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9

One domain is $\{\rho_1(x,y)<1\}$, the other is $\{\rho_2(x,y) <1\}$, where $\rho_1(x,y) =\sqrt{x^2+y^2}$, and $\rho_2(x,y) = \max(|x|,|y|)$. Find a map $F\colon (x,y) \mapsto (x',y')$ so that $\rho_1(x,y) = \rho_2(x',y')$. It should be linear on each line through the origin. One can take: $$F(x,y) = \frac{\rho_1(x,y)}{\rho_2(x,y)} \cdot (x,y)$$ that ...


7

HINT: Start by finding an injection $f$ from $\Bbb R$ to the set of positive real numbers. Then the map $$g:\Bbb R\times\{0,1\}\to\Bbb R:\langle x,i\rangle\mapsto\begin{cases} f(x),&\text{if }i=0\\ -f(x),&\text{if }i=1 \end{cases}$$ is an injection. For $f$ consider an exponential function.


7

Note that $g(n)\neq f_n(n)$ for every $n\in\Bbb N$. Therefore, $g\neq f_n$ for every $n\in\Bbb N$.


5

I'd read $A=\varnothing$ as $A$ is empty or $A$ is the empty set.


5

Hint If we denote $$S := (-1, 1) \times (-1 ,1) \qquad \text{and} \qquad D := \{(x, y) : x^2 + y^2 = 1\},$$ we can easily write down a homeomorphism $f: \partial S \stackrel{\cong}{\to} \partial D$ between their boundaries in $\Bbb R^2$ by projecting points in $\partial S$ along rays from the origin; explicitly, this map is: $$f(x, y) := \frac{1}{\sqrt{x^2 + ...


5

As written, $\emptyset$ is an element of the given set (and of course it is also a subset as it always is). The four elements you list are indeed the four subsets of the given two-element set, so you're right.


5

Indeed, a function is a subset of $A \times B$ with the following very important property: for every $x \in A$, there exist a unique $y \in B$ such that $(x,y) \in A \times B$. Intuitively, this tells you that a function cannot take an element $x$ into several distinct values $y$ - that wouldn't be a function anymore, but somehing called "binary relation" ...


5

HINT: If there is a surjection from $A$ onto $B$, then there is an injection from $B$ into $\mathcal P(A)$. This leaves the case where $n=0$, but that's easy.


4

No. Of course not. Your proof is perfectly valid. This is also mentioned in the errata for the book.


4

If $A$ has $18$ elements, and $B\subseteq A$ has fewer than $9$ elements, then $B^c\subseteq A$ has more than nine elements. Also, if $B$ has more than $9$ elements then $B^c$ has fewer than $9$ elements. That means the number of subsets of $A$ with fewer than $9$ elements equals the number of subsets of $A$ with more than $9$ elements. Of course, the ...


4

The fact that you found one injective function which is not surjective does not mean that there are no other functions which are injective but not surjective. For example, $f(n)=n+2$ is an injective function from $\Bbb N$ into $\Bbb N\setminus\{0\}$, but it is not surjective. Does that mean there are not bijections? No, for example $g(n)=n+1$ is a ...


4

A is the set of reals integers $x$ such that $x^2<x<10$. $x^2<x$ is equivalent to $0<x<1,1<10$. Thus there is no integer ('whole number') solution. This is coherent with the given answer since no real number (be it integer or not) verifies $x^2+1=0$.


4

Probability is not just defined on a set. It's also defined on a $\sigma$-algebra of subsets. Namely, a probability structure is made of three parts: The underlying set, say $X$. The sets to which you can assign probability. The function which assigns probability. There are several axioms which we require of these structures, such as the sets to which ...


4

If you wonder why so far only the "trivial" example of a Dirac measure has been mentioned, there is actually a good reason for this. The young Ulam wondered whether there is an uncountable set $X$ for which there is a probability measure $$ \mu \colon \mathcal P(X) \rightarrow [0,1] $$ that vanishes on singletons (i.e. $\mu(\{x\})= 0$ for all $x \in X$). He ...


4

Take a set of rational numbers $A$ such that $$ \sum_{a\in A}=x<\infty. $$ If $A$ is finite, then $x$ is rational, which is not necessarily true if $A$ is infinite. This doesn't have as much to do with sets really, but shows how being closed under finitely many operations does not determine behavior under arbitrarily many operations. You could play this ...


