# Tag Info

2

Apart from the fact that your proof could benefit from some better wording, it is perfectly correct. I would still suggest you try to reword it more clearly as that will help you later on. Make it more clear what you want to do. Something like: We wish to prove that $R$ is reflexive i.e. that for every $x$, $xRx$ is true: $$\forall x: xRx$$ Let $x_0\in ... 2 In practice you just check it by brute force. You can make it a bit faster to conduct by drawing a diagram with nodes indicating the elements 0, 1, 2, 3 and arrows between elements that are related (and such a diagram is good to do anyway in order to train your intuition). Then, instead of checking every combination of pairs for transitivity, you can just go ... 1 As your question suggests, you can indeed use a matrix to visualize the relation. In your example, the matrix is $$A = \left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1\\ 1 & 0 & 0 & 1 \end{array}\right].$$ Note that there is a$1$in the$(i,j)$-th cell if and only if$(i,j)\in R$. ... 0 We say$a$divides$b$, denoted by$a | b$, if$b$is a multiple of$a$(ie,$b$is an integer multiple of$a$). Equivalently,$a |b$iff$b=ka$for some integer$k$. To remember what "$2$divides$6$" means, perhaps you can remember the phrase "$2$divides$6$into$3$parts". Hence,$2 | 6$. Note that$2 | 0$because$0$is an integer multiple of$...

1

Symmetric: $xRy \iff yRx$ Is it fair to say that the statements "$x$ lives in the same house as $y$" and "$y$ lives in the same house as $x$" are equivalent? If so, then the relation is symmetric. Reflexive: $xRx$ Is it always true that "$x$ lives in the same house as $x$"? If so, then the relation is reflexive. Transitive: $xRy \wedge yRz \implies xRz$ ...

1

Given two integers $a$ and $b$, we say $a$ divides $b$ if there is an integer $c$ such that $b=ac$. Source. This is what $a$ divides $b$ means. The shorthand notation is $$a|b$$. In your example, $$a|a^2\iff a\leq a^2$$ since by definition there exists $c$ such that $a^2 = ac$, namely $a = c$.

1

$a$ is said to divide $b$ if there is an integer $c$ such that $bc=a$

1

Let $R$ be a relation on set $A$ such that $R^{-1}\circ R=\left\{ \left\langle a,a\right\rangle \right\}$ for some $a\in A$ and $R\circ R^{-1}=\triangle_{A}$. Then $\left\langle x,b\right\rangle \in R\implies\left\langle x,x\right\rangle \in R^{-1}\circ R\implies x=a$ . So $R=\left\{ a\right\} \times S$ where $S$ is a non-empty subset of $A$. Based on ...

1

You appear in effect to be assuming that $\succsim$ is the same as ordinary $\ge$. This need not be the case. In fact, the whole point of the argument is that the lower contour sets for any continuous preorder on $X$ are closed, so that in this sense all continuous preorders on $X$ behave like the familiar natural order $\ge$. Of course the steps of the ...

3

The pre-order being complete has no topological meaning, but purely set-theoretic. It means that for any two points $x,y$ in the domain of the pre-order, we must have $x \succsim y$ or $y \succsim x$ (or possibly both here, because we have a pre-order, so we can have both at the same time (invariant goods (?), or some such thing, economics is not my field, ...

1

I think you have not yet got a hold of the difference between minimum and minimal and maximum/maximal. A minimum of a partially ordered set is an element smaller than all other elements. An element is minimal if there is no element that is strictly smaller than it. For a totally ordered set, these are the same. But for a partially ordered set there can be ...

2

Clearly the first is not transitive: $1+6=7$ and $\operatorname{rest}(6,7)+\operatorname{rest}(8,7)=7$ but neither $1=8$ nor $\operatorname{rest}(1,7)+\operatorname{rest}(8,7)=7$ is true. Now consider the second relation: in fact we can write $$n\ \beta\ m\iff n^2\equiv m^2\pmod7.$$ Then can you start from this definition and prove that it is both symmetric ...

1

Symmetry should be obvious: just switch $m$ and $n$ in the definitions, and see what you get. As to transitivity, think $n = 1$, $m = 6$, $q = 8$. For one of the two relations (which one?), say $\gamma$, you have $n \gamma m$ and $m \gamma q$ but $n \not\gamma q$.

