# Tag Info

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Since I like to represent alternative set theories, I thought I'd list some examples of proper classes in NF, a set theory that does have a universal set (and therefore provides examples besides the "you can form $V$ from it" types). The Russel class, obviously. $\iota$ -- the singleton function $x\mapsto\{x\}$. In NF if $\iota$ exists you can prove ...

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Let $(X,\le)$ have a maximum $1$ and let $$A=X\setminus\{1\}$$ Let $f$ be as said, we have $$f(A)=A<1$$ since if for some $a\in A$, $f(a)=1$ then $a=1\notin A$ by 3. $1$ is not in $A$; by 4, $f(1)\ne 1$ so $f(1)\notin A$. But $f(1)\in X$ so $f(1) =1$. So a false statement is proved, from which you can deduce anything you desire include the a proposition ...

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In two-sorted first order set theory with proper classes (such as NBG) it is an axiom: $$\forall x\, \exists X\, \forall z\, [ z \in x \Leftrightarrow z \in X ]$$ where lower-case letters denote type-0 objects (sets) and upper-case letters denote type-1 objects (classes). This would mean that there are coclasses for every set: $$\forall x\, \exists \bar ... 0 Consider X=\Bbb R with the usual order and A=\Bbb Q. It is easy to show that every function satisfying the first three requirements must be the identity. 0 I assume you mean sets, and not groups. There are several things to note: An element is either a member of a set, or not. But never both things at the same time. Two sets are equal if they have the same elements. That is to say, each is a subset of the other. Besides that, you should review the definitions of all those symbols, \cap, -,\varnothing. 1 With your definition, elements of A^3=A\times A \times A are couples of the form ((x,y),z) with x,y,z \in A. Notice that from this it follows that A\times (A\times A) \neq (A\times A)\times A and your definition does not apply to the cases n=1 and n=0. It seems better to define A^n as the set of n-uples of elements of A. An n-uple is ... 2 Note that equivalence classes ARE NEVER EMPTY. If anything \pi\in[\pi]_S. You can show that [x]_S is always countable by construction a bijection between [x]_S and \Bbb Q. The quotient set (also group, but you mentioned that you don't want algebraic arguments) is not made of rational numbers, but rather of an element from each equivalence class. By ... 2 First note that equivalence classes R are subsets of A. Because the size of any subset of a countable set is countable, all equivalence classes are countable. For part 3, suppose all equivalence classes are countable. The order of a union countable sets is also countable, so the order of A (which is the order of the union of all equivalence classes) ... 0 I have searched and searched for an answer to this question that makes intuitive sense and have yet to find one. So, after some thought of my own, this is what I came up with. Suppose that the countable ordinals were countable. Let f be a one-to-one correspondence between the natural numbers and every well-ordering of the natural numbers. For instance: ... 0 We have: \{ x : P(x) \}  \equiv \{  all element x such that the statement (or proposition) P(x) is true for the element x\}  \equiv The set of all element x such that the statement (or proposition) P(x) is true for the element x Or, Simply { element of the set : condition for element } For more information about this, ... 0 Here, f: A \rightarrow B is in green and \{S_i\} = S_i for S = A, B and i = 1, 2. Ignore \{b_2\} = B_2 in this picture. This picture proves that  f(A_1) - f(A_2) \neq f(A_1 - A_2). Incidentally, the same picture works for Is this proof correct for : Does F(A)\cap F(B)\subseteq F(A\cap B)  for all functions F? 1 Sure. Take the order type \omega. (To be clear: Here \emptyset has a supremum but not an infimum, and the whole set has an infimum but not a supremum.) 4 There are exactly four completions of this theory (for infinite models), and you've got them all. The theory is known as the theory of a dense linear order, and Cantor's back-and-forth argument shows that any two countable dense linear orders, with the same endpoint situation, are isomorphic. It follows that each of those four theories is complete, since ... 1 First question: A smallest element of A is an element x\in A such that x\le y holds for all y\in A (or equivalently: x<y for all y\in A\setminus\{x\}). Some care must be taken and we distinguish between smallest and minimal if the set is not totally orderd (i.e. there may exist elements a,b such that neither a=b nor a<b nor a>b), ... 0 Use a fat Cantor set, and show that it has positive Lebesgue measure. Then, remember that all countable sets are null. 0 transitive: R^+ has it. reflexive: R^+ may not have it, R^* has it. Part of the assumptions. symmetric: Take < on any nontrivial totally ordered set. It's already transitive, and <^* = ≤, which is not symmetric. irreflexive: of course R^+ does not have it. antisymmetric + asymmetric: use a nonstrict partial order, like < on ... 0 Being @potato's answer completely correct, I try to clear up the fundamental misunderstanding. A is the set \{a,b,c\}. Let's define a function g:A\to\mathcal{P}(A) such that:$$ g(a) = A_a = \emptyset \\ g(b) = A_b = A \\ g(c) = A_c = \{a,b\} $$The rest of the A_x do not enter at all in this definition. Now you need to find B, and finally show ... 0 The confusion arises from mixing up the domain (A) of f and the image of a specific point x\in A under f. What the theorem says, reworded, is the following: Suppose B is the set of all subsets of A, and suppose that f:A\to B is a surjection. It doesn't matter what's f, the only important thing is that for any subset Z of A (as such, ... 0 (a,b,c) = {{a},{a,b},{a,b,c}}? [(a,b,c) = (d,e,f)] <==> [[a=d & b=e] & c=f] [ [a=d & b=e] & c=f ] ==> [ {{a},{a,b},{a,b,c}} = {{d},{d,e},{d,e,f}} ] Now suppose instead that [[a=d & b=f] & c=e]: {{a},{a,b},{a,b,c}} = {{d},{d,f},{d,e,f}} The set {{d},{d,f},{d,e,f}} is not the same as the set {{d},{d,e},{d,e,f}} because the set ... 