# Tag Info

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The "sensible" statement to consider is: Conjecture. For all sets $X$, the power set of $X$ satisfies the following identity. $$A∩(B∪C)=(A∩B)∪C$$ This is false. E.g. take $A=\emptyset_X$ and $C=X$. Then $A \cap (B \cup C) = \emptyset_X$, while $(A \cap B) \cup C = X$. However, it can be salvaged by adjoining the condition that $C \subseteq A,$ ...

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Hints: $A \cap(B\cup C^*)=(A\cap B)\cup (A\cap C^*)$ So if $A \cap C=\emptyset$, then $A \subset C^*$ and the equality is holds.

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I have already given this answer elsewhere. Let me repeat this here. To prove the countability of $\bigcup _{n=1}^\infty A^n$, it suffices give an injective function $\phi$ from this union to $A$. Without loss of generalty take $A$ to be the set of positive integers and for the sake of this proof regard them as written out in base 10, which makes use of ...

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Here is the idea: Suppose $Q=\{q_n\}$ is a countable collection of distinct elements, and $R=\{r_1,...,r_k\}$ is a finite set of other (distinct) elements. Define $\phi(r_i) = q_i$ and $\phi(q_i) = q_{i+k}$. Then $\phi: Q \cup R \to Q$ is a bijection. That is, if we have a countable collection, we can 'absorb' a finite number of elements (in fact, we can ...

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The elements of a set are separated by commas. In (C), you see that "$2$" is the first element, and "$\{2\}$" is the second element. However, the elements of (D) are "$\{2\}$" and "$\{\{2\}\}$" (which are sets), neither of which is simply the number $2$.

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Let $|A|= a,\quad |B|= b,\quad |C|= c \quad$ then $0 \le |A \cap B|=n_1 \le a$ or $b$ $\quad$ And $\quad$ $a$ or $b \le |A \cup B|=u_1 \le a+b$ $\implies n_1 \le u_1$ $0 \le |B \cap C|=n_2 \le b$ or $c$ $\quad$ And $\quad$ $b$ or $c \le |B \cup C|=u_2 \le b+c$ $\implies n_2 \le u_2$ $0 \le |A \cap C|=n_3 \le a$ or $c$ $\quad$ And $\quad$ $a$ or ...

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The term on the right hand side can be written as $$A\cap(B\cup C)=(A\cap B)\cup(A \cap C)$$ Thus, the given relation becomes $$(A\cap B)\cup C=(A\cap B)\cup(A \cap C)$$ Now, for the first direction (of the iff statement) assume that $C\subseteq A$. Then $$A\cap C=C$$ and it is immediate that the above equality holds. For the other direction (of the iff ...