1

IMO your proof is incomplete: you only showed that for any $x\in Z$ you have $xR y$, so that $Z\subset \text{Seg}_R(X,y)$ and inclusion in the other way is not shown. In other words, how do you know (given your proof) $xRy$ implies $x\in Z$? For intuitive appreciation, I think it's probably better (well at least for me) to write $<$ for $R$ in this case, ...

0

Since $Z$ is an $R$-section of $X$, we know that $Z\subsetneq X$ since we additionally assumed $X\neq Z$. So then $X\setminus Z\neq\emptyset$ and $X\setminus Z\subseteq X$ so we know $X\setminus Z$ has a $R$-least element since $X$ is well ordered by $R$. $\require{cancel}$ So $\exists\ y\in X\setminus Z$ st. if $y\neq z \implies yRz \land\ y\cancel{R}z ... 0 I think you should specify the set on which you define the empty relation. Suppose A,B are two sets where A is empty and B is non-empty.Define the empty relation$ R_1$(respectively$R_2$) on A (respectively on B). Since for every x∈A,(x,x)∈$R_1$,$R_1$is reflexive. Since there exists x∈B (you can choose any x in B) such that (x,x)∉$R_2$,$R_2$is not ... 2 Your proof is correct. For reflexivity, note the following: If$R$and$S$are reflexive relations on a set$X$and$x\in X$. Then,$(x,x)\in R$and$(x,x)\in S$so that$(x,x)\in R\cap S$. I.e.,$R\cap S$is reflexive. 1 Continuity of this relation is defined to be that if it is preserved in limits: if for any pair of sequences$(x^n,y^n)$converging to x and y respectively and$x^n\succsim y^n\forall n$, then$x\succsim y$. This statement easily implies for any sequence of points$\{y^n\}$with$x\succsim y^n\forall n$and$y^n$converging to$y$, we have$x\...

0

Let me divide it into two bits: (1) what are the upper bounds of $S$ and (2) is there a least one? First part: $u$ is an upper bound of $S$ if $s \propto u$ for all $s$ in $S$. Given $s$, the expression $s \propto u$ means that $u$ lies in $\{s, 4s, 4s+1, 4s+2, \ldots\}$. Call this set $U(s)$. For instance, $2 \propto u$ means that $u$ lies in $U(2) = \{2, ... 1 A maximal element of$S$must be a member of$S$. An upper bound need not be, but it has to be comparable to all members of$S$.$10$is maximal in$S$because it is not$\propto$any member of$S$. It is not an upper bound as$12 \not \propto 10$. Neither is$12$an upper bound as$10 \not \propto 12$The "in$\Bbb N$" part just means you should consider ... 0 G. Sassatelli has already sketched the solution in a comment. If the image is$S\subseteq[n]$with$|S|=k$, the images of the elements in$S$are fixed (they have to be mapped to themselves), and the remaining$n-k$elements can independently be mapped to any of the$k$elements in$|S|$. There are$\binom nk$subsets with$k$elements. Thus the number of ... 2 The relationships amongst the elements of$B$don’t actually matter. For the infimum you’re looking for the largest subset of$U$that is a subset of every member of$B$, and for the supremum you’re looking for the smallest subset of$U$that contains each member of$B$as a subset. (Here largest and smallest refer to the subset relation:$X$is smaller than ... 1 Let$A$be a partially ordered set. For$B\subseteq A$we say that$a=\sup B$if for every$b\in B$,$b\leq a$, and for every$c\in A$, if every$b\in B$satisfies that$b\leq c$, then$a\leq c$. Namely, amongst the set of upper bounds for$B$,$a$is the minimum. The infimum is defined similarly, reversing the order and taking the maximum of lower bounds. ... 1 Your answer for the example in your last paragraph is correct. The "inclusion relation" on the set of subsets of a set defines just a partial order, not a total order. Many (most) pairs of sets aren't related. Hint for a): what is the largest set that's a subset of both$\{1\}$and$\{2\}$? What is the smallest subset that has both as subsets? Similarly ... 1 Any relation$R$generates an equivalence relation$E$wich is by definition the intersection of all equivalence relations that contain$R$as a subset. Any equivalence relation$E$on a set$X$corresponds with a partition$P$on$X$that has the equivalence classes of$E$as its elements. Denoting the equivalence class represented by$x\in X$as$[x]$we ... 0 You know that the transitive closure is$<$. If a relation is reflexive then it must also contain$(x,x)$. Transitivity means it becomes$\le$. if a relation is symmetric then if it contains$(x,y)$then it must contain$(y,x)$. Transitivity means it becomes$\neq$. 2 Hint: Work through some examples. Fix one integer and figure out everything it's related to. For example, consider$7$. We know that$(7, 8) \in t(R)$. But since$(8, 9) \in t(R)$, we know by transitivity that$(7, 9) \in t(R)$. But we can repeat this argument with$(9, 10) \in t(R)$to get that$(7, 10) \in t(R)$. A simple induction argument would show that ... 1 In general: if$R$is a relation then$S:=\bigcup_{n=1}^{\infty}R^n$is its transitive closure. If$T$is a transitive relation with$R\subseteq T$then with induction it can be shown that$R^n\subseteq T$for each$n\in\{1,2,\dots\}$, so that$S\subseteq T$. Conversely it can be shown that$S$is a transitive relation with$R\subseteq S$. -1 For congruence or an equivalence relation, you need 3 properties: reflexive ($aRa$) symmetric ($aRb \Leftrightarrow bRa$) transitive ($aRb, bRc \Rightarrow aRc$) In your case,$a-a=0 \forall a \in \mathbb{R}$, so$R$is reflexive. If$a,b \in \mathbb{R}$and$a - b \in \mathbb{Z}$then$b-a = -(a-b) \in \mathbb{Z}$, so$R$is symmetric. Can you prove$...

