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$F = \{ (n,\mathbb R) \mid n \in \mathbb N \}$ is a set (or family) of infinite size. But $\{F(n) \mid n \in \mathbb N \} = \{ \mathbb R \}$ has only a single element. The author probably provided this simple example to allow his reader to reflect on the definition of family.

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It depends on your framework, but in structuralist mathematics, sets and predicates are very different. Here's the low-down: Idea 0. Sets form a category. Its called $\mathbf{Set}$. Idea 1. Given an object $X$ of the category $\mathbf{Set}$, we define a new category $\mathrm{Sub}(X)$ as follows: an object $\mathfrak{a}$ consists of a set ...

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I'm gonna give you the names of five celebrities. The first name is Michael Jackson. The second name is Michael Jackson. The third name is Michael Jackson. The fourth name is Michael Jackson. The fifth name is Michael Jackson. How many names have I given you? In one sense, I have given you five names: to each number $n$ between $1$ and $5$, I have ...

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Set theory is a form of logic, and logic is a form of set theory. In my opinion this is made most clear in the form of category theory. However, I would not expect the "typical" mathematician to be sufficiently fluent in formal logic to deal with this comfortably. In my opinion, the typical mathematician works in first-order logic on top of (some fragment ...

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We want to find the total number of relations on an $n$ element set which are both equivalence relations and partial orders. Suppose the Hasse diagram of a poset contains two distinct elements $a$ and $b$, with element $a$ below $b$ and joined to $b$ (so that $a \le b$ in the poset). Then $(a,b)$ is in the relation but $(b,a)$ cannot be in the relation ...

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No. Who said anything about Zorn's Lemma? Zorn's Lemma has nothing to do with this. You're supposed to show that If $C$ is a chain of sets with respect to inclusion, then for every finite subset $F$ of $\bigcup C$, there is $X\in C$ such that $F\subseteq X$. The suggested proof is by induction on $\lvert F\rvert$.

