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11

No, take $A=D=\{1\}$ and $B=C=\{2\}$. $$\begin{array}{c} \begin{array}{c|cc} 2&\bullet&\bullet\\ 1&\bullet&\bullet\\ \hline &1&2\\ \end{array}\\\\ (A\cup C)\times(B\cup D) \end{array} \qquad\qquad \begin{array}{c} \begin{array}{c|cc} 2&\bullet&\\ 1&&\bullet\\ \hline &1&2 \end{array}\\\\ (A\times B)\cup(C\times ...


9

Let $A=[0,1]=B$ and $C=[1,2]=D$ Here the first figure represents $(A\times B) \cup (C\times D)$ while the second one represents $(A\cup C) \times (B\cup D).$


7

This claim is clearly false: it means that if three set have no common element, a pair of them have no common element. Counter-example: $\,A=\{1,2\},\enspace B=\{2,3\},\enspace C=\{3,1\}$.


6

It doesn't; $f$ is not injective: $f(2, 4) = \frac{2}{4} = \frac{1}{2} = f(1, 2)$


5

Hint: consider $A=\{1,2\}$, $B=\{2,3\}$ and $C=\{1,3\}$. When you have a statement of the form If $A\cap B\cap C=\emptyset$, then $A\cap B=\emptyset$, $A\cap C=\emptyset$ and $B\cap C=\emptyset$ you can't prove it true by example. You can prove it false by showing a counterexample.


5

$S \in P(A) \cap P(B)$ if and only if $S \in P(A)$ and $S \in P(B)$ if and only if $S \subset A$ and $S \subset B$ if and only if $S \subset A \cap B$ if and only if $S \in P(A \cap B)$.


5

You can define a function $f : \mathbb{N} \to (-1,1), \ n \overset{f}{\mapsto} \sin(n)$. It will not be a surjection because there are only countably many points in the image and $(-1,1)$ is uncountable. To make it a surjection, we can just define the image set to be $T$, the points that are the images of $f$. The set $T$ looks a lot like the rationals in ...


5

It doesn't hold that $B \subseteq A$, but $A \subseteq B$. In order to show this, we pick a $x \in A$. Then $x$ is of the form $x=4n+3$ for some $n \in \mathbb{N}$. $x=4n+3=2(2n+1)+1$. $n \in \mathbb{N}$, so $2n+1$ also belongs to $\mathbb{N}$. So $x$ is of the form $2m+1, m \in \mathbb{N}$ and thus $x \in B$.


4

Alright, so let's begin by thinking about what this shape looks like. Lets say that we set $z$ to some constant value and see what the cross-section looks like. If $-1\le z\le 1$, then this clearly is empty. If $\left|z\right|>1$, then we get the open disk $x^2+y^2<z^2-1$ (interior of a circle). Now, lets look at those questions: a) Convex: No. ...


4

No it's not surjective because $ \emptyset $ is not in the range of $f$


3

There is no surjective map from $A \to 2^A$ as Cantor has taught us with his diagonal argument.


3

define $x_0 = 0$ and $x_n = 1/n $ for $n \geq 1$ ... define $f:[0,1] \to (0,1]$ $f(x)= x $ if $ x\neq x_i , i\geq 0$ and $f(x_i) = x_{i+1} , i \geq 0$


3

Comment: I thought this problem looked familiar--I encountered it some time ago at the beginning of a real analysis class (surveying proof techniques, basic set theory, etc.). Although you may have encountered this problem elsewhere, it is problem 1.9.25 in Witold Kosmala's book A Friendly Introduction to Analysis. There, it occurs as a review problem where ...


3

HINT: Take $A$ to be an infinite set; $\Bbb N$ will work well. It’s impossible with a finite set.


3

I'd use less symbols and more words. The aim is to show that, if $x\in D^c$, then $x\in A^c$. So, let $x\in D^c$. By hypothesis, $x\in B^c$, that is, $x\notin B$. Since $A\subseteq B$, we also have $x\notin A$ (otherwise $x\in B$). So $x\in A^c$. Your proof is good, anyway.