1 Here is a similar situation. Consider G=\{x\in\mathbb R\mid x^2+x=6\}. How would you check that \color{red}{\bf42} belongs to G or that \color{red}{\bf42} does not belong to G? You would have to determine whether \color{red}{\bf42} satisfies the defining property of G or not. To be in G, one must be a real number (and \color{red}{\bf42} is) ... 4 f \colon A \to \mathscr{P}(A). Thus every x\in A is mapped by f to a subset of A, f(x) \subset A, or f(x) \in \mathscr{P}(A). Now for every particular x\in A, the question arises whether that x is an element of the set that is the image of x under f, whether x \in f(x). That is different from the question whether x belongs to the ... 3 The general strategy can work, but you have to be more careful in choosing the sequence v_n. As you have it, v_n might be eventually monotonic, and it might converge to an endpoint of one of the closed intervals, which would contradict your implicit claim that \lim_n v_n\in J_\infty. To avoid this possibility, try constructing v_n in such a way that ... 0 I am not quite sure how you are obtaining your answer. I will try to describe how I thought about this. Hopefully it is helpful. We want the determine B. The set B is defined as the set of all elements in A satisfying a certain property: x is not contained in A_x. Let us examine each element of A in turn and see if it is in B. Since A_a is ... 2 Methinks you most likely failed to break it down into enough little pieces to wrap your mind around. Suppose that there exist a \in A and b \in B such that y = f(a) + f(b). Then f(a) \in f[A] and f(b) \in f[B], so y\in f[A]+f[B]. Suppose that y \in f[A]+f[B]. Then there exist p\in f[A] and q\in f[B] such that y = p + q. By the ... 3 You don't get nothing -- every infinite sequence of "left" and "right" that doesn't eventually end in all "left"s or all "right"s will map to a sequence of nested open intervals where both right and left endpoints move towards each other. They therefore have a point in common that never gets removed. In fact, if you use closed intervals instead of open ... 1 Show that the relation \sim' defined in 2. is also an equivalence relation and of course it contains R, and any other equivalence relation E that contains R must contain this \sim'. 1 Products and coproducts are dual in the following sense: Let \mathcal{C} be a category and \{X_i\} be a family of objects in \mathcal{C}. Let \{X'_i\} be the same family but considered as a family of objects in \mathcal{C}^{op}. Then: The product \prod_i X_i in \mathcal{C} is the same as the coproduct \coprod_i X'_i in \mathcal{C}^{op}. 4 Yes, although this is often found to be surprising. To see it, replace \mathbb{N} with \mathbb{Q}, which has the same size, and consider the collection of Dedekind cuts in the rationals. 2 Coproduct is the dual of product in the categorical sense, i.e. we have to reverse all occuring arrow in the definition. While product in most concrete categories of structures is indeed realized on the Cartesian product of the underlying sets (because the forgetful functor usually admits a left adjoint), the coproduct may vary from category to category. ... 2 The error lies in a confusion between taking the pre-image of f on a point, versus taking the pre-image of a subset of the target. This is a subtle point that often arises with ordinals, since every ordinal \beta below another ordinal \alpha is both an element of \alpha as well as a subset of \alpha, and the pre-image of \beta under a function ... 0 The family of circles centered at a given point is an uncountable family of pairwise disjoint circles. 7 Hint: Every disk of non-zero radius contains a point with rational coordinates. For circles, geometry will give you an easy uncountable collection. 3 HINT: Consider the case where both x and \{x\} are elements of y. 2 For the first question, you have to prove that [x]+[y]=[x']+[ y], for all x,x' with [x]=[x']. Same for [x]\cdot[y]=[x'][y]. For the second question, i think that is just part of the definition of a ring. You could also just bruteforce this part by checking it for alle equivalency classes [a], [b] and [c] The third part: You know that ... 2 Looks good. It is the unique equivalence class containing the function \cos x \in \mathbb{R}^{[0,1]}. [\cos x] = \{f \in \mathbb{R}^{[0,1]} \mid fR\cos x\} = \{f \in \mathbb{R}^{[0,1]} \mid f(0) = \cos 0 = 1\}. You're close. I think you have the correct equivalence classes, but there are a lot more than you've said. I don't really understand how what ... 3 Your answer to part 1 is fine. The notation means "functions from the interval [0, 1] to the reals"; the interval [0, 1] is the set of all real numbers between 0 and 1, inclusive. For part 2, the notation [\cos x] means "the equivalence class of the function x \mapsto \cos x. What functions are equivalent (under this relation) to cosine? Ones ... 2 I believe A is equal to the set of all functions that map from [0, 1] \to \mathbb R . (2) The equivalence class of f(x) = \cos x are all those functions g(x)\in A that share the same value at x = 0, as does f(x) = \cos x. Note that f(0) = \cos 0 = 1, and hence, all functions g(x) \in A such that g(0) = 1 belong to the equivalence class ... 10 One version of the Baire category theorem says that if X is a complete metric space, and \{G_n:n\in\Bbb N\} is a countable family of dense open sets in X, then \bigcap_{n\in\Bbb N}G_n is dense in X. \Bbb R with the usual metric is complete. Suppose that \Bbb Q=\bigcap_{n\in\Bbb N}G_n, where each G_n is open in \Bbb R. \Bbb Q is dense in ... 0 Notice the set of all finite subsets of \mathbb{Z} can be rewritten as a countable union.$$\Big\{\; A \subseteq \mathbb{Z}\;:\;|A| < \infty \;\Big\} = \bigcup_{n=0}^\infty \Big\{\;A \subseteq \mathbb{Z}\;:\; \max\big\{\; |x| \;:\; x \in A \;\big\}\le n\;\Big\} $$Now the sets of subsets on R.H.S have cardinality 2^{2n+1} for each n, the set of all ... 0 Every expression for a finite subset S \subseteq \mathbb{Z} is a base-14 integer expression, using four new digits:$${\{} = 10, \quad {\}} = 11, \quad {,} = 12, \quad {-} = 13$$so for example$$\begin{align} \{1,-2,9,51\} &= 11+1\cdot14+5\cdot14^2+12\cdot14^3+9\cdot14^4+12\cdot14^5+2\cdot14^6\\ & \qquad ...

