# Tag Info

2

No. It's not true. Taking "equivalent" and "matching indexes" as "equipotent", or "having the same cardinality" meaning "there exists a bijection" or a "a one-to-one correspondence" between the sets. If $X$ is any set, then $\mathcal P(X)=\{A\mid A\subseteq X\}$ cannot be put in a one-to-one correspondence with $X$. In particular if $X$ is infinite, its ...

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The idea of having a "first element" is related to the idea of being "well-ordered." You might wish to look this latter term up for more information. As a specific example, the following two infinite sets cannot be matched up in a one-to-one and onto way (i.e., bijectively): The integers, $\mathbb{Z}$, and the set of all subsets of the integers, ...

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Recall that if $\eta\leq\zeta$ then $\eta\subseteq\zeta$ as sets; and that $\sup A=\bigcup A$ when $A$ is a set of ordinals. That should be enough to fill in the gap that you (rightfully) feel exists in your proof. It should also be noted that if you consider the definition of ordinal addition by partitions, that is $\alpha+\beta=\gamma$ if and only if ...

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$\{x\in A : \phi(x)\}$ is by definition the set whose members are exactly anything that is a member of $A$ and satisfies $\phi$. So $y\in\{x\in A:\phi(x)\}$ means just $y\in A\land \phi(y)$. You don't need any quantifier here. You want the variable $y$ to be free in the unfolding, just as it is free in "$y\in \cdots$". The $\{n^2:n\in N\}$ case is a ...

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If every $A_i$ is nonempty, you have described a partition of the set $U$. The definition holds for any set, not just for the current universal set. If the union of disjoint (non-empty) sets equal any set $X$, we say in the same way that we have a partition of $X$.

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One way to look at it: define the symmetric difference $A \oplus B = (A-B) \cup (B-A)$. This defines a (commutative) group operation on all subsets of the universe set, where every set is its own inverse and $\emptyset$ is the neutral element. Your statement is then $A \oplus B = A \oplus C \rightarrow B = C$, which follows because we cancel by adding $A$ ...

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Assume that $x$ is an element that is in $B$ but not in $A$. Then $$x\in B-A=B\cap A^c$$ and therefore $$x \in (A-B)\cup (B-A)$$ Due to the given equality we must also have that $$x \in (A-C)\cup (C-A)$$ but since $x \notin A$ by assumption we have that $$x \in C-A$$ or $x \in C$. The same holds for every element in $C$ witht that property, so that we have ...

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Consider the set $O = \{(x,y) \in \mathbb{R}^2: xy > 0 \}$, which is open (it's just the union of two open quadrants in the plane, or we see it as $g^{-1}[(0,\rightarrow)]$ for $g(x,y)= xy$ from the plane to $\mathbb{R}$, which is continuous). Then your set $A$ is just the points of $\overline{B}$ that are in $O$ as well, so $A = O \cap \overline{B}$ and ...

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Assuming that $\overline{C}$ is the complement of $C$: $$A = A \cap (C \cup \overline{C}) = (A \cap C) \cup (A \cap \overline{C}) \subseteq (B \cap C) \cup (B \cap \overline{C}) = B \cap (C \cup \overline{C}) = B$$

