New answers tagged

0

$A$$∪$($A$$∩$$B$)=$(A∪A)∩(A∪B)$=$A∩(A∪B)$=$A$, the first equality using the distribution law and the last equality since $A\subset A∪B$.


2

Assume that $A$ and $B$ are subsets of some universal set $X$. Then $$\begin{align*} A\cup(A\cap B)&=(\color{red}A\cap X)\cup(\color{red}A\cap B)\\ &=\color{red}A\cap(X\cup B)\\ &=A\cap X\\ &=A\;. \end{align*}$$


1

HINT You want to reflect lines $ x= \pm 0.5, y= \pm 0,5 $ about $ y = m x , m= \tan \theta .$ Instead of that consider four among the eight rotated lines $$ \pm x \cos \theta \pm y \sin \theta = 0.5 ; \, \pm x \sin \theta \pm y \cos\theta = 0.5 $$


1

Let $ABCD$ be your square. Let's call $G$ and $H$ the points where $l$ intersects, respectively, the sides $AB$ and $CD$. We have the two trapezia $GBCH$ and $AGHD$ that both have the same area (half of the original square, so $1/2$). We now want to reflect $AGDH$ around $l$, let's call $L$ and $K$ the images of $A$ and $B$ after the reflection. The new ...


1

You could take an arbitrary x that belongs to one side, and show that it belongs to the other side. let $x \in A$ 2 possibilites: $x \in B$ then $x \in A\cap B$. Then $ x \in (A \cap B) \cup (A-B) $ $x \notin B$ then $x \notin A \cap B$ but $ x \in (A-B)$ Then $ x \in (A \cap B) \cup (A-B) $ Then $ A \subset x \in (A \cap B) \cap (A-B) $ let $ x \in x ...


0

$A=A\cap (B\cup\bar{B})=(A\cap B)\cup (A\cap\bar{B})=(A\cap B)\cup(A\setminus B)$


2

Let $a \in \bigcup X$. This means that $a \in x$ for some $x \in X$. Since $x$ is transitive, $a \subseteq x$. Since $x \subseteq \bigcup X$, we have $a \subseteq \bigcup X$. So, $\bigcup X$ is transitive. Let $a \in \bigcap X$. This means that $a \in x$ for all $x \in X$. Since every such $x$ is transitive, $a \subseteq x$. So, $\forall x \in X\ : a ...


1

The following example has always been very enlightening to me as it shows how important the underlying topology is. Consider $\mathbb R$ with the standard topology and $[0,1)$ with the subspace topology (induced by the inclusion) of $\mathbb R$. Have a look at $$ i: [0,1) \to \mathbb R$$ the inclusion map. $[0,1) \subset \mathbb R$ is not open as $1$ is ...


2

Sure: let $\mathbf{Y}$ be any space where $y_0\in \mathbf{Y}$ is not open, and consider the constant function $f\colon x\mapsto y_0\colon \mathbf{X}\to\mathbf{Y}$; then $f^{-1}(\{y_0\}) = \mathbf{X}$.


2

Consider any set $X$ with $|X| \geq 2$. Then if $\mathcal{T}$ is the indiscrete topology on $X$ we have: $id_X: (X,\mathcal{P}(X)) \rightarrow (X,\mathcal{T})$ is continuous and the pre-image of every subset is open in $(X,\mathcal{P}(X))$. I'm not sure what is meant by 'more than one pre-image value'? If $f:X\rightarrow Y$ is a function, then for any ...


2

If $A=\bigcup_{a\in D}F_a$, then $A^c=\bigcup_{a\in \{0,1\}^n\setminus D}F_a$.


1

This isn't true: Let $\Omega = \{0,1\}$, $\Lambda = \{2,3\}$, $A(2) = \{0\}$, $A(3) = \{1\}$, $B(2)=B(3) = \{0,1\}$. Then $\bigcup_{\lambda \in \Lambda} A(\lambda) \times B(\lambda) = \{0\} \times \{0,1\} \cup \{1\} \times \{0,1\} = \{0,1\} \times \{0,1\}$. So $\{0,1\} \times \{0,1\} \subseteq \bigcup_{\lambda \in \Lambda} A(\lambda) \times B(\lambda)$, but ...


0

Example of subsets can be chosen from the smaller sets $\{1,2,\ldots 30\}$. Let $A$ be the set of dates in a calendar month when Alice is going to have pizza for dinner, and let $B$ be the set of dates Bob is going to have pizza. Now $A\subset B$ means, on all the dates Bob had pizza Alice had it too. Now you can draw up a pizza eating schedule for both of ...


0

Yes. For instance, $A=n\mathbb{Z}$ and $B=m\mathbb{Z}$, with $gcd(m,n)=1.$


1

Any two subsets with a non-empty symmetric difference ($A \setminus B) \cup (B \setminus A$) satisfy your requirements.