3

For any cardinal $\kappa$, you can define the product measure on $2^{\kappa}$ where $0, 1$ are assigned one half measure each. This measure is defined on the sigma algebra generated by clopen subsets of $2^{\kappa}$. Let us call the corresponding measure algebra $\text{Random}(\kappa)$. The class of measure algebras thus obtained forms the building block for ...


3

That symbol usually denotes the set of all functions $B\to A$. I suspect the notation is suggested by the fact that the size $|A^B|$ of this set is $|A|^{|B|}$. Addendum: To see this, consider some simple examples. Suppose $A$ consists of a single point, say $A=\{a\}$. Then the only function $f:B\to \{a\}$ is the constant function with $f(b)\equiv a$ for ...


3

The ordinals below $\epsilon_1$ are those that can be written in Cantor normal form when a symbol for $\epsilon_0$ is allowed. More precisely, they are the values of the terms that can be built from the following symbols: $0$, $\omega$, $\epsilon_0$, $+$, and $\uparrow$ (where the last of these means ordinal exponentiation). So you can code each such ...


3

What does $$x\in \bigcup_{n \in \mathbb N} \bigcap_{k \geq n} A_k$$ mean? It means that there exists some $n\in\mathbb N$ such that $x\in \bigcap_{k\geq n} A_k$, which is the same as saying that $x\in A_k$ for all $k\geq n$. That is, $x$ is contained eventually in all of $A_n$, $A_{n+1}$, $A_{n+2}$, etc. In other words, $x$ is contained in $A_k$ for all but ...


3

Outline: Map the origin to itself. Now consider points $P$ in the disk with polar coordinates $(r,\theta)$, where $r\gt 0$ and $0\le \theta\le \frac{\pi}{4}$. Map $P$ to $\phi(P)$, where $\phi(P)$ has polar coordinates $(r\sec\theta,\theta)$. Do the geometrically same thing for the remaining $7$ sectors of the disk.


3

The equation you quote is the definition of the $f^{-1}$ notation on its the left-hand side. If $f$ is a function $X\to Y$, then it defines a function called $f^{-1}:\mathcal P(Y)\to\mathcal P(X)$. Note that this is a different $f^{-1}$ than the inverse function $f^{-1}:Y\to X$ which you can define if (but only if) $f$ is a bijection. The ambiguity usually ...


3

In this context, $f^{-1}(B)$ is not itself a function. We read this as the inverse image of $B$ under $f$, and it is the set that contains every element in $X$ that maps into $B$ under $f$.


3

For $a,b\in\Omega$ say $a\sim b$ if there exist finitely many elements $x_0=a,x_1, \ldots , x_n=b$ such that $x_{i+1}\in \mathcal P_1(x_i)\cup \mathcal P_2(x_i)$ for $0\le i<n$. Then $\sim$ is an equivalence relation on $\Omega$ and the corresponding partition has the desired property. Also, any other partition with the desired property must be coarser ...


3

That the function is not defined on $a$, and thus it's not defined on $A$.


3

Most introductory books on axiomatic set theory directly define set theoretical concepts in the language of first order logic. Once you know these definitions, all you need to do is translate, translate and translate until you are satisfied. One such book that I use can be found here: ...


2

Well, there are $2^{18}$ subsets all in all. Of these, $\dfrac{18!}{9!\cdot 9!}$ have exactly $9$ elements. The rest either have more than $9$ or fewer than $9$. Now, by symmetry, the number of subsets with more than $9$ elements equals the number of subsets with fewer than $9$ (pair each subset with its complement). Thus the number of subsets with more ...


2

Since $f(x)=y\in Y$, we have $x\in f^{-1}(Y)$ by the definition of $f^{-1}(Y)$. Also, $x\in X$ by construction. Therefore $x\in X\cap f^{-1}(Y)$, so...


2

If you think about a partition as being induced from an equivalence relation, then we know that if $E_1,E_2$ are two equivalence relations then there is no reason for $E_1\cup E_2$ to be an equivalence relation. However, $E_1\cup E_2$ is a reflexive and symmetric relation. So if we take $E$ to be its transitive closure, then $E$ is the smallest equivalence ...


2

Notice that $B \cap (A \cup C) = (B \cap A) \cup (B \cap C) = \emptyset \cup \emptyset = \emptyset$.


2

Your method is a good general method, but I think it is not terribly likely to yield results in this case. The trouble is that we need to respect the fact that this works over $\mathbb N$ - that is, we don't care if there are pairs of pairs of real numbers such that $f(i,j)=f(u,v)$, so long as they aren't a distinct pair of natural numbers - but there are ...



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