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1

Yes you should group the columns in the following sense: Hint let's take an example: $$S_3=\begin{pmatrix}1 & 1+2 & 1+3\\ 1 & 1+2^2 & 1+3^2 \\ 1 & 1+2^3 & 1+3^3 \end{pmatrix}$$ now if we made the operations $C_2\leftarrow C_2-C_1 $ and $C_3\leftarrow C_3-C_1 $ you will get: $$S_3\sim \begin{pmatrix}1 & 2 & 3\\ 1 & 2^2 ...


2

Everyone is giving generalized arguments for the most part, but to disprove a conjecture, all you must do is find one counterexample, and this is what Ittay was hinting at in a comment. Thus, I'll give you one example of many that invalidates your claim. Consider the following sets: $A=\{1,2\}$ $B=\{2,3,4\}$ $C=\{1,4,5\}$ $U=\{1,2,3,4,5\}$ $C^C=\{2,3\}$ ...


2

$A=\{1\}, B=\{2\}, C=\{2,3\}$. Then $A\cup B\not\subseteq C^c$.


1

No. If $A,B$ are sets, then $A$ need not be a subset of $B$ or $B^c$. For example, take $A=B\cup B^c$ where $B$ and $B^c$ are both nonempty.


1

The claim is false as currently stated. To see this consider the real number line with subsets $A = [0,1], B=[1,2]=C$. Instead of being concerned on whether or not $A\cup B \subset C$, the key thing is that $(A \cup B) \cap C = \emptyset$. If that is true, then $A\cup B \subseteq C^c$. Otherwise if $(A \cup B) \cap C \neq \emptyset$, then $A\cup B ...


1

HINT: Consider the case where $A\subseteq C$ and $B\cap C=\varnothing$. What happens then? General piece of advice, if you have an hypothesis just try and prove it. If you failed, try to use this failure to create a counterexample; if you succeed go over your proof carefully several times, show it to someone else, and if you're finally sure it works, then ...


5

A falsity implies anything. And the condition is always false. Due to Cantor's theorem, $|P(S)| > |S|$ always. But $P(S) \subseteq S$ implies $|P(S)| \leq |S|$ (which is false!), and a falsity implies literally everything, including $\mathcal{P}(S) \in S$.


7

If $ \mathcal{P}(A) \subseteq A $, then $ \mathcal{P}(A) \in \mathcal{P}(A) \subseteq A $, because $ \mathcal{P}(A) $ is a subset of $ A $ and hence an element of $ \mathcal{P}(A) $. Therefore, $ \mathcal{P}(A) \in A $. Note: The Axiom of Regularity is necessarily violated here (thanks, Trevor, for pointing out the need to rephrase).


1

You can partition any set $X$ as $X=\displaystyle\bigcup_{x\in X}\{x\}$.


0

A 1-1 correspondence is another word for bijective. Thus, it is one-one and onto. A one-one function is a mapping such that f(x1) = f(x2) if and only if x1 = x2. An onto mapping is a mapping such that every element is mapped over. So suppose we have f = x + 1 and our set is [1,2,3,4]. Then our new set would be [2,3,4,5] so we see that this is a 1-1 ...


0

If you want to show that $A\sim A$, all you have to do is find a function from $A$ to itself which is both injective and surjective. For example, the identity function defined by $f(a)=a$.


2

Suppose you were given an injective function $f : X \to Z$, and you knew from the definition that $$\text{If } x \neq y, \text{ then }f(x) \neq f(y).$$ Suppose now that somebody hands you the image under $f$ of two things in $X$, and tells you that $f(x_1) = f(x_2)$. You would be forced to conclude that $x_1 = x_2$, wouldn't you? Otherwise, if they were ...


0

A one to one correspondence indicates that there is a map $f: A\to B$ that is onto, so that for every $b\in B$ there is an $a\in A$ such that $f(a) = b$, and also one to one, meaning that the $a\in A$ such that $f(a) =b$ is unique.