5

Yes. For instance $$ A = \{1\}, B = \{2\}\\ A = \text{All the negative integers},B = \text{All the positive integers}\\ A = \{2n \mid n \in \Bbb Z\}, B = \{2n+1\mid n \in \Bbb Z\}\\ A = \Bbb Z \setminus \{1\}, B = \Bbb Z \setminus \{2\}\\ A = \{5n \mid n \in \Bbb Z\}, B = \{3n \mid n \in \Bbb Z\} $$


1

First, here is my direct answer to your question: this proof is structurally fine, but its wording can be improved on two points. First, $\;\forall \lambda a \in \lambda A\;$ is formally incorrect and also confusing: you just mean $\;\forall a \in A\;$. Second, "If the above holds $\;\forall \ldots\;$" should be "Since the latter inequality holds ...


2

You can definitely have an ordered pair (or ordered tuple) with the same element, and you should definitely write that element twice (or more). Tuples are not sets in the sense that order and repetition matter. So you can't just ignore these like you do with sets.


1

I think the fact that, in the first sentence, you refer to $(a,a)$ as an 'ordered set' is where the confusion lies. In the context of cartesian products just use 'ordered pairs', then since $(a,a)$ is not actually a set, the repeated elements rule does not apply and so you would keep it as $(a,a)$


1

Uniqueness part (in words) For all A1,A2 in P(U), if { [for all B(B in P(U) implies A1 union B = A1] and [for all B(B in P(U) implies A2 union B = A2] } then A1 = A2. Now to symbolize this is direct.


2

Sorry, I accidentally deleted my response. It's not possible for the union to be finite and the collection infinitie. The point here is that, we can note that if $U=\cup C$ then $C\subseteq P(U)$. The power set of S is finite if and only if S is finite..


1

Hint: What if $A=\{a\}$, $B=\{b\}$, and $C=\emptyset$?


2

As pointed out by @RobertIsrael in the comments, in this particular problem, it's easier to do the induction on $m$ because it's easy to assign an extra element in $E$ to an element in $F$ while it's not that straightforward the other way around. I must admit though, this problem can be easily solved without the help of induction. Note that you can assign ...


0

Lemma: Denote the cardinality of a set,$A$, by $|A|$. Suppose $A$ and $B$ are sets with $|A| = 2$ and $|B|=1$, then $A\not = B$. Proof: Let $A=\{a_1,a_2\}$ with $a_1 \not = a_2$, and $B=\{b_1\}$, then if $b_1\not \in A$ we have $A\not = B$ and we are done, so assume that $b_1\in A$. Without loss of generality we can say $b_1=a_1$. But in this case we ...


0

$A\cup B\cup C$ can be partitioned into three different sets. Namely, a set $X$ such that $x\in X \implies x$ is an element of only one of $A$, $B$, or $C$. A set $Y$ such that $y\in Y \implies y$ is an element of exactly two of $A$, $B$, and $C$. A set $Z$ such that $z\in Z \implies z$ is an element of $A$, $B$, and $C$. We know that $|Y|=6$. The rest ...


0

HINT Note that $|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|B \cap C|-|C \cap A|+|A \cap B \cap C|$. If $f(A)$ denoted the elements of $A$ that were only in $A$ note that $|f(A)|=|A|-|A \cup B|-|A \cup C|+|A \cup B \cup C|$. Try to use $|A \cap B \cap C|=0$. This implies that $|f(A)|+|f(B)|+|f(C)|=|A\triangle B \triangle C|=|A|+|B|+|C|-2|A \cap B|-2|B ...


1

To understand lattices first you need to understand partially ordered sets. A partially ordered set is a set with an ordering operation $ \leq $ that sometimes works. For example the set of all people ordered by ancestry is a partially ordered set (poset). If your mother is your ancestor and your mother's mother is your ancestor then your grandmother is ...


1

Consider a partially ordered set $(A,\leq)$, and a non-empty subset $B$. A lower bound is an element $l$ such that for every $b \in B, l \leq b$. A greatest lower bound is an element $l_0$ such that for every lower bound $l$, $l \leq l_0$. An example would be $A = \mathbb{R}$ with the standard order, $B$ the subset $(0,1) \cup (2,3)$. Any element less than ...


1

A lattice is an algebraic structure, generalizing each of the following pairs of (binary) operations: $$\min,\ \max$$ $$\inf,\ \sup$$ $$\bigcap,\ \bigcup$$ $$\mathtt{and},\ \mathtt{or}$$ $$\gcd,\ \mathrm{lcm}$$ A partially ordered set can be naturally equipped with an algebraic lattice structure whenever every pair of elements $a,b$ has a greatest lower ...


-1

I am sure that in the context of set theory a lattice is the concept from order theory. It has a load of interesting properties, being simultaniously a type of order and an algebraic structure. The set of all subsets of some set with union as join and intersection as meet form a lattice (in fact, it is even a complete lattice).


1

So you are correct in stating that $(A\cap B^c)$ and $(A^c\cap B)$ are disjoint. This is obvious because $(A\cap B^c)\cap(A^c\cap B)=\{0\}$ By the properties of probability, $P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)$. So, $P((A\cap B^c)\cup (A^c\cap B))=P(A\cap B^c)+P(A^c\cap B)-P((A\cap B^c)\cap(A^c\cap B))=P(A\cap B^c)+P(A^c\cap B)$ Logically, $P(X\cap ...