3

You obviously do not yet fully understand Cantor's argument and its implications. Where you are correct: Cantor's argument indeed relies on the fact that there exists a decimal representation of numbers. Where you are wrong: It is not true, as you are implying, that Cantor's argument only works if we represent numbers in a particular way. Cantor's ...


3

$$A=\{1,2\}, B=\{2,3,4,5\}, C=\{1,5\}$$ then: A∩B∩C= ∅ A∩B={2} A∩C= {1} B∩C= {5}


3

Assuming that you mean Lebesgue integrable, note that if a function is non-zero only on a subset of the Cantor set, then it is integrable. This alone gives you $2^{2^{\aleph_0}}$ integrable functions whose integral is $0$. (Note that this works for Riemann integrable functions as well, since on $[0,1]$ the function is integrable, since the points of ...


2

Assume $\alpha=\beta+1=\beta\cup\{\beta\}$ where $\beta$ is an ordinal (note that this is applicable for all nonzero finite ordinals, but also for many infinite ones). Then $x\in\bigcup\alpha$ iff $x\in y$ for some $y\in \alpha$. And this is equivalent to $x\in\beta$ or $x\in y$ for some $y\in \beta$. Since $x\in y\in \beta$ implies $x\in \beta$ as well, we ...


2

Consider the following sets $A$={$1,2,3,4,5$}, $B$= {$1,2,3$}. Then we can see that every element (an element is a member of the set) of $B$ is also in $A$ Then we say that $B$ is a subset of $A$. More generally speaking, if every element of set $B$ is in a set $A$ then we say $B$ is a subset of $A$ and denote it by $B \subset A$ With regards to your ...


2

$|B|=|C|$ means that there is a bijection, $g,$ from $B$ to $C$ so $g \circ f$ is the injection you want.


2

You can write a union using complements and intersections. Namely, you have to check that $A \cup B = (A^c \cap B^c)^c$. This is obvious because the complement of the union is those elements which are in neither $A$ nor $B$, ie the elements in both $A^c$ and in $B^c$.


2

The statement is false, assuming that "$f^{-1}$ exists" means "$f^{-1}$ is a function" and not $f^{-1}:B\to A$; for example the identity function $f:\Bbb N\to\Bbb R$ defined by $f(x)=x$ has an inverse function $f^{-1}:\Bbb N\to\Bbb N$ as $f(x)=x$, but it is not surjective onto $\Bbb R$. If we assume that $f^{-1}$ is a function with domain $B$, then it is ...


2

The special role of the naturals is that their cardinality is the smallest that is not finite. That this cardinality bears the special name "countable" is of course owed to the fact that we use the naturals for counting :)


2

You are right on both counts. Suppose $g:A\to B$, $f:B\to C$. For the first question, suppose $C$ has only one element and $B$ has more than one element. Let $g$ be the function that takes every element of $A$ to a fixed element of $B$: $g(x) = b_0$ for all $x\in A$. Then $g$ is not surjective, but $f\circ g$ and $f$ both are. For the second question, let ...


2

All right, here’s the solution to your problem. Let $A$ be a set with an injection $f$ into $\Bbb N$. If $A$ is finite, it’s countable according to your definition. If, on the other hand, $A$ is infinite, then you need to show that $A$ has a subset in one-to-one correspondence with $\Bbb N$. That is, you want an injective map $g:\Bbb N\to A$. To get such a ...


2

You are correct; the answer should be $2^{36}-2^{35}=2^{35}$. Another way of looking at it: $x$ must be in any such subset, and $y$ must not be. Thus, we have $35$ "free" elements that could either be in or out of the subset. So, there are $2^{35}$ ways of forming that subset.


2

You are making the common mistake to confuse between a number and its decimal representation. An easy way to see that $\Bbb R$ is uncountable, regardless to how we can or cannot represent real numbers, is to see there is an injection from $\mathcal P(\Bbb N)$ into $\Bbb R$, defined by $\displaystyle A\mapsto\sum_{n\in A}\frac1{3^{n+1}}$. This function ...


2

To say that $\{\varnothing,\{\varnothing\}\}\subseteq A$ is the same as saying $$\varnothing\in A\text{ and }\{\varnothing\}\in A$$ While it is true that $\varnothing\subseteq A$, it is certainly not necessarily true that $\varnothing\in A$. And in this case, it is indeed false.



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