1

An equivalence relation is just a relation which is reflexive, transitive, and symmetric. A congruence relation, however, is a bit more: it's a relation which respects some structure. Note that this means the phrase "congruence relation" by itself is vague: I have to tell you what structure I want it to respect. Here's the right picture: I have some ...

0

There is a category sometimes called $\mathbf{Rel}$ whose objects are sets, morphisms $r:A\to B$ are arbitrary subsets of $A\times B$, and composition is given by ordinary composition of relations. In categorist lingo, it's an example of what's called a power allegory. Just a few of its features: For all $A,B$, $\hom(A,B)$ comes with a Boolean algebra ...

0

I don't think you are thinking about it the right way, but maybe I am thinking about it a way that a person who is right wouldn't think. Anyway. The words domain and codomain are usually reserve for functions. A function is a relation, and so the answer could be that some relations have domains and codomains. So, no I wouldn't use domain and codomain if I ...

2

Absolutely it makes sense. The domain of the relation R is the set of arguments i.e. first members of each ordered pair and the codomain is the set of which the values i.e. second members of all the ordered pairs is a subset of. In other words, the range of the relation is a subset of the codomain. Set-theorists distinguish between the range and codomain of ...

0

Fix the pair $(a,b)$, and start with any $(c,d)$ it's equivalent to. It's clear that $(c+1,d+1)$ is also going to be equivalent, and distinct from $(c,d)$. So we can take the function $f$ with $f(n)=(c+n,d+n)$ and get a bijection between $\mathbb{N}$ and a subset of the equivalence class.

0

Hint: when is $(m,0)\mathrel{R}(n,0)$?

3

if $\alpha>0$ then $\alpha+\frac{1}{\alpha}\ge\,2\,\,$. We know $x^{2006}\ge 0$ then $x^{2006}+\frac{1}{x^{2006}}\ge\,2\,\,$ as aresult $$y=x^{2006}+\frac{1}{x^{2006}}+5\ge\,2\,\,+5\ge7$$ i.e $R_f=[7,\infty)$

0

$$\lfloor x\rfloor\{x\}=1$$ $\lfloor x\rfloor\not =0$. Let $\lfloor x\rfloor=k, k\in Z$. Then $$\{x\}=\frac1k$$ $0\le\{x\}<1 \Rightarrow k>0$ T Then $$x=\lfloor x\rfloor+\{x\}=k+\frac1k, k\in \mathbb N$$