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Let $a_n$ be a convergent sequence with its terms in $\bar{A}$. If it has finitely many distinct elements, then it has to become constant after a certain index, and therefore its limit is in $\bar{A}$. Otherwise, the sequence $a_k$ contains arbitrarily small elements of $A$, so there is a subsequence $b_k$ converging to $0$. Then, we have $... 1 Here is another way to do it which I find easier: Since$\overline{A}$is defined as the set$A$and all its limit points, by definition$\overline{A}=A\cup L$, where$L$is the set of limit points of$A$. Now you only need to show that$L=\{0\}$, i.e. that$A$only has only$0$as its limit point. The result then follows immediately. 1 It suffices to show that$\overline{A}-A=\{0\}$. To show that$\overline A-A \subseteq \{0\}$Suppose there exists some$a \in \overline{A}-A$so that$a \neq 0$. Suppose$a<0$. Then$\mathcal{B}(a,\frac{a}{2})$shows a contradiction. Supposes$a\geq 1$, and you will find a contradiction for the same reason. Finally, suppose that$a \in (0,1)$. Then ... 8 Your set is actually equal to$(2, \infty)$, since$(0,0)=(1,1)=\varnothing$, but even the set$\{ 0, 1 \} \cup (2, \infty)$is not equal to$\mathbb{R}^{\ge 0}$: for example,$\frac{1}{2}$and$\frac{3}{2}$are not elements of this set. 0$x \in f^{-1}(Y\setminus B)\Leftrightarrow f(x) \in Y\setminus B\Leftrightarrow f(x)\notin B\Leftrightarrow x\in X=f^{-1}(Y)$and$x \notin f^{-1}(B)\Leftrightarrow x \in f^{-1}(Y)\setminus f^{-1}(B)=X\setminus f^{-1}(B). $0 Observe that$\{A\times X|X\in\mathscr B\}$is a collection of sets$A\times X$so$\langle x,y\rangle\notin\{A\times X|X\in\mathscr B\}$because$\langle x,y\rangle$is an element of some element of the collection but not an element of the collection. $$A\times X\in\{A\times X|X\in\mathscr B\}\land\langle x,y\rangle\in A\times X\implies \langle ... 1$$\langle x,y\rangle\in A\times X\tag1$$is not the same thing as$$\langle x,y\rangle\in\{A\times X\mid X\in\mathscr B\}\tag2$$as you seem to think. Note e.g. that the elements of \{A\times X\mid X\in\mathscr B\} are sets of the form A\times X and not ordered pairs like \langle x,y\rangle. However if X\in\mathscr B then (1) allows you to conclude ... 1 Your unfolding of x\in \bigcup(\mathscr F\setminus\mathcal P(B)) is wrong -- it should be$$ \exists C\in\mathscr F : C\notin\mathcal P(B) \land x\in C $$which is the same as$$ \exists C \in\mathscr F : C\not\subseteq B \land x\in C $$Note that this has C\not\subseteq B where you wrote x\not\subseteq B. And this is easy to establish for the D ... 2 I will denote \sim such a relation. If a\sim b, then b\sim a (symmetry), and then a=b (antisymmetry). The order is partial, but each time you can compare two elements with each other, they must be the same. So the only element you can compare with a (for any a) is a itself. Therefor there is only one such relation, which is trivial: ... 0 Proof: for any y\in B, g(y) \in C; since g \circ f is onto, then there exists x \in A such that g \circ f (x) =g(f(x))=g(y). Note that g is one to one, therefore f(x)=y; this shows that f is onto. 1 Your definition of onto mixes up the domain and codomain, and you don't seem to be proving that f is onto. Choose any b \in B. We seek some a \in A such that f(a) = b. To this end, let c = g(b). Since g \circ f is onto, we know that there is some a \in A such that g(f(a)) = c. In other words, g(f(a)) = g(b). But since g is one-to-one, ... 1 Let A_1 = \{1,2\}, A_2 = \{2,3\}. Then K = \{1,2,3\}. Then, j(1) = 1, j(2) = 1, so this is not injective. Question for you: What if each A_i contains at most 1 element? Then what can you say? 1 Take K= \mathbb{N}, A_1= \left\{1,2 \right\}, A_2= \left\{3,4 \right\},.... Then j: \mathbb{N} \rightarrow \mathbb{N} and j(1)=j(2), so no j is not necessarily injective. 1 Well, S=\{ (-\infty, b) : b \in \Bbb{R} \} \cup \{ (a, \infty) : a \in \Bbb{R} \} consists of elements from two sets - since it is the union of two - which is$$ S_a:=\{ (a, \infty) : a \in \Bbb{R} \} $$and$$ S_b:=\{ (-\infty, b) : b \in \Bbb{R} \} $$which are both uncountable sets of open intervals. If we now take the union S:=S_a\cup S_b this means ... 4 When in doubt, try to work out smaller examples. If S is the union of \{\{1\}\} and \{\{2\}\}. Which one of the three is correct? S=\{1,2\}, S=\{\{1,2\}\}, S=\{\{1\},\{2\}\}. Not sure? Let's try an even simpler example. S is the union of \{A\} and \{B\}, what is S? If you answered \{A,B\}, you're correct. Let's kick it up a notch. ... 3 Note that if A\cap B=\varnothing, then A\cup B=A\mathbin\triangle B. So if you can show that A\setminus B\in S, then A\cup B=(A\setminus B)\mathbin\triangle B. Next, observe that A\cap(A\mathbin\triangle B)=A\cap((A\setminus B)\cup(B\setminus A))=A\setminus B. 3$$A\cup B=((A \triangle B) \cap A)\triangle B$$2$$ \left(A\triangle B\right)\triangle\left(A\cap B\right)=A\cup B\;. $$1 Personally I would advice you to think of a family (A_{\lambda})_{\lambda\in\Lambda} as a function f that has \Lambda as its domain and has the set \{A_{\lambda}\mid\lambda\in\Lambda\} as its codomain. The function is prescribed by \lambda\mapsto A_{\lambda}. Note that for \lambda_1,\lambda_2\in\Lambda you can have: ... 0 I believe both examples you quoted say that families are sets whose members are sets, the difference being the first example is indexed. For your example of sets with repeated elements like {a,a,a}, this is a multiset. See https://en.m.wikipedia.org/wiki/Multiset 2 For a collection C of subsets of a set S, the intersection of C is the set of elements of S which belong to each element of C. So if C is empty, it would be the collection of x such that, for each set E in C, we have x \in E. Since there are no such sets E, one can say that IF E \in C THEN x \in E, because the part just after the IF ... 3 No, it's like the empty product or the empty sum.$$\prod_{x\in\emptyset}x=1\sum_{x\in\emptyset}x=0\bigcap_{x\in\emptyset}x=G$$and through similar reasoning one would get$$\bigcup_{x\in\emptyset}x=\emptyset$$as the operation on the empty set is always the neutral element of that operation. The reason is such that you can say "any collection" ... 