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3

A way to do this is to take a denumerable set in $D \subset (0,1)$, to define a bijection as the identity outside $D$, to map the first element of $D$ to $1$ [and the second to $0$] and to "push" the remaining elements of $D$ down by $1$ [or $2$] to get a bijection with $(0,1]$, and $[0,1]$.


2

By "almost equal", you mean "almost everywhere equal". Which means that the two functions agree except possibly on a set of measure zero. You can kind of visualize the measure of a subset of the real line as being its "length". And in fact, this is exactly what the measure is for intervals: that is, the measure of $(a,b)$ is $b-a.$ Now, what we call ...


2

That's a very creative idea. I don't know of an established set theory symbol for this (e.g. http://www.rapidtables.com/math/symbols/Set_Symbols.htm doesn't even define such an "operation") and it doesn't seem to be a standard symbol in $\LaTeX$. One problem I can see is that the symbol $\ominus$ is quite widely used to denote the symmetric difference of ...


2

It's not awful, at least as an idea, but I would certainly say it's unnecessary, and I strongly dislike your particular notation choice. Basically it's a symbol that's too different and unintuitive to have to remember for too unimportant and uncommon a situation to be worthwhile. In my experience, once one gets to mathematics classes, textbooks, etc. that ...


6

A picture is worth a thousand words. $A-(B \cup C)$ is everything in A that is not also in either B or C. $A-(B \cap C)$ is everything in A that is not also in both B and in C.


1

You can't "similarly run backwards". If you try to detail that part, you'll see there is a problem. If $x\notin B\cap C$, it does not mean that $x\notin B\cup C$. For exemple if $x\in B$ but not $C$, you have $x\notin B\cap C$ but you don't have $x\notin B\cup C$.


1

"And not in at least one of B,C. So $ x \in A$ and $x \in $ both A and B except one." That's the heart of the problem. $x\notin (B \cup C) \implies x \notin B $ and $ x \notin C$. x is in neither B nor C. Do you see where you went wrong now?


2

If something is not in $B \cup C$, then it is in neither $B$ nor $C$. Because if it was in $B$, then it is in $B$ or $C$, which is $B \cup C$. So the first mistake is in the second sentence. You could show that $A - (B \cup C) \subseteq A - (B \cap C)$. Because the former one is what is in $A$, but not in $B$ nor $C$. The latter is what is in $A$, but not ...


2

$B \cup C$ is all elements that are in either $B$ or $C$. So $A - (B \cup C)$ are all elements in $A$ but not in either $B$ or $C$, so in $A$ but in neither $B$ nor $C$. $A - (B \cap C)$ would be all elements in $A$ but not in $B$ and $C$. $(A - B) \cap C$ would be all elements that are in both $A$ and $C$ but not in $B$. $(A - B) \cup C$ would be all ...


1

For $x\in\Bbb R$ let $k(x)$ be the greatest integer such that $[2k(x)+1]\pi\le x$. The function $f(x)=(k(x),\tan x)$ is a bijection between $\Bbb R\setminus\{(2k+1)\pi:k\in\Bbb Z\}$ and $\Bbb Z\times \Bbb R$. Now it suffices to show that $\Bbb R\setminus \Bbb N$ and $\Bbb R$ have the same cardinality.


6

You can do it in two ways (at least): You can note that $\Bbb{|R|\leq|R\times Z|\leq|R\times R|}$, and apply some general theorems on cardinals. You can note that $\Bbb R$ and $[0,1)$ have a bijection between them, so you're really looking at $[0,1)\times\Bbb Z$ in terms of cardinality. Now use your imagination to write down a bijection with $\Bbb R$.


2

Context is everything. What is the set $\{x\mid x\neq 0\}$? If you are working in $\Bbb Q$ the set is different from the set you obtain if you work in $\Bbb R$ or in $\Bbb Z$ or in $\Bbb C$. If you are working inside "the mathematical universe", and you can "access" every mathematical object, then this is not a set anymore, since the mathematical universe ...


2

In your post you only proved that $B = \overline A \cup X \implies A\cap B \subseteq X \subseteq B$, although you're missing the parenthesis in the expression $(\overline A \cap \overline X)\cup \overline A \cup X$. After parenthesizing it, your equality still holds: \begin{align} (\overline A \cap \overline X)\cup \overline A \cup X &= ...