4

The best explanation I've heard for the meanings of "injective" and "surjective" are $f:X\to Y$ is injective if for each $y\in Y$, there is at most one $x\in X$ with $f(x)=y$ $f:X\to Y$ is surjective if for each $y\in Y$, there is at least one $x\in X$ with $f(x)=y$ Naturally, combining them gives $f:X\to Y$ is bijective if for each $y\in Y$, there is ...


3

In general, if you have two sentences $P$ and $Q$, then $$(P \implies Q) \iff (¬Q \implies ¬ P).$$ So: $$(f(x)=f(y) \implies x = y) \iff (x \neq y \implies f(x) \neq f(y)).$$ The idea, however, is just what you said: given a point in the image, there is only one pre-image. Examples: $f: \Bbb R \to \Bbb R$, given by $f(x) = x^2$ is not injective ...


4

This refers to the set $\mathbb{R}$ , but without $0$, i.e. the complement of $\{0\}$ in $\mathbb{R}$.


5

Real numbers that aren't $0$. \ is set difference.


3

It just means that $A_1,\ldots,A_n$ are sets. It is often the case that they are distinct, but if it was not mentioned then it might be the case that $A_1=A_2=\ldots=A_n$. The subscript is just an index. And formally it means that there is a function $A$ whose domain is $\{1,\ldots,n\}$ and we write $A_i$ for the set obtained by $A(i)$. Any additional ...


0

Since there is a bijection $f\colon\mathbb N\to\mathbb Z\setminus\{0\}$ and the function $g\colon \mathbb Z\setminus\{0\}\to\{1/n:n\in\mathbb Z\setminus\{0\}\}$ defined by $g(n):=1/n$ is a bijection, we know that the composition $g\circ f$ is a bijection, so the set $\{1/n:n\in\mathbb Z\setminus\{0\}\}$ must be countable. Remark. If you have no defined the ...


0

You can see the book "Book of Proof" of Richard Hammack; it have many diagrams and pics. The chapter about cardinals is very educational. P.S.: Machine leaning is more about Linear Algebra and Probability Theory.


1

Let me solve one of the directions. I would like to prove that $A\subseteq B$. How do I do it? I take some arbitrary element of $A$ and show that it must also be an element of $B$. So take some $x\in A$. By definition of $A$, I can write $x=2n+1$ for some integer $n$. If I can somehow also write $x$ as $x=2m-21$ for an integer $m$, then I know that $x\in B$ ...


0

Do you know any number theory? You can prove the sets are equal if you can prove that $x = 2n + 1$ and $x = 2m - 21$ have the same solutions. Consider the diophantine equation: $$x - 2n = 1$$ By Bezout's lemma, this has a solution as long as $\operatorname{gcd}(x,-2) | 1$. In other words, as long as $x$ is odd. Etc.


0

If $x\in A$, $x=2n+1=2(n+11)-21\in B$ $\Rightarrow \forall x\in A, x\in B \Rightarrow A\subseteq B$. If $x\in B$, $x=2m-21=2(m-11)+1\in A$ $\Rightarrow \forall x\in B, x\in A \Rightarrow B\subseteq A$.


0

$2n+1=2n+1+(21-21)=2n+22-21=2(...)-21$ so $A\subset B$. Other side is similar.


1

The proposition of the exercise tells you to consider that $A \subseteq B$. Power set of A is defined as follows: if X is in $\mathscr{P}(A)$, then $X \subseteq A$. But as stated before, $A \subseteq B$, so $X \subseteq B$, too. Thus, as $X \subseteq B$, it follows that X is in $\mathscr{P}(B)$.If X is in $\mathscr{P}(A)$ implies that X in $\mathscr{P}(B)$, ...