0

Notice that no matter what number is $x$ we have that $\lfloor x\rfloor\in\Bbb Z$. Equating then $\{x\}\in\{1/z:z\in\Bbb Z\setminus\{0\}\}$, and notice that we can write $$\{x\}=x-\lfloor x\rfloor$$ Then we can write the equation system $$x-\lfloor x\rfloor=\frac1z\quad\text{ and }\quad\lfloor x\rfloor=z$$ then we get that $$z=x-\frac1z\quad\to\quad x=z+... 6 Suppose [x]\{x\}=1 . x can not be an integer, or else \{x\}=0. Also, x>0 or else [x] < 0 and \{x\} > 0. Therefore, let [x] = a and \{x\} = b, where 0 < b < 1. Then ab = 1, so b = \dfrac1{a}, so x = a+b =a+\dfrac1{a} . 0 Onto means that y = x^{2006} + x^{-2006} must hit every value in the range, i.e. \mathbb{R}. But x^{2006} > 0, x^{-2006} > 0, so clearly y > 5, and the range is \mathbb{R}_{>5}. 3 Hint: x^{2006} is always non-negative. (Why?) Furthermore, \frac{1}{x^{2006}} is always positive when x\neq 0 (Technically, your proposed "function" is not a function at all since it is undefined for x=0. Regardless, looking at the domain instead as \Bbb R\setminus\{0\} it will still have problems with being onto) 0 If for every x\in X there is some y\in X such that xRy, then yRx because R is symmetric. Then, xRy and yRx, so xRx since R· is transitive. But the condition "for every x\in X there is some y\in X such that xRy" is essential. 1 That the product has period \pi is easily seen, once we prove$$\cos x\cos2x\cos3x =\frac{1}{4}\left(\cos6x+\cos4x+\cos2x+1\right).$$There are at least two quick ways to do this. One uses \cos nx =\tfrac{1}{2}\left(z^n+z^{-n}\right) with z:=e^{ix}; the other uses \cos A\cos B=\tfrac{1}{2}\left(\cos\left(A+B\right)+\cos\left(A-B\right)\right). Edited ... 0 S is called the transitive closure of R, often written R^+. If we write a\rightsquigarrow b for (a,b)\in R, then (a,x) is in S if there is a path$$ a \rightsquigarrow b \rightsquigarrow c \rightsquigarrow \cdots\rightsquigarrow v\rightsquigarrow w \rightsquigarrow xFor example, if R is \{(x,y)\in\mathbb R^2\mid x+x = y\} (never mind ... 1 Think of R as defining a graph. x and y are related by S if x and y if there is a path between x and y, i.e. if x and y are in the same connected component. You just need to apply the definition of equivalence relation to S. (i) Show that there is a path between x and x via R. (ii) Show that if there is a path from x to y ... 1 \begin{align} \cos(3x - x) &= \cos 3x \cos x + \sin 3x \sin x \\ \cos(3x + x) &= \cos 3x \cos x - \sin 3x \sin x \\ \hline \cos x \cos 3x &= \dfrac 12 \cos 2x + \dfrac 12 \cos 4x \\ \cos x \cos 3x &= \dfrac 12 \cos 2x + \dfrac 12 (2 \cos^2 2x - 1) \\ \cos x \cos 3x &= \dfrac 12 \cos 2x + \cos^2 2x - \dfrac 12 ... 2 Transitivity of a relation R is the statement that (a,b) \in R \wedge (b,c) \in R \implies (a,c) \in R, where \wedge is the "and" symbol and \implies is the "implies" symbol, in terms of logical representation. This is sometimes simplified to the notation aRb \wedge bRc \implies aRc. In words, this essentially means that given element a and b ... 1 R is a transitive relation if for every a,b,c\in A,(a,b),(b,c)\in R$$then$$ (a,c)\in R$$When A=\{1, 2, 3\} and R=\{(1, 1),(2, 2), (1, 2), (2, 1), (1, 3)\} You have (2,1),(1,3)\in R, but (2,3)\notin R, so R is not transitive. 1 "Onto" is essentially a representation of surjectivity. A function f : A \rightarrow B is surjective if and only if$$\forall y \in B. \exists x \in A. f(x) = y.$$In words, it means that every element of y can be mapped to by f using some element in the domain of f, A. The way one can prove that a function is surjective is by taking an arbitrary ... 0 For any nonempty set X, the sets$$ \Delta_X=\{(x,x):x\in X\}  and $X\times X$ are both equivalence relations. They are the same only in a very special case.

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