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Counterexample: Let $f: \omega \rightarrow \omega$ be defined by $f(0) = f(1) = 0$ and $f(n) = 1$ if $n \geq 2$. For $n \geq 0$, let $C_n = \{0, 1\} \bigcup \{n+2, n+3, \dots\}$. Then $C_n$'s are decreasing and $f[C_n] = \{0, 1\} \subseteq C_n$ for each $n$. Let $C = \bigcap_{n \geq 0} C_n = \{0, 1\}$. Then $\bigcap_{n \geq 0} f[C_n] = \{0, 1\} \neq f[C] = ... 0$(A \cap B) \cup C = A \cap (B \cup C) \Longleftrightarrow C \subseteq A(\Rightarrow)x \in C \Rightarrow x\in (A \cap B) \cup C \Rightarrow x \in A \cap (B \cup C) \Rightarrow x \in A\therefore C \subset A(\Leftarrow) x \in A \cap (B \cup C) \Leftrightarrow x \in (A \cap B)\cup(A \cap C) \Leftrightarrow x \in (A \cap B) \cup C$This last ... 0 You are right - but, as it turns out, the two notions are equivalent. There are lots of ways to define a metric on$\Bbb R^k$. The "natural" Euclidean metric is$d(x,y) = \|x-y\|_2$, where$\|z\|_2 = \sqrt{z_1^2+\cdots+z_k^2}$. Another way is to set$d'(x,y) = \|x-y\|_\infty$, where$\|z\|_\infty = \max\{|z_1|,\dots,|z_k|\}$. (And there are many others ... 3 Why should$B$not be well founded (or rather: properly defined)? The comprehension used to define$B$, i.e. to select elements from a given set by a specific property, is the main method to describe a set. Here,$B$is the subset of$A$that consists of precisely those elements o$x$of$A$for which the deus-ex-machina property$x\notin x$holds. (Btw. ... 1 The inclusion-exclusion principle states that for sets$A, B, C$and universal set$U$:$|A\cup B\cup C| = |A| + |B| + |C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|$In our example and letting$A$be the set of students studying Japanese,$B$being the set of those studying Polish, and$C$being the set of those studying Arabic, plugging in what ... 1 HINT: Let$n\in\Bbb N$be arbitrary, and fix$x\in A_n$. If$y\in\Omega\setminus A_n$, there is an$S_y\in\Sigma$such that$A_n\subseteq S_y\subseteq\Omega\setminus\{y\}$. Since$\Omega$is countable, ... 1 See the Principle of Inclusion-Exclusion for more information and a faster (but slightly less intuitive) way to solve the problem. Basically, you have$100$on the left hand side of your equation, and on the right hand side:$(65-12-25-x)$- Japanese only$(42-25-7-x)$- Polish only$(45-12-7-x)$- Arabic only$25$- Japanese and Arabic only$12$- ... 1 You just need to prove that$A_j \in \Sigma$. Suppose$A_j = \{x_1, x_2, \cdots\}$, then$\exists B_1 \in \Sigma$such that$A_j \subset B_1$. Suppose$\{y_1, y_2, \cdots \}$are elements in$B_1 $that are not in$A_j$, then for each$y_n$we can either find$C_n \in \Sigma$such that$y_n \in C_n$but all$x_n$'s are not in it, or find$D_n \in \Sigma$... 1 Let$f\colon A\to A$be a surjective map. Let $$\mathcal A=\{\,X\subseteq A\mid f|_X\colon X\to A\text{ is injective}\,\}.$$ Then$\mathcal A$is not empty because$f|_\emptyset$is injective. Let$B\in\mathcal A$be maximal (how?). Then$f|_B$is onto as otherwise we'd find$b\in A$with$f(b)\notin f(B)$and then$B\cup\{b\}$would conflict with$B$'s ... 10 Let$f: A\to A$be any function. For each$a\in A$, let$N(a)$be the number of elements in$A$that are mapped to$a$. Then$\sum_a N(a) = n = |A|$because every element of$A$is mapped to some element of$A$. (This is the statement that the fibers of$f$partition$A$.) If$f$is surjective, then$N(a)\ge 1$for all$a$. If$f$is not injective, then ... 2 The trick, when you do this by induction, is that you replace$f$by a function$g$which maps the$n+1$-th element to itself. Without loss of generality, we can assume$A=\{0,\ldots,n\}$. And$f\colon A\to A$is surjective. First we can assume that$f(n)\neq n$, for if it were then by simply taking$k$to be some element such that$f(k)=0$and redefining ... 0 Countable Finite:$A=\mathbb{R}^+B=\mathbb{R}^-$Countable Infinite:$A=\mathbb{R}^+\cup\mathbb{N}B=\mathbb{R}^-\cup\mathbb{N}$Uncountable Infinite:$A=\mathbb{R}B=\mathbb{R}$1 You could go for$R\times\{0\}$and$\{0\}\times R$for the first. For b) roughly$ R\times N$and$N\times R$. For c)$ R\times R$and$R\times R$0 a)$A = \{x: x \in \mathbb{Q^c} \text{and} x \geq \sqrt{2}\}$, and$B = \{x: x \in \mathbb{Q^c} \text{and} x \leq \sqrt{2}\} \to A \cap B = \{\sqrt{2}\}$- a finite set. b)$A = \left([0,1]\cap \mathbb{Q^c}\right)\cup \{\frac{1}{n}: n \in \mathbb{N}\}$, and$B = \left([-2,-1] \cap \mathbb{Q^c}\right)\cup \{\frac{1}{n}: n \in \mathbb{N}\}$, then$A \cap B = ...