2

By definition, $h^{-1}(w)$ is the set: $$\{v \in \{a,b\}^* \mid h(v) = w\}$$ Depending on your definitions, this can be an abuse of notation, because $h^{-1}$ might not be a mapping. However, it is convenient to omit the inner braces in the formally more correct $h^{-1}(\{w\})$. As to the set itself: Because there is no possibility for $h$ to yield three ...


0

You know that there exist injective functions $f, g$ $$A \overset{f}{\hookrightarrow} B \overset{g}{\hookrightarrow} C$$ since $A \prec B \prec C.$ I assume you have the Cantor-Bernstein theorem, that $P \sim Q$ if and only if $A$ and $B$ inject into each other. This means that, since $B$ strictly dominates $A$, and $C$ strictly dominates $B$, any ...


0

It may be helpful to break it down into several claims to be proved: $\boxed{\text{(a) and (b) cannot both hold}}:$ Suppose they did. Since $A \neq \varnothing$, there is some $a \in A$. But $A \subseteq B$, so $a \in B$. But then $a \in A \cap B = \varnothing$, a contradiction. $\boxed{\text{(a) and (c) cannot both hold}}:$ Similar reasoning as above. ...


0

Hints: $A\cup B\in T$ since $T$ is assumed to be a topology, you have for possibilities that one of which must hold - for example if $A\cup B=X$ then $B=X\setminus A$, what if $A\cup B=A$ or $A\cup B=B$ ? can $A\cup B=\emptyset$ ? Now, can both of those three options hold simultaneously ? for example if both (b) and (c) are true then $A=B$, but you know ...


1

A way to approach it is to consider constructing the transitive closure in stages - what relation is formed if we take the set of $(x,y)$ such that either $(x,y) \in D_r$ or there is some $z$ such that $(x,z),(z,y) \in D_r$? Then what happens if you include longer chains like this? Slightly more detailed hint (but still only hinting):


1

Commutativity of sets under set union gives us $$(Y\cup X)' = (X\cup Y)'$$ Furthermore, it is also true that sets commute under set intersection: $A\cap B = B\cap A$.


0

From the definition of intersection of two sets , what it means to do in simple language is that you have to locate all the similar elements in two set, here is an example. $\left\{1,2,3,\underbrace{4,5,6},\right\}$ ,$\left\{\underbrace{4,5,6},7,8,9\right\}$ = $\left\{4,5,6\right\}$


0

For the first example - just look at the numbers which are members of both sets. For the second - find the members of each set by solving the equation the defines it. the set is the set of all these members. And i did not understand the last question. Please make an effort and show that you tried something, after that I'd be glad to give more details.


0

Hint: Let $X$ be a topological space, and let $B\subset X$ be a subspace. Show that point $x\in X$ is such that any open set containing $x$ intersects $B$ if and only if $x$ is contained in every closed subset of $X$ that contains $B$. Now try applying this with $B=A$ and $B=X-A$.


2

Consider a collection of an arbitrary number of open sets in this topology. Either this collection only contains the empty set (so there is no union to take), or this collection contains some nonempty $U$ of the topology. Since $U$ contains $A$, then the union over all open sets in this collection also contains $A$ and is therefore in the topology as well. ...


3

It's not really correct to put an equal sign there. What would make more sense is to construct a bijection. Elements of $\{0,1,2\}^{\mathbb{Z}^2}$ are functions $f : \mathbb{Z}^2 \to \{0,1,2\}$. Elements of $(\{0,1,2\}^\mathbb{Z})^{\mathbb{Z}}$ are functions $g : \mathbb{Z} \to \{0,1,2\}^{\mathbb{Z}}$. The bijection is this: $f \leftrightarrow g$ if and only ...


3

Not quite. But your intentions are probably correct. $\{x\}$ is a set with a single element. Therefore $\{x\}^\Bbb Z$ is the set of all functions from $\Bbb Z$ into that singleton, and there is only one function like that: $f(k)=x$ for all $k$. If, however, you mean $\left(\{0,1,2\}^\Bbb Z\right)^\Bbb Z$, then this is not equal to, but isomorphic to. You ...


