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1

It's easiest to show this component-wise. Using infix notation to avoid parentheses, I argue for symmetry. If $(x,y)R'(w,v)$ then it implies $xRw$, which by symmetry of $R$ implies $wRx$; likewise the definition of $R'$ implies that $ySv$, which in turn implies $vSy$. Therefore $((w,v),(x,y))$ is a pair that satisfies the conditions of $R'$. Reflexivity and ...


1

Yes, it seems correct. It can also be done without referring to elements, using only the set operations and their properties. $\def\1{{\bf1}}$ $x\in (A_1\cup A_2)\oplus(B_1\cup B_2)\ \iff\ \1_{A_1\cup A_2}(x)\ne \1_{B_1\cup B_2}(x)$. Now use that $\1_{A_1\cup A_2}(x)=\max(\1_{A_1}(x),\,\1_{A_2}(x))$ to conclude that at least one of the inequalities ...


0

The membership table should look like this: $$\begin{array}{ccc|cc}A & B & C & A\setminus(B\cap C) & (A\setminus B)\cup (A\setminus C)\\\hline 1&1&1&*&*\\ 1&1&0&*&*\\ 1&0&1&*&*\\ 1&0&0&*&*\\ 0&1&1&*&*\\ 0&1&0&*&*\\ 0&0&1&*&*\\ ...


0

Hint Just the first two: The set of all boys is the set of all people who are children and are males. Therefore, a person is in the set of boys if and only if the person is in the set of children and the person is in the set of males. Which set operation is connected with the and logical operator? The set of all girls is the set of all people who are ...


2

Remember that when you're looking for overlapping elements of two groups, you're looking for their intersection. This is a big upside down U in set notation. An apostrophe signifies a "not". So answer a here essentially says, "every element of the men group that is also in the children group". a) $M \cap C$ b) $M' \cap C$ c) $M' \cap C'$ d) $C' \cap H ...


3

Suppose that $A\cap B\neq \emptyset$. Then there exists $x\in A\cap B$. If $f(X)=(A,\emptyset)$, $X\cap B=\emptyset$, so $x\not\in X$. But $X\cap A=A$ so $A\subseteq X$ and $x\in X$. This is a contradiction. For the other direction, assume that $A\cap B=\emptyset$. Let $(C,D)\in P(A)\times P(B).$ Then if $X=C\cup D$, $f(X)=(C,D).$


2

Hint: Prove the contrapositive. For example, if $A \cap B \neq \emptyset$, then it is impossible to get $(A_1, \emptyset)$ for some $A_1 \subset A$ with $A_1$ non-empty. Explicitly, take $E = \{1,2,3\}$ and $A = \{1,2\}$ and $B=\{2,3\}$. Then there is no $X \in \mathscr{P}(E)$ so that $f(X) = (\{2\}, \emptyset)$. Recall that for any set $R$, $\emptyset \in ...


1

Let it be that $\lambda.x_1+\mu.x_2=0$ and $(\lambda,\mu)\neq(0,0)$ (or equivalently $\lambda\neq0\vee\mu\neq0$). If e.g. $\lambda\neq0$ then $x_1=\alpha.x_2$ where $\alpha=\frac{\mu}{\lambda}$. Then $x_1\in S_2$ so that $x_1\in S_1\cap S_2$ and a contradiction is found. Likewise you can find a contradiction in the other case.


1

Let $x_1$ and $x_2$ are dependent. So $\alpha x_1+\beta x_2=0$. If $\alpha\ne0$, $x_1=-\frac{\beta}{\alpha}x_2$ and $x_1\in S_2$. If $\alpha=0$ then $x_2=0$ which contradicts that $x_2\in S_2$


0

$$[a,b]=\{a,a+1,\cdots b\}.$$ $$(a,b]=\{a+1,a+2,\cdots b\}.$$ $$[a,b)=\{a,a+1,\cdots b-1\}.$$ $$(a,b)=\{a+1,a+2,\cdots b-1\}.$$


2

You can't write out the elements of an interval on $\mathbb{R}$, since it is uncountable. However, one can represent an interval using set builder notation like so: $$ [a, b) = \{ x \mid a \leq x < b \} $$


1

Ahh, this is Cantors Theorem. Listed here: Non-existence of a Surjective Function from a Set to Its Subsets (Cantor's theorem) Thanks for the help everyone!