1

There seems to be one step missing: ordered pairs (the elements of the cartesian products) can be reduced to sets Following Kuratowski we can define $$ (x,y):=\{\{x\},\{x,y\}\}$$ which may look somewhat arbitrary, but conveys the essetnial notion of ordered pair: $(x,y)=(u,v)\iff x=u\land y=v$.


0

A sequence is a function $f:\mathbb Z^+ \to \mathbf S$. $f:\mathbb Z^+ \to \mathbf S$ is a special case of a relation $f \subset \mathbb Z^+ \times \mathbf S.$ The cartesian product $\mathbb Z^+ \times \mathbf S$ is the set $\{(x,y): x \in \mathbb Z^+$ and $y \in \mathbf S \}$ I'm not sure what you mean by "basic". It seems that you are trying to use the ...


0

Therefore relations makes use of the cartesian product operator, which also makes use of sequences. What.   Ah... Yes, that is a just wee bit recursive, isn't it now? The Cartesian product of sets $\rm A, B$ is defined as: $\mathrm A{\times}\mathrm B = \{\langle a,b\rangle \mid a\in\mathrm A \,\wedge\, b\in\mathrm B\}$ That is, it is the set of ...


5

The set of cats is not a rational number. So $x$ could be the set of cats. That's probably not a real number. In slightly more serious terms, where do these $x$ belong? You need a set to draw these from. Read about dedekind cuts for one approach that derives the reals from subsets of rationals. Cauchy's completeness argument defines them uses sequences of ...


1

If 271 belongs to A ,let B be the empty set.If 271 does not belong to A, let B=X.


1

This is partly a question of quantifiers: You are trying to prove "For all A, there exists a B such that ...". From the definition of $A \Delta B$, you should be able to prove that $B = (A \Delta B) \Delta A$ So for any $A$, you are free to choose an $A \Delta B$ that contains $271$ (or any element, or elements, you choose), and the appropriate $B$ is ...


2

The order of the quantifiers is important. The statement as you have interpreted it would be written $$(\exists B\in P(X))(\forall A\in P(X))(271\in A\Delta B).$$ That is, there exists a fixed set $B$ such that for every set $A$, we have $271\in A\Delta B$. In other words, first you fix $B$, then you have to check it against all $A$. But the statement as ...


0

it's really simple , but first let's simplify it ; let's grant numbers to these sets , elements in fact . Universal set(U):{1,2,3,4,5,6} , A:{1,2} ,B:{1,5,6} ,C:{6,3,4}. As I am not certain if the image that forms your question is the result , primarily expected or the way that they were organized in a way to know its intersections ,let's start! A/(B^C) ...


0

I think the easiest way to prove your claim is by making two observations at the outset: $$ (R\setminus S)\setminus T=(R\cap S^C)\cap T^C\tag{1} $$ and $$ R\setminus(S\setminus T)=R\cap(S\cap T^C)^C=R\cap(S^C\cup T)=(R\cap S^C)\cup(R\cap T)\tag{2}. $$ Now your element-chasing proof is extremely easy: \begin{align} x\in (R\setminus S)\setminus T&\implies ...


0

An element in $(R\smallsetminus S)\smallsetminus T$ is an element of $R$, not in $S$ nor in $T$. An element in $R\smallsetminus (S\smallsetminus T)$ is an element in $R$, not in $S\smallsetminus T$. This means it is not in $S$, except if it is also in $T$. Thus, isn't the inclusion clear?


1

You are only dealing with three sets, so you can easily draw a Venn diagram, similar to this one. Instead of $A, B,$ and $C$ you have $R,S,$ and $T$. All that remains for you is to label the various intersections appropriately. This will help you understand what it going on. In order to actually write the proof, all you must do is show that any element of ...


0

Here is a brass tacks approach. Let $x \in (R - S) - T$. So $x \in R - S$ and $x \not\in T$. Hence, $x \in R$ and $x \not\in S$ and $x \not\in T$. We must show that $x \in R - (S - T)$, so we must show $x \in R$ and $x \not\in S -T$. We know $x \in R$, so we must show $x \not\in S - T$. At this point, you can show it by contradiction, or by a certain ...


3

Hint: Use two steps and show $$(R\setminus S)\setminus T\subseteq R\setminus S\subseteq R\setminus(S\setminus T)$$


0

It's not reflexive. $0\times 0\not\gt 0$. It IS transitive. If $ab \gt 0$ then $a,b$ have the same sign. If $bc\gt 0$ then $b,c$ have the same sign and thus $a,c$ have the same sign and $ac\gt0$.