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Some detailed hints: Both examples can be constructed using subsets of the real numbers. For (a), I would look at certain intervals of the real line. Can you find two real intervals which intersect in a finite set? For (b), I would consider two sets $C \cup E$ and $D \cup E$, where $C$ and $D$ are disjoint uncountable subsets of real numbers and $E$ is a ...

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To understand the $\aleph$ numbers properly you need to understand the ordinal numbers first, at least a little bit. The idea of an ordinal is to model the notion of a length of a queue to the bathroom in a party. There is an empty queue, then there is one person waiting, then another, and so on. But here we can talk about infinite queues as well. At some ...

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This is an issue about which mathematicians who haven't studied set theory beyond what they actually use are often confused. $\aleph_1$ is the cardinality of the set of all countable ordinal numbers. $\aleph_2$ is the cardinality of the set of all ordinal numbers of cardinality $\le\aleph_1$. $\aleph_3$ is the cardinality of the set of all ordinal numbers ...

1

$(a,b) \, R \, (c,d) \iff a+b=c+d$ Every relation in a set $X$ defined by $x \sim y \iff f(x) = f(y)$, where $f:X \to Z$, is an equivalence relation because equality is an equivalence relation. There is nothing really left to prove after you make this observation. In your case, $Z= \mathbb Z$ and $f(a,b)=a+b$.

2

First point: your answer has an "if" but no "then". It therefore is ungrammatical, makes no sense, and cannot possibly be the right answer. Second point: you can do this by symbolic logic if you want, but why? IMHO it merely adds extra work. (That's in this case: in a more complicated question it might be a good idea.) The negation of "if $P$ then $Q$" ...

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Hint: Try proving that any element $x$ of the left hand side must be an element of the right hand side, and vice versa. You can't use associativity; you're being asked to prove this associativity property.

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Choose any $(x,y,z) \in A \times B \times (A \cap C)$ so that $x \in A$ and $y \in B$ and $z \in A \cap C$. Then $z \in A$ and $z \in C$, so $(x,y,z) \in A \times B \times A$ and $(x,y,z) \in A \times B \times C$. But then $(x,y,z) \in (A \times B \times A) \cap (A \times B \times C)$, as desired.