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Consider an element $a \in A \cup B$. The left-hand side counts the number of such elements. If $a \in A \cup B$, there are three possibilities: $a \in A$ and $a \in B$: in this case $a \in A$, $a \in B$ and $a \in A \cap B$, so $a$ is counted $1+1-1=1$ times in the right-hand side. $a \in A$ but $a \not\in B$: in this case $a \in A$, $a \not\in B$ and $a ... 1 Let$a \in A$be arbitrary. We will show that$a \in B$. Suppose$a \in C$. Then$a \in A \cap C$, hence$a \in B \cap C$, which implies$a \in B$. Similarly, if$a \in \overline{C}$it follows that$a \in B \cap \overline{C}$, hence$a \in B$. Since$a \in B$for all$a \in A$, we conclude that$A$is a subset of$B$or$A \subseteq B$. 0 Answer: I was being too strict. 3 For$\;n=0,1,2\;$the case is clear. Suppose it is true for$\;n\;$, show for$\;n+1\;$. So suppose$\;A:=\{a_1,...,a_n,a_{n+1}\}\;$is our set, and out$\;B:=\{a_1,...,a_n\}\;$. The gist of all this business is to observe that any subset of$\;A\;$is either a subset of$\;B\;$or a subset of$\;B\;$union the singleton$\;\{a_{n+1}\}\;$(why?!), and ... 0 To add to what others have said, arbitrary definitions do not create referrants. The paradox shows there is no referrant to that definition in a normal system of set theory, which then had to be made rigorous. The definition is not "gibberish" as one said, as there are nontraditional set theories that permit the Russell set, but give up something else. We ... 2 This definition is not for a natural number, but rather for an ordinal. Let's see that$s=\omega$satisfies this definition.$\omega$is transitive. If$y\in\omega$then$y$is transitive. If$x\subseteq\omega$and transitive, and$z\subseteq x$and transitive (in particular, both$x$and$z$are natural numbers), then$z\subseteq x\cup\{x\}$. If$x$is a ... 1 Please, note that the non existence of the Russell's set is a theorem of logic; to prove it no axiom of set theory is necessary. See for example George Tourlakis, Mathematical Logic (2008), page 185 [we assume as available the symbol$\bot$(i.e.falsum) and the tautology$\lnot A \rightarrow (A \rightarrow \bot)$; without$\bot$we may use a formula$B$... 0 This paradox can be rephrased as: "The fact you can describe something using mathematical concepts doesn't mean that it should exists" or in other words " math is not waving a wand in the air and yell some magical (read mathematical) formula to summon an object" Why? Russel shows that the set of all set, which is a sentence which use only mathematical ... 1 From the wikipedia page for Russel's paradox: According to naive set theory, any definable collection is a set. Let R be the set of all sets that are not members of themselves. If R is not a member of itself, then its definition dictates that it must contain itself, and if it contains itself, then it contradicts its own definition as the set of all sets ... 0 Hint 1: Show that if$[a]\cap[b]\ne\emptyset$then$[a]=[b]$. 1 You can also see this by a counting argument if$U$is finite. If$|U|=n$, then$|\mathcal P(U)|=2^n$, and if$|A|=k$, then$|\mathcal P(A)|=2^k$. Then $$|\mathcal P(A^c)| = 2^{n-k}, |\mathcal P(U)-\mathcal P(A)| = |\mathcal P(U)|-|\mathcal P(A)| = 2^n-2^k=2^k(2^{n-k}-1)$$ These are different unless$(n,k)=(2,1)$- in other words, your sets are almost ... 2 $$\emptyset \in \mathcal{P}(A^c)$$ $$\emptyset \in \mathcal{P}(A) \implies \emptyset \notin \mathcal{P}(U) - \mathcal{P}(A)$$ So: $$\mathcal{P}(A^c) \neq \mathcal{P}(U) - \mathcal{P}(A)$$ 3 Let$U=\{a,b,c\}$and let$A=\{a\}$. Then$\mathcal{P}(U)-\mathcal{P}(A)$contains the set$\{a,b\}$, which is not part of$\mathcal{P}(A^c)$. Therefore, the answer to your question is no. -1 The real numbers between 0 and 1 can be thought of as the set of all functions from N to N minus a few elements to ensure unique representation. 3 It is proven in R. Grigorchuk, Degrees of growth of finitely generated groups, and the theory of invariant means. Math. USSR, Izv. 25, (1985) p. 259-300 that there is a continuum of pairwise nonisomorphic 2-generated periodic groups of intermediate growth (Theorem A of the paper). Therefore, it follows that${\mathcal N}(F_n)$has cardinality of ... 0 Let f(0)=1. Let f(n)=2^n if n is positive and f(n)=3^(-n) if n is negative. I leave verifying 1-1 to you, and note that the function is onto in the sense that all functions are onto their range. The range is a set of positive integers. 0$A\in \mathcal P (A)\rightarrow A \in \mathcal B\rightarrow A\subset BB\in \mathcal P (B)\rightarrow B \in \mathcal A\rightarrow B\subset A$Using the conclusions of each line$A=B$7 You can prove: $$\bigcup \mathcal{P}(A)=A$$$$\bigcup\mathcal{P}(B)=B$$ and by hypothesis you have$\mathcal{P}(A)=\mathcal{P}(B)$therefore $$B=\bigcup\mathcal{P}(B)=\bigcup\mathcal{P}(A)=A$$ 1 The power-set$\mathcal P(A)$is the set of subsets of$A$. So here we have $$\mathcal P(A)=\bigl\{\varnothing,\{0\},\{5\},\{\pi\},\{0,5\},\{0,\pi\},\{5,\pi\},\{0,5,\pi\}\bigr\}$$ -3 If we know$A$we can determine unequivocally$\mathcal{P}(A)$and if we know$\mathcal{P}(A)$we can determine unequivocally A, as element of$\mathcal{P}(A)$which includes any other subset , so if$A=B$then$\mathcal{P}(A)=\mathcal{P}(B)$and if$A\neq B$then$\mathcal{P}(A)\neq\mathcal{P}(B)$what completes the proof of your theorem. Another way of ... 