1

Hint You must supplement your textbook with some book on set theory, like e.g. : Patrick Suppes, Axiomatic set theory (1960 - Dover reprint). See page 22 for the usual definition of the "inclusion" relation : $A \subseteq B \leftrightarrow \forall x ( x \in A \to x \in B)$ and page 47 for the definition of the power set : $\mathcal P(A) = \{ B : B ...


9

The objects in set theory are sets. Only sets in $\sf ZFC$ and its related theories. This means that if you want to interpret a mathematical object in set theory you need to assign it a set. Of course you are free to assign to it any set that you wish, as long as you have Tue axiom of replacement set theory is more or less interpretation agnostic (in the ...


-3

You have N≠ℵ0, Infact one is the set (N), while ℵ0 is a transfinite number. Asking ℵ0=N, is somewhat.. you are asking.. Is 2={1,2} ? which is obviously not.


11

There's nothing deep going on here. Its just that: Its often convenient to identify $\mathbb{N}$ with the least infinite ordinal $\omega$. Its often convenient to identify each well-orderable cardinal number $\kappa$ with the least ordinal $\alpha$ such that $|\alpha| = \kappa$. This is called the Von Neumann cardinal assigment. Under these ...


1

When an event, call it $T$, is a tail event, it is in $\cap_{n=1}^{\infty}\sigma(X_{n},X_{n+1},...)$, so it is an element is every one of the $\sigma$-algebras $\sigma(X_{n},.X_{n+1},...)$. The reason we can interpret this as not depending on any finite set $\{X_{1},X_{2},...,X_{n}\}$ is because while the event is in $\sigma(X_{1},X_{2},...)$, it is not ...


1

While sample set could also be worthwhile terminology, it is worth recalling that any space in mathematics is always a set with some sort of operations defined. For example, in the theory of stochastic processes, we could let $\Omega = \mathcal{D}(\mathbb{R}_+, \mathbb{R})$, where $\mathcal{D}(\mathbb{R}_+, \mathbb{R})$ is the Skorokhod space of real-valued ...


2

I think is a matter of definitions. Let $≤$ be a preorder. We define the induced equivalence by $x \sim y \iff x ≤ y$ and $x ≥ y$. We also define the induced strict variant by $x < y \iff x ≤ y$ and $x \nsim y \iff x ≤ y$ and $x \ngeq y$. Now we can prove the following general theorem: If $≤$ is a preorder, then $<$ is a strict order (i.e. an ...


3

We say "$A$ injects into $B$", and may write $f: A \hookrightarrow B$.


7

Probably the most standard term of the sort you're looking for is "equipotent", though it isn't used particularly often (more often, people will just say two sets "have the same cardinality" or "are in bijection"). People who think about categories a lot sometimes say "isomorphic as sets".


1

No, you're not missing anything. The correct answer $\{x\mid 0<x\le 12\}$ may also be written more compactly as $(0,12]$ or $]0,12]$.


0

If $\alpha \in X \cap (Y \setminus Z)$ then $\alpha \in X$ and $\alpha \in (Y \setminus Z) \implies \alpha \in Y, ~ \alpha \not\in Z$. So $\alpha \in (X \cap Y).$ But $\alpha \not \in Z$ so $\alpha \not \in X \cap Z$. Therefore $\alpha \in (X \cap Y) \setminus (X \cap Z).$ Now if $\alpha \in (X \cap Y) \setminus (X \cap Z)$ then $\alpha \in X$ and $\alpha ...