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Any time you’re asked to prove a statement of the form $X\subseteq Y$, you should think first of the element-chasing approach: let $x$ be an arbitrary element of $X$, and show somehow that this forces $x$ to be an element of $Y$. If you can do that, you’ve proved that $X\subseteq Y$. In your case that means starting out like this: Let $x\in A\times ... 1 HINT: For convenience let$D_n=A_n\setminus A_{n+1}$for$n\in\Bbb N$, and let$C=\bigcap_{n\in\Bbb N}A_n$. Use the hypotheses to show that$\mathscr{P}=\{C\}\cup\{D_n:n\in\Bbb N\}$is a family of pairwise disjoint sets whose union is$U$. (In other words, they form a partition of$Uexcept that some may be empty.) A generalization of the desired result can ... 2 hint: to get from $$\{x|x⊆A∩B\}$$ to $$\{x|x∈P(A) \text{ and }x∈P(B)\}$$ The property $$X\subset Y \iff X\cup Y = Y$$can be useful. solution: \begin{align} \{x|x∈P(A) \text{ and }x∈P(B)\} &= \{x|x\subset A \text{ and }x\subset B\}\\ &= \{x|x\cup A = A \text{ and }x\cup B = B\}\\ &= \{x|x\cup A \cup B = A\cup B\}\\ &= ... 2 You are going in the right direction. \mathcal{P}(A) = \{x | x \subseteq A \}, so just by replacing the symbols we see \mathcal{P}(A \cap B) = \{x | x \subseteq A \cap B\}. So you are being asked to relate\{x | x \subseteq A \cap B\}$$To \mathcal{P}(A) \cap \mathcal{P}(B) which is$$\{x | x \in \mathcal{P}(A) \text{ and } x \in \mathcal{P}(B)\}$$... 0 I think you only need to prove A_i \backslash A_{i+1} and A_j \backslash A_{j+1} are disjoint for i<j. This follows from for i+1\leq j, we have A_{j}\subset A_{i+1}. Then the statement holds by intersects \Big(\bigcup_{n=0}^\infty (A_{2n} \backslash A_{2n+1})\Big)^c on both sides of your assumption 2. 0 No, in general this is wrong! Consider the Borel algebra generated by the cofinite topology:$$\Sigma:=\{E\subseteq[0,1]:\#E\leq\aleph_0\lor\#E^c\leq\aleph_0\}=\sigma(\mathcal{T}_{cofinite})$$and the finite Borel measure:$$\lambda(\#E\leq\aleph_0):=0\lambda(\#E^c\leq\aleph_0):=1$$Then the atoms are all the uncountables but no singleton is a point ... 2 The example you give actually is a \sigma-finite Borel measure. Equip [0,1] with the cofinite topology (in which a set is open iff it is either empty or its complement is finite). Then your \Sigma is the Borel \sigma-algebra of the cofinite topology (it is a nice exercise to verify this). However, there is the following result: Proposition. Let ... 1 To show that X \cap Y = Y, it suffices to show that Y \subseteq X. So all we need to do is prove that \overline C - A \subseteq \overline A \cup B. Indeed:$$ \overline C - A = \overline C \cap \overline A \subseteq \overline A \subseteq \overline A \cup B $$1 Hints: f must one-to-one for the inverse function to exist. Consider what happens if it is onto S_2 or not. (See Bijection, injection, surjection) 2 Modern set theory conceives of a set as an abstraction of a property. Two properties might seem different, but be essentially the same because they are true of the same objects. For example, the property \mathcal O_1 of being a natural number of the form 2n+1, and the property \mathcal O_2 of being a natural number that is the difference of two ... 1 In the case of two empty sets, \emptyset_1, \emptyset_2, we have that \forall x(x\in\emptyset_1\Leftrightarrow x\in \emptyset_2) is vacuously true. 1 As$$\forall x(x\not\in\emptyset_i)$$is obvious that$$x\in\emptyset_1\iff x\in\emptyset_2$because both conditions are false. 3$\mathbb{N}$has a minimal element;$\mathbb{Z}$doesn't. An order isomorphism preserves minimality of an individual element. Can you show this? 2 You can't prove that there is no bijection between$\Bbb N$and$\Bbb Z$, because there is one. On the other hand, you can certainly prove that their natural orders are not isomorphic. Simply ask yourself, if$f$is an order preserving injection from$\Bbb N$to$\Bbb Z$, and$k=f(0)$, what could$f^{-1}(k-1)$be? 1 If$R$is anti-symmetric then any subset of$R$is anti-symmetric, also$R\cap S$. I think the normal way of doing such things is: If$(a,b),(b,a)\in R\cap S\implies (a,b),(b,a)\in R\implies a=b$. 0 If your new equivalence relation$\underset{S}{\sim}$is a constriction from the whole set$E$, then the answer is yes. Equivalence relation have to satisfy 3 axioms: Reflexivity.$a \underset{S}{\sim} a$. It is true for new relation because of the fact that we can look at the element$a$like an element of the whole$E$and use fact that$a ...

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Yes. But you should give your reasoning if this is an exercise. Try arguing that X have a subset of every order and so assuming it have a finite cardinality contradict this fact (how many subsets can a set of cardinality n have?)

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