \in B \leftrightarrow \exists D(c \in D \land D \in A)$. In your example : $A= \{ ...


4

In ZFC set theory, everything is a set, and in order to speak about numbers you need to chose some set to represent each number. The most commonly used representation of the natural numbers is to choose to represent the number $n$ by the set $\{0,1,\ldots,n-1\}$. Under this representaion (the Von Neumann ordinals), the number $0$ is represented by the ...


5

No, this is not equivalent. It is consistent that $A\notin A$ for all $A$, but there is $A,B$ such that $A\in B\in A$. The proof is quite difficult, but we can start with a model with atoms, and an extensional relation on the atoms, then realize that relation as $\in$. With this we can start with two atoms, and the relation will simply be ...


0

+1 for an interesting question. Here's an idea to get you started. Let $\frak{U}$ denote the class of all models of $\mathbf{ST}$. Write $x \in \frak{U}$ to mean that $x$ is a family of elements that assigns to every model of $M$ of $\mathbf{ST}$ some element $x_M \in M$. Write $f : \mathfrak{U} \rightarrow Y$ to mean that $f$ is a family of functions that ...


1

I would beg to disagree. Image processing is quite formal. Images are treated as discrete functions in 2D, and are manipulated using very formal methods from calculus, linear algebra, and statistics.


1

Mumford got interested; http://en.wikipedia.org/wiki/David_Mumford and http://www.amazon.com/Pattern-Theory-Stochastic-Real-World-Mathematics/dp/1568815794/ and many vision pdfs at http://www.dam.brown.edu/people/mumford/vision/papers/ Anyway, theory, such as you request, and practice are currently very far apart. You might look up "persistent homology" ...


2

Your Question is much more complicated than it looks. First some philosophy: Are all proof equivalent? What do you mean by equivalent in relation to proofs? (not just in relation to this proof but to any proof at all) Euclidean geometry: Euclid was a bit lacks with his Postulates and Common Notions, The axioms of his geometry were only found in late 19 ...


0

All proofs I have seen so far trackback to the parallel axiom, or the axioms of similarity(which I think are equivalent). I'd be interested to see these "algebraic" proof though. Anything proved by pure algebra alone must be universally true, which is not so in the case of the Pythagorean theorem. It must use (euclidean) geometry in some form or the other.


11

Sure, the Pythagorean theorem is an item in the theory of Euclidean geometry, and it can be derived from the modern axioms of Euclidean geometry. A full set of Euclidean geometry axioms contains the information about similarity and area that are sufficient to prove the Pythagorean theorem "synthetically," that is, directly from the axioms. The algebraic ...


2

Pythagoras is equivalent to the parallel postulate.


3

Here's another attempt at doing the whole thing, without the possibly confusing notation. Let's define a kind of object that we'll call a "prn". A "prn" is defined as an ordered pair $(a, b)$ (with some additional structure that we'll define below), and to make it clear that it's a "prn" and not just an ordered pair, we'll denote $(a \star b)$. We can ...


0

The main motivation of defining the rationals this way is of course the attempt to make sense of division, since none of the integers except $1$ have a multiplicative inverse. So we start with definition 2, but the symbols used are misleading, since they suggest that division is already taking place, which is not the case. Remember, until now we only know ...


1

[EDIT: I miscalculated local exponents in the previous revision of this answer. Local cartesian closedness is, obviously, unnecessary.] We have a Yoneda lemma whenever we can speak of categories relative to any "sets" that form a category. In particular, it suffices that the category of "sets" has pullbacks. If $\mathbb{C}$ has pullbacks, then for every ...


0

Alternative idea of axiom: property verified by the structure we are interested in.


0

A formalist perspective is that if something didn't satisfy the axioms of Euclidean geometry, then we wouldn't call it a Euclidean plane. If something doesn't satisfy Peano's axioms, then we won't call it a collection of natural numbers. And so forth.


3

Since by definition no axiom can be proved using only the other axioms, the choice of axioms is somewhat arbitrary as you can always decide to work within a system with less or more axioms. Mathematicians therefore choose axioms based on how useful the results based on those axioms can be. For instance, if we chose not to use the axiom of choice, we could ...


0

Axioms are the things that are taken as basic, unchallenged assumptions. Depending on what area of mathematics you are working within, these may change. For instance, all of the elementary results of arithmetic are usually taken as unstated axioms in higher branches of mathematics (so you don't prove $1+1=2$ in a calculus course, but might in a certain other ...


0

If mathematics were a chess game, propositions are the possibile chess positions. Inference rules are the valid moves. Postulates (or axioms) is the initial position of pieces. Theorems are the positions you can reach in a game by applying moves to the initial position.


1

Nope. Suppose your only axiom for an algebraic structure with a binary operation * is $\forall$x $\forall$y (x*y)=(y*x). Two constructions (if I've understood your term correctly) which satisfy this axiom are: 1 a b a a b b b a and 2 a b a b b b b a But clearly these structures are not isomorphic.


5

Usually we can't assume that a set of axioms only has one model up to isomorphism. In fact, if a set of first-order axioms has just one infinite model, then it is a consequence of the Löwenheim–Skolem theorem that it has models of all infinite cardinalities. Since all isomorphisms are bijections, it follows that there must be two non-isomorphic models. The ...


3

The complete ordered field example "cheats", because it implicitly invokes set theory in addition to ordered field theory. You may find the theory of real closed fields to be interesting: it is the theory of complete ordered fields without "cheating", and there are many non-isomorphic examples. The smallest example is the field of all numbers that are both ...


4

Consider, for example, the group axioms. It is not the case that any two structures that satisfy the axioms are isomorphic -- in other words, not all groups are isomorphic. It is a quite unusual property of the axioms for complete ordered fields that they allow exactly one realization up to isomorphism.


1

This two equations can be taken from field axioms: $$x.(0+ 0)= x. 0$$ $$x.(0+ 0)= x. 0 + x. 0$$ thus, we have $$x. 0 = x. 0 + x. 0$$ and with this, we have proved that $x. 0 = 0$.


2

Hint: In any ring we have $0=0+0$ and distributivity.



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