1

$$(X \cap Y) \setminus (X \cap Z) = (X \cap Y) \cap (X \cap Z)^C = (X \cap Y) \cap (X^C \cup Z^C) \\ = (X \cap Y \cap X^C) \cup (X \cap Y \cap Z^C) = X \cap(Y \cap Z^C) = X \cap (Y \setminus Z) $$


0

There's a great visual proof for this, have you seen it? It's simple enough that I can just describe it, and perhaps you can draw it yourself: Take a circle $C$ of circumference $|a-b|$ (this will be our interval $(a,b)$, after we remove a single point), and place it above a straight line $L$ of infinite length having no endpoints (this is our stand-in for ...


2

Hint: First find an easy bijection between any two given open intervals (you can even find a continuous one), and then use a function and certain interval that you can show a bijection to ${\mathbb R}$ with (for example, tangent function).


3

The function $y = \tan(x)$ is bijective on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. We would like to shift/stretch it so that it's bijective on $(a,b)$. The period should be $b-a$, so we would at least have $y = \tan\left(\frac{\pi}{b-a}x \right)$. Now translate it.


0

The (admittedly, not very revealing) answer is $$B=\left\{b\in\Bbb N:\forall a\in A\bigl(\lvert b-a\rvert\in C\bigr)\right\}.$$ By definition, any set with the property you described will be a subset of this one.


1

Sure, just let $B=\{a+c:a\in A,C\in C\}\cup\{a-c:a\in A,c\in C\text{ and }a>c\}$.


0

If I understand your question correctly, you ask how $x\in E$ and $x\notin P(E)$ can be true at the same time. Well, that is easy to see with an example: Be $E=\{1,2,\}$. Then $P(E)=\{\emptyset,\{1\},\{2\},\{1,2\}\}$. Clearly in this case none of the elements of $E$ is in $P(E)$. Note that $1\ne\{1\}$. However note that in other cases, elements of $E$ can ...


1

$A=\{x\in E,\, x\notin f(x)\}\in P(E)$. So if $f$ is surjective, then there exists $y\in E$ such that $f(y)=A$. Try to show that this leads to a contradiction. See also: https://en.wikipedia.org/wiki/Cantor's_theorem


1

It’s generally easiest to start with the more complicated expression, which in this case is the indicator function corresponding to the righthand side, $$(1_A+1_B-1_A\cdot 1_B)\cdot(1_A+1_C-1_A\cdot 1_C)\;.$$ If you multiply this out, you get $$\begin{align*} 1_A\cdot(1_A&+1_C-1_A\cdot 1_C)+1_B\cdot(1_A+1_C-1_A\cdot 1_C)-1_A\cdot ...


2

Suppose $f$ were not surjective. Then there is $b \in F$ such that $f(a) \not = b$ for all $a \in E$. That is, there is $b \in F$ such that $f^{-1}(\{ b \})$ is empty. What is another set which has empty preimage? Suppose $f^{-1}$ were not injective. Then there are $A, B$ distinct such that $f^{-1}(A) = f^{-1}(B)$. Equivalently, $\{ x \in E \vert f(x) \in A ...


23

For this kind of question, it's often best to make a table. Here's the data when put into a table. $$ \begin{array}{c|cc|c} &\text{ Is Reasonable } & \text{ Not Reasonable } & \text{ Total} \\ \hline \text{Is Worse} &363 & ??? &487 \\ \text{Is Not Worse} & ??? & ??? & ???\\ \hline \text{Total} &756 & ??? & ...


1

First part: $A \cup B=E \Rightarrow f$ is injective. Suppose that $f(A_1)=f(A_2)$ for some $A_1,A_2 \in \mathcal{P}(E)$. We have to prove that $A_1=A_2$. But: $$f(A_1)=(A_1 \cap A, A_1 \cap B)$$ $$f(A_2)=(A_2 \cap A, A_2 \cap B)$$ So: $$A_1 \cap A=A_2 \cap A$$ and $$A_1 \cap B = A_2 \cap B$$ So: $$(A_1 \cap A) \cup (A_1 \cap B)=(A_2 \cap A) \cup ...