1

To elaborate on Thomas Andrews' comment: Addition is well-defined: For all $a,b,c,d\in\mathbb{R}$, if $a=b$ and $c=d$, then $a+c=b+d$. The well-defined nature of addition in $\mathbb{R}$ is taken as an axiom. Hence, if $a=b$ and $c=d$, we have $$ a+c=b+d\Longleftrightarrow a+c=b+c, $$ but I think there is something more interesting you could show, namely ...


1

You start with $$ a \oplus c = x \quad (*) $$ where $x$ is the resulting value, then note that $a = b$, so you replace $a$ with $b$ in equation $(*)$. This gives $$ b \oplus c = x $$ Of course this means $$ a \oplus c = x = b \oplus c $$ and thus $$ a = b \Rightarrow a \oplus c = b \oplus c $$ Let us redo this for variables $a, b, c \in \mathbb{R}$ and the ...


0

Define $(a,b)$ to be the set $\{a, \{a,b\}\}$, and define $A \times B$ to be the set $\{ (x, y) : x \in A \land y \in B \}$. Then $\mathbb{R} \times \emptyset = \emptyset$, and $\mathbb{R} \times A$ for any non empty set at results in a set of ordered pairs. This begs the question, is $\mathbb{R}^1$ a set of ordered pairs? I would say not, as this could make ...


0

No. All limit ordinals are infinite,1 but not all infinite ordinals are limit ordinals. To point out, note that every ordinal has a successor, which is a strictly larger set (inclusion-wise, not cardinality-wise). So every limit ordinal has a successor ordinal which is also infinite. Cardinals are ordinals which have the property that no smaller ordinal has ...


0

No: $\omega+1=\omega\cup\{\omega\}$ is infinite, but it is not a limit ordinal, because it has a largest element, namely, $\omega$. In fact $\omega+n$ is not a limit ordinal for any positive integer $n$; the next limit ordinal after $\omega$ is $\omega+\omega$, the limit of the successor ordinals $\omega+n$. Any ordinal that can be written in the form ...


0

Limit ordinals are not the same as infinite ordinals. $\omega+1$ is an infinite successor ordinal.


1

HINT: Given $f\in F$, consider the following functions: for each $m\in\Bbb N$ let $g_m\in F$ be defined by $$g_m(k)=\begin{cases} f(k),&\text{if }k\ne m\\ f(k)+1,&\text{if }k=m\;. \end{cases}$$


0

The Cartesian Product of two sets $A$ and $B$ is defined as $$ \{(x,y): x \in A \text{ and } y \in B\} $$ (0,2) and (0,4) would not be in the Cartesian product because $0$ is not in $A$.


0

Let's start building up a way of describing the set of all lines in the plane. There are two types of lines: vertical lines and non-vertical lines. Any vertical line can be represented by an equation of the form $x=c$, for some unique constant $c \in \mathbb R$. So if we call $S$ the set of all vertical lines, there is a clear one-to-one correspondence ...


-1

This question has several answer. Strictly speaking, there exists a bijection but it is somehow a very complicated one and not very useful either. It is not continuous, meaning that two points that are close to each other can correspond to lines that are very far. In fact, both sets, lines in the plane and points in the plane, are also in bijection with... ...


0

Assuming AC the answer is yes. (I'm not sure what the answer is in ZF). Hint: Think about even and odd ordinals, and how you might use their existence for your purposes. This actually proves a bit more namely that you can split any infinite set into two sets with cardinality equal to the original set.


1

Here is a sketch: The left hand side: and the right hand side: This is something you really could do by yourself! There are two ways to finish your proof: 1) You take this sketch as a help and try to proof this equation formally 2) Explain why making a sketch like this is indeed enough in this case (we only have 3 sets) to prove the equation.



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