7 The nifty thing about sets is that they are things. We can talk about functions whose values are sets, or functions that take sets as inputs, or make sets of sets, or sets of sets of sets ... This is quite useful, and in fact absolutely ubiquitous in higher mathematics. The downside of this is that when sets are "things", there can't be a set for any ... 0 It depends which axioms you are using. In ZF, the predicate S(x) = \text{True}, \forall x does not correspond to a set. 0 Any legitimate question depends on whether the terminology is understood correctly. The problem is, the definition of a "point" is not understood correctly. Your teacher believes a point has no size because most, if not all, of the geometry books, old and new, say so. It is an error of immense proportions, wasting people's time and even warping their ... 1 For the algebraic structures you are considering, an isomorphism is the same thing as a homomorphism which is a bijection. If f: X \rightarrow Y is a function between finite sets, and X and Y have the same number of elements, then f is (i) bijective \iff (ii) injective \iff (iii) surjective. Putting these two facts together gives you what ... 2 This is a case where everything follows directly from the very definition. we have$$ A\setminus B:=\{x:x\in A\wedge x\notin B\} $$and for A=B we have therefore$$ A\setminus A:=\{x:x\in A\wedge x\notin A\} $$which means, that for any x\in A we also need that x\notin A which of course holds for no x\in A. Therefore the set of elements, which lie in ... 0 Assume there exists x \in A \setminus A. Then x \in A but x\notin A. This is a contradiction, thus there exists no x \in A \setminus A, which means that the set is empty 2 Let a_1=0 and a_n=a_{n-1}+2n for n\ge 2. Thus,$$\langle a_n:n\in\Bbb Z^+\rangle=\langle 0,4,10,18,\ldots\rangle\;.$$Let A=\{a_n:n\in\Bbb Z^+\}\cup\{-a_n:n\in\Bbb Z^+\}. Then A-A=2\Bbb Z\setminus\{-2,2\}: every even difference greater than 2 in absolute value is \pm(a_{n+1}-a_n) for some n, and every non-zero difference of two members of ... 1 Hint: Can there be a point a with both a\in A and a\notin A? 1 Here’s one way to think about it. Let \mathscr{P} be a family of non-empty subsets of a set X. Then \mathscr{P} is a partition of X if it satisfies two conditions: every x\in X belongs to at most one of the sets in \mathscr{P}, and every x\in X belongs to at least one of the sets in \mathscr{P}. The first of these is just an informal ... 3 Yes, you are right. Generally, we have 2^\kappa > \kappa for any cardinal number \kappa. 0 A straightforward example of a partition is the following: Let X = \{1, 2, 3 \}. Then \mathcal{P}_1 = \{ \{1\}, \{2\}, \{3\} \} forms a partition of X. Do you see why this is a partition? If so, could you give another partition of X? 0 If you consider a tuple and a equivalence relation, well, then you ought to define them first: a tuple is a collection (?) of objects (?) and an equivalence relation is a subset (?) of the product (?) of the tuple with itself satisfying some axioms... What you're quoting is an intuitive definition of "set", to grasp the idea that an element that belongs to ... 2 The word "distinct" here refers merely to the fact that there is no sense in which one element can be in the set "more than one time". (It doesn't express this idea very well in my opinion, but it is nevertheless what it is meant to express). A more precise -- but possibly less intuitive -- attempt at a definition would be A set is something that ... 1 Let I be the greatest ideal which does not contain X. We shall show that \forall A\subset X, A\in I \iff X\setminus A\not\in I. So suppose A\in I. If X\setminus A\in I, then X=I\cup (X\setminus A)\in I, a contradiction. Thus X\setminus A\not\in I. Conversely, if A\not\in I, then adding X\setminus A to I generates an ideal containing I. ... 1 Yes, a function is well-defined if each element of its domain is mapped to exactly one element of the range. As long as this is true, it does not matter how exactly you describe this mapping. 1 Consider the general case, where A=\{1, 2, 3, \dots, k\} for any positive integer, where k is not restricted to be of the form 3n+1 (however, so that I can write \left\lfloor\frac{k}{3}\right\rfloor later, I will restrict k\geq 3). I will show by induction that there is no subset B of A where no three distinct elements of B sum to another ... 1 If you are going to talk about a "last" element, you have to put an order on the set. So instead of thinking of S as a set, you want to think of it as a tuple \langle 23,45,32,56\rangle, which is really a function \sigma:3\to \mathbb R by \sigma(0)=23, \sigma(1)=45, \sigma(2)=32, and \sigma(3)=56. If you want everything but the last element, ... -1 Actually, all of these are the subsets of S=\{23,45,32,56\}:\{23\},\{45\},\{32\},\{23,45\},\{23,32\},\{23,45,32\} 0 Since \cap and \cup are distributive over each other we have A\cap (B\cup C)=(A\cap B)\cup (A\cap C)\subset (A\cap B)\cup C. 2 Suppose A and B are disjoint sets in X with complements A^c, B^c. You know that each x \in X that isn't in B is in B^c. You also know that a \in A \Rightarrow a \not\in B because A and B are disjoint. This implies that A \subseteq B^c. Because A \cup A^c = X, B^c \cup A^c = X . For simple set theory, the visualization of sets ... 2 While it is important to understand "automatic" things like De Morgan's laws, and to be able to apply them in as many situations as possible, it is also important to be able to solve problems like this one "from first principles" to make sure De Morgan's laws and such don't blind us to the real meaning of things. In this case: Suppose A \cap B = ... 2 If we assume that$$A \cap B = \emptyset,$$Then by the rule of complements,$$\iff (A \cap B)^c = (\emptyset)^c = X \setminus \emptyset = X\iff (A \cap B)^c = A^c \cup B^c = X,$\$ by De Morgan's laws.

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