5 Everything in your answer is correct until the implication "Therefore$A \subseteq B$and$A=B$". The second implication (i.e. that$A=B$is not yet justified). So keep only the first, i.e. that $$A \subseteq B \tag 1$$ Reason in exactly the same way to deduce that $$B \subseteq A \tag2$$ Now combining$(1)$and$(2)$yields the result $$A=B$$ 3 The proof is almost okay. You need to argue why$B\subseteq A$. But it follows from the same argument. 0 This question does not need transfinite induction other than maybe in the definition of addition. Recall that$\alpha+\beta$is the order type of a well-ordering whose underlying set is the disjoint union of$\alpha$and$\beta$. So if$\alpha,\beta$are countable, their disjoint union is and hence$\alpha+\beta$is too. Similarly the product of two ... 2 HINT:$\alpha+\beta+1$is$\alpha+\beta\cup\{\alpha+\beta\}$. 2 It's a simple equivalency proof :$ E \in \mathcal P (A\cap B) \Leftrightarrow E\subset A\cap B \Leftrightarrow \left[E\subset A \textrm{ and } E\subset B\right] \Leftrightarrow \left[E\in\mathcal P (A) \textrm{ and } E\in \mathcal P(B) \right]\Leftrightarrow E\in\mathcal P (A) \cap \mathcal P(B)$You should maybe show carefully this equivalence : ... 5 Yes, this statement is true. Let$x \in \mathcal{P}(A \cap B)$. Then, by the definition of the power set,$x \subseteq A \cap B$, which by definition of intersection implies that$x \subseteq A$and$x \subseteq B$. So by definition of power set,$x \in \mathcal{P}(A)$and$x \in \mathcal{P}(B)$. Hence,$x \in \mathcal{P}(A) \cap \mathcal{P}(B)$. We ... 1 Hint (Surjecticity) Can you solve this $$g(m,n)=(p,q)\iff\left\{\begin{array}\\ m+n=p\\ m-n=q \end{array}\right.\;\;?$$ and what happens if$p+q$is odd? . (Injectivity) In the case$p+q$is even express$m$and$n$with$p$and$q$. What you can deduce? 1 For this kind of questions, you really need to stick to the definitions as much as possible. Especially if you don't know where to start. You can't do anything without the definition. So let's begin. First, a function is defined to be bijective if it is at the same time surjective and injective. So we'll get an answer to (c) as soon as we have an answer to ... 0 Hint on surjectivity: if$(x,y)=g(m,n)$then$x-y$is even. 0 Hint:$2m=(m+n)+(m-n), 2n=(m+n)-(m-n)$. 0 If you accept well-ordering, then consider X as a well-ordered set and then the functions from X to Y represent sequences of the elements of Y according to the well-ordering of X. So, if X is the ordinal numbers then the functions represent sequences of elements of Y extending through the ordinals. At a more mundane level, if X is specifically the positive ... 2$(\Rightarrow)$Suppose$M \neq N$and$M \subset N$. Since$M \subset N$, every element of$M$is also an element in$N$, that is,$a \in M \rightarrow a \in N$. However, since$M \neq N$, we can find an element that is in one set but not the other set. BUT, we know that every element in$M$is also in$N$, so this means that we must be able to find an ... 1 The term "vacuously" means that there are no counterexamples. That is when we have an implication "If ... then !!!", but the "..." part never happens. But this is not the case, it's just happened that there are no interesting cases to verify. This is because the definition of a transitive relation is not "For every$x,y,z$distinct we have ..." but rather ... 1 Assume that$M \subset N$, i.e. if$x \in M$then$x \in N$. We have$M=N$if, and only if, for all$x \in N$we have$x \in M$. Hence$M \neq N$if, and only if, there exists an$x \in N$for which$x \notin M$. If there exists an$x \in N$with$x \notin M$then$x \in N\backslash M$and hence$N\backslash M \neq \emptyset$2 If$M\subset N$then$M=N$if and only if they have the same elements. Therefore$M\neq N$if and only if there is an element of one which is not in the other. Since all elements of$M$are in$N$this implies that there is an element of$N$which is not in$M$. 2 HINT: Recall that$M\neq N$if and only if$M\nsubseteq N$or$N\nsubseteq M$. 3 Wikipedia article about subset says: If$A$and$B$are sets and every element of$A$is also an element of$B$, then:$A$is a subset of (or is included in)$B$, denoted by$A \subseteq B$, or equivalently$B$is a superset of (or includes)$A$, denoted by$B \supseteq A$. 2 I would give you some hints here. For the first one, you can use the fact that if$f, g\in C [0,1]$satisfies$f(x) = g(x)$for all$x\in \mathscr D$, where$\mathscr D$is a dense subset of$[0,1]$, then$f\equiv g$. Here you can choose$\mathscr D$to be countable. Then think of$|\{ f:\mathscr D \to \mathbb R\}|$. (The set of all real-valued function ... 2 Paragraph added after reading the OP’s previous question about renaming: It’s really helpful for (aspiring) mathematicians to understand a couple of things that are more often taught in computer science than mathematics: a) the difference between bound and free variables in mathematical expressions, and b) the fact that the logical statements mathematicians ... 1 His claim in the first sentence is just that $$\exists{y}:\phi(x)$$ means the same as $$\phi(x),$$ because the existentially quantified variable ($y$) doesn't appear in the formula being quantified. (The first form is confusing, though, because it makes you wonder if it's a typo and should read$\exists{x}:\phi(x)\$ instead.) This part of the paragraph ...

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Taking from mookid, so we have an answer, yes.

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