1

Take $X=A=[0,1]$, and let $\succsim$ be simply $\ge$; clearly identity function on $A$ represents $\succsim$ on $A$. Let $B=\left[0,\frac12\right]$, and let $v:B\to[0,1]:x\mapsto 2x$; $v$ represents $\succsim$ on $B$, but clearly $v$ cannot be extended to a representation of $\succsim$ on $A$.


3

If $p$ is false then $p\rightarrow q$ is true for any statement $q$. This becomes evident by realizing that $p\rightarrow q$ is actually the same as $\neg p\vee q$. So defining $p$ as the (false) statement $x\in E\cap E^c$ and $q$ as the statement $x\in\varnothing$ we arrive at the conclusion that $x\in E\cap E^c\rightarrow x\in\varnothing$ is a true ...


3

You did prove that the implication $x \in E^c \cap E \implies x \in \varnothing$ is true, which means that $E^c\cap E \subseteq \varnothing$, but there's an easier way to prove that $E^c\cap E = \varnothing$: you could try proving it by assuming that there is an element $x$ in $E^C\cap E$ and reaching a contradiction. Using this technique, you could have ...


1

You seem to be missing the lattice where you have $a<c<e$ and $a<b<d<e$ but $c$ is incomparable to $b$ and $d$. I.e. there are two paths from the bottom to the top but the one is of pathlength two and the other is of pathlength three.


0

Let define the set $T$ bit differently, $T=\{m\in \mathbb{N}: ml=nl \ \ \text{implies} \ \ m=n \}$. We shall prove that $T$ is inductive. First we show that $1\in T$. Assume that $m=1$, and $1\cdot l=n\cdot l$. If $n=1$, then $m=n$, and thus $1\in T$. If $n\neq 1$, then $n$ is in the range of function $S:\mathbb{N}\to \mathbb{N}$ (that is the function that ...


1

Hope you've solved it by now, but here is a complete proof for others who may find themselves looking for answers Showing the first inclusion is straight forward def. $f(A) = \{f(x) : x \in A \}$ def. $f^{-1}(B) = \{x : f(x) \in B \}$ $x \in A \subset X$ $f(x) \in f(A)$ $f^{-1}(f(A)) = \{x : f(x) \in f(A) \}$ $x \in f^{-1}(f(A))$ $A \subset ...


0

First show that $\{ 3 \mathbb{Z} \} \cup \{ 3\mathbb{Z} + 1 \} \cup \{ 3\mathbb{Z} +2 \} \subset \mathbb{Z}$ let $x \in \{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\}$. then $x = 3r$ for some $r$ in $\mathbb{Z}$, or $x = 3s+1$ for some $s$ in $\mathbb{Z}$, $x = 3t+2$ for some $t$ in $\mathbb{Z}$ Since in any of these three cases $x$ is ...


1

I find doing proofs this way (especially in analysis) very helpful in understanding the details of what is happening. Here is a complete proof for the forward direction; of course I would shorten in considerably if I were to tun it in as an assignment or just thinking through it for myself. $f: (X, d) \rightarrow (Y, \rho)$ is uniformly continuous $E ...


0

There's more than logical inference to be done in this step. I think that, in effect, you need the following lemma: For all real numbers $r$, if for every $p,q \in E$ we have $d(p,q) \le r$, then $\operatorname{diam}(E) \le r$. The proof of this lemma essentially involves looking at the definition of $\operatorname{diam}(E)$, which is defined as a ...


1

Enthdegree gave a recursive solution in the comments. This can be used to derive the explicit form $$ V\left(\bigcup_iX_i\right)=\sum_iV(X_i)-\sum_iV\left(\left(\bigcup_{j\lt i}X_j\right)\cap X_i\right)\;, $$ of which your equations for $n=2$ and $n=3$ are special cases.


3

In general, if $A\subseteq B$, then $\mathscr P (A)\subseteq \mathscr P (B)$ because every subset of $A$ is a subset of $B$. More formally, if $a\in \mathscr P (A)$, we need to show that $a\in \mathscr P (B)$. But this is trivial, since if $x\in a$, then $x\in B$ which implies that $a\subseteq B$ which is the same as $a\in \mathscr P (B)$. Now take ...