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Yes, the empty set is a subset of every set. All tuples $(z,\emptyset)$ will be included including $(\emptyset,\emptyset)$

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It's $\mathfrak{P}(M)=\{ \emptyset, \{\emptyset\}, \{a\}, \{\{a\}\}, \{\emptyset, a\}, \{\emptyset, \{a\}\}, \{a, \{a\}\}, \{\emptyset, a,\{a\}\} \}$ where $M=\{\emptyset, a,\{a\}\}$.

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No, $|[h]|$ is countably infinite. For each $n\in\Bbb N$ let $$H_n=\left\{f\in\Bbb N^{\Bbb N}:f(k)=h(k)\text{ for all }k\ge n\right\}\;;$$ then $[h]=\bigcup_{n\in\Bbb N}H_n$, and each set $H_n$ is countable. In fact there is a very natural bijection between $H_n$ and $\Bbb N^n$. By the way, uncountability of $|[h]|$ does not imply that its cardinality is ...

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You're wrong, on two accounts really. The fact that the equivalence class is not countable doesn't mean its size is $\aleph_1$, unless you assume $\sf CH$ or so. $|[h]|$ is in fact countable. Note that if $(f,h)\in R$ then as a subset of $\Bbb{N\times N}$, the symmetric difference $f\mathop{\triangle} h$ is finite. How many of those are there?

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That's a difficult question to answer well without knowing more about why you're interested or what you hope to accomplish. Logic and set theory are both very broad areas, though they don't look like it from the outside. My suggestion would be something like Set Theory and Logic by Robert Stoll. This is available as a Dover reprint or a Kindle book, for ...

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Note for nerds: the following avoids the axiom of choice, but is not intuitionistically valid. Let $f\colon \Bbb N \to \Bbb N$ be a decreasing function, where $\Bbb N = \{0,1,\dotsc\}$ (I am including $0$). Let $n-1$ be the smallest number at which $f$ takes on its smallest value. Then we can represent $f$ as a finite sequence, an $n$-tuple, ...

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There is a nice bijection between $\mathbb{Z}$ and the set $\mathbb{N}$ of nonnegative integers. From this we can get a bijection between the set of finite subsets of $\mathbb{Z}$ and the set of finite subsets of $\mathbb{N}$. Thus it is enough to show that the set of finite subsets of $\mathbb{N}$ is countable. For any finite subset $A$ of $\mathbb{N}$, ...

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Daniel Velleman's How to Prove It (2nd Edition). It's basically written for exactly your purpose, and it's quite good. http://amzn.com/0521675995

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HINT: Every [strictly] decreasing sequence of natural numbers is finite. Observe that a decreasing function is fully determined by a finite set of natural numbers, and conclude the countability.

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In the search for a bijection, consider an "infinite hotel" map where $2p\over q$ is mapped to $p\over q$ and $2p+1\over q$ is mapped to $-p\over q$. Proving whether this mapping has surjective and injective properties should be fairly straightforward. Note that $p\over q$ is not required to be in reduced form. Edit: Efforts to rectify the shortcomings ...

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