3

Here's one direction of part a). Suppose $A\cup B=E$, then for any $X\subseteq E$, $X=(A\cap X)\cup (B\cap X)$. If $f(X)=f(Y)$ then $(A\cap X,B\cap X)=(A\cap Y,B\cap Y)$ which implies $A\cap X=A\cap Y$ and $B\cap X=B\cap Y$. So, $$X=(A\cap X)\cup(B\cap X)=(A\cap Y)\cup(B\cap Y)=Y.$$ Hopefully this will get you thinking along the right track.


1

Hints on a) If $A\cup B\neq E$ then $E-(A\cup B)$ has more than one subset. These are all mapped to? If $A\cup B = E$ then $X=(A\cap X)\cup(B\cap X)$ Hint on b) If $A\cap B=\varnothing$ then $X=A_0\cup B_0$ where $A_0\subseteq A$ and $B_0\subseteq B$ is sent to?... If $x\in A\cap B$ then can you find a set $X$ such that $x$ in element of exactly one ...


1

You are correct. if a [a,d]-[b,c]=[a,b)\cup(c,d] because "[b" more than b, including it, "c)" means less than c including it.


1

Yes. (PS: These are not intervals, but union of intervals!)


0

For each $b \in B$ let $F(b)$ be the subset of $A$ such that $f(a) =b$ for all $a \in F(b)$. Then since $f$ is surjective each $F(b)$ is non empty. Axiom of choice says I can form a set $F'$ by picking one element from each $F(b)$. Now define $g: B \to A$ by $g(b) = a$ where $f(a) = b$ with $a \in F'$


1

An element in that set must satisfy both conditions. There are no elements that simultaneously satisfy both conditions. So there are no elements in the set.


2

If $x+6=6$, then $x=0$ and it's the only one solution... And $0^2\neq 16 $, hence the set $A$ is empty... Where's the problem? The $x$ in the set must satisfy both conditions, that is $x+6=6$ and $x^2=16$, but there are no such numbers.


0

The Axiom of Choice (AC): If $ F$ is a set of non-empty sets then there is a function $g$ with $dom(g)=F$ such that $ g(x) \in f$ for each $x \in F$ . Such a function $g$ is called a choice function for $ F$. So given a surjection $f:A \to B$, the axioms of Separation and Comprehension give us $ F$ , the set of fibers of $f$, that is $ F=\{f^{-1} \{ b \} : ...


0

for $b\in{B}$ since $f$ issurjective $f^{-1}(b)$ isnot empty. using the axiom of choice $g(b)\in{f^{-1}(b)}$.


1

Consider the mapping $$f: P(\mathbb{N}) \to \{0,1\}^{\mathbb{N}}$$ where $f(A) = (a_n)_{n \in \mathbb{N}}$ is the sequence of zeros and ones such that $a_n = 0$ if $n \notin A$ and $a_n = 1$ if $n \in A$. Now you can show that this is a bijection.


0

Just to add a bit to Colm's very good answer. You want to prove $$ \forall a (a\in \varnothing \implies a \in B) $$ Membership in the empty set can be hard to get your head around. So first, look at the implication. The statement $p\to q$ is equivalent to $\neg p \vee q$. (Think of it as “it is never true that $p$ is true and $q$ is not also true.” ...


0

Here is a way to calculate all examples, in some sense.$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} $ First, we try to simplify ...


0

The simplest example would be $f:\{1\}\to\{1,2\}$ with $f(1)=1$ and $B=\{2\}$.


2

Let $X=Y=\Bbb R$, $f(x)=x^2$ and $B=(-1,\infty)$. Now, $$f(f^{-1}(B))=f(\Bbb R)=[0,\infty)\subsetneq (-1,\infty).$$



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