2

$X\subset Y$ implies every element of X is an element of Y, so subsets of X are subsets of Y, so $\mathcal{P}(X)\subset\mathcal{P}(Y)$. Finally, for $Y=\mathcal{P}(X)$ you have $\mathcal{P}(X)\subset\mathcal{P}(\mathcal{P}(X))$.


8

Your result is an immediate consequence of the following proposition. Proposition. Suppose $X\subseteq Y$. Then $\mathscr P(X)\subseteq\mathscr P(Y)$. Proof. Let $E\in\mathscr P(X)$. Then $E\subseteq X\subseteq Y$ so that $E\subseteq Y$. Hence $E\in\mathscr P(Y)$. This proves $\mathscr P(X)\subseteq\mathscr P(Y)$. $\Box$ Do you see how your problem is now ...


4

HINT: Don't work hard, work smart. Recall, or prove, that ${}^C({}^BA)\sim{}^{C\times B}A$. Now use the facts that $\mathcal P(X)\sim{}^X2$ and that $\Bbb{Z^+}\sim\Bbb{Z^+\times Z^+}$.


0

Let $S$ be some set. Then $|2^S| > |S|$, where $2^S$ denotes the power set of $S$, and the set of functions from $S$ to ${0,1}$ has an obvious bijection with the power set of $S$. Basically it is a one-step reduction to cantor's theorem -- the set of functions from $\mathbb{R}$ to $\{0,1\}$ is equal in size to the power set of $\mathbb{R}$. $|2^S| > ...


0

Take $A=\{a\}$, $B=\{b_1, b_2\}$ and $C=\{c\}$ for the sets. And $R=\{(a,b_1),(a,b_2)\}$, $S=\{(b_1,c)\}$ and $T=\{(b_2,c)\}$ for the relations. Then $S \circ R=\{(a,c)\}$, $T \circ R=\{(a,c)\}$ and $S \setminus T=S$. Hence $(S◦R) \setminus (T◦R)= \emptyset$ while $(S \setminus T)◦R=\{(a,c)\}$.


0

Let $R$ be the full relation (that relates every element of $A$ to every element of $B$), and let $S$ and $T$ be two different bijections between $B$ and $C$. Then both $S\circ R$ and $T\circ R$ are the full relation from $A$ to $C$, so their difference is the empty relation, whereas the left-hand side is not the empty relation since $S\neq T$.


0

ajotatxe gave the easiest way to prove it. There's also another way of proving it. Define the relation $\sim$ on $\mathbb{Z}$ by:$$\forall a,b\in\mathbb{Z},\,a\sim b\Leftrightarrow3|a-b$$ $\sim$ is an equivalence relation on $\mathbb{Z}$ and you can easly check that $\{3t\}$, $\{3t+1\}$ and $\{3t+2\}$ are its equivalence classes. So they form a partition of ...


0

Just apply the Euclidean division by $3$. For any integer $n$ there exist some $t\in\Bbb Z$ and some $r\in\{0,1,2\}$ such that $n=3t+r$. Furthermore, the pair $(t,r)$ with these properties is unique.


0

The answers for $d$ and $e$ are incorrect. Remember, $A\setminus \emptyset$ is composed of all elements which are in $A$ and are not in $\setminus$. Likewise, $A\setminus\{\emptyset\}$ contains all elements that are in $A$ and are not in $\{\emptyset\}$.


5

Let $S=\{f:\Bbb R\to\{0,1\}\}$ and let $\alpha:\Bbb R\to S$ be surjective. Define for each $x\in\Bbb R$ $$f(x)=1-\alpha(x)(x)$$ Note that $f\in S$. Since $\alpha$ is surjective, there exists some $y\in\Bbb R$ such that $\alpha(y)=f$. Then $$f(y)=1-\alpha(y)(y)=1-f(y)$$ which is a contradiction. Remark: Note that $\Bbb R$, or even its cardinality, doesn't ...


1

For finite sets, cardinality is just the number of elements in the set. In your case, you have two elements in your set: the element $1$, and the element $\{2,3\}$ (which happens to be a set itself). So you are right, the cardinality is $2$.


1

That set has two elements. One of the elements is $1$ and the other is $\{2,3\}$. Think in a set like it were a box. This box has a ball in which is written "$1$" and a smaller box. No matters what is inside the smaller box, the bigger box has two things: the ball and the box. Remark: Actually, every element of a set is a also set. Although, we "forget" ...



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