0

Notice that $(A\cap B^c)\cup (A^c\cap B) \;\equiv\; (A\cup B)\setminus (A\cap B)$ Since $(A\cap B)\subseteq (A\cup B)$, then since probability is a $\sigma$-algebraic measure. $$\mathsf P((A\cup B)\setminus (A\cap B)) = \mathsf P(A\cup B)-\mathsf P(A\cap B)$$ The rest is just $(A\cup B) \equiv A \cup (B\setminus(A\cap B))$ so...


0

To clarify the grouping, the set in question is $$A \cap \left[(A \cup B \cup C) \setminus A\right].$$ Consider $(A \cup B \cup C) \setminus A$. In words, from $(A \cup B \cup C)$, we remove every element in $A$. Therefore, no element of $A$ is in $(A \cup B \cup C) \setminus A$. Therefore, $A$ and $(A \cup B \cup C) \setminus A$ have no elements in common. ...


0

If $x\in A\cap(A\cup B\cup C)\backslash A$, then $x\in A$ (because of the "$A\cap$") and $x\not\in A$ (because of the "$\backslash A$"). There can be no such $x$, so $A\cap(A\cup B\cup C)\backslash A=\emptyset$.


0

So we must prove: $$P((A \cup B^c) \cap (A^c \cup B))=P(A) + P(B) -2P(A \cap B^c).$$ Let us try writing the set whose probability we are measuring on the LHS in simpler terms: \begin{align*} (A\cup B^C)\cap(A^C\cup B)={}&(A\cap A^C)\cup(A\cap B)\cup(B^C\cap A^C)\cup(B^C\cap B)={} \\ {}={}&\varnothing\cup(A\cap B)\cup(A\cup B)^C\cup\varnothing. ...


0

Hint: First we have $$ (A \cap B^c) \cup (A^c \cap B)=A\bigtriangleup B= A\cup B-A\cap B $$ where $A\bigtriangleup B$ is known as symmetric relation. Also $$ A\cap B\subset A\cup B \quad\text{and }\quad P(A\cup B)=P(A)+P(B)-P(A\cap B) $$ Thus $$ P(A\cup B-A\cap B)=P(A\cup B)-P(A\cap B)=P(A)+P(B)-2P(A\cap B) $$


0

For the second step, you don't want to plug the decompositions of $A$ and $B$ directly into your equation. Instead, you want to take the equation $P(A)=P(A\cap B^c)+P(A\cap B)$ (which you get directly from $A=(A\cap B^c)\cup (A\cap B)$ since the two sets are disjoint) and rearrange it to get $P(A\cap B^c)=P(A)-P(A\cap B)$, and then plug this expression for ...


1

$C=\{(0,0)\}$ will work. If it suits your needs to keep $C=\{0\}$, then let $D:= A\times B \cup C\times C$, which you can write $D:= (A\times B) \cup (C\times C)$ if the former looks ambiguous to you.


1

I'll denote $\overline{A}=A^c$. By De Morgan's Law: $$\overline{\overline{B}\cap \overline{B\cap A}}=\overline{\overline{B}}\cup \overline{\overline{B\cap A}}$$ By the double complement or Involution law: $$=B\cup(B\cap A)$$ By the Absorption Law it equals $B$.


2

By De Morgan's Law: $(B^c \cap (B \cap A)^c)^c = B \cup (B \cap A) = B$


0

Just for visual representation, the shaded region is the set you are looking for:


1

$B = \{ x \mid 4 \leq x \leq 7\}$ and $C^c = \{x \mid x < 2 \text{ or } x \geq 6\}$, so $$B \cap C^c = \{x \mid 4 \leq x \leq 7 \text{ and } ( x < 2 \text{ or } x \geq 6)\} = \{6,7\}.$$


0

No, $B \cap C^{c} \neq \mathbb{Z}$. Write out the definition of $C^c$ and look for the overlap between that and $B$. (Also, $C$ isn't a subset of $B$; $2 \in C$, but $2 \not \in B$.)


5

Remember that the difference between a partial order and a total order is that in a partial order, you can have incomparable elements. Can you think of sets $A$ and $B$ such that $A\not\subseteq B$ and $B\not\subseteq A$? To clarify: such an example will merely show that $(\mathcal{P}(\mathbb{Z}), \subseteq)$ is not a total order. You still need to show ...


1

Hint : consider $A_m = \{0,1,…,m\}$, $A_m' = \{0\}$. However, if all the $A_m$ are equal, then it is true : it is sufficient to prove that $$A^{\mathbb N\setminus\{1,...,n\}} = A^{\mathbb N}$$ To do this, pick $X=(x_{n+1},x_{n+2},…) \in A^{\mathbb N\setminus\{1,...,n\}}$, that means $x_j \in A_j := A_0$ for all $j ≥ n+1$. Then, you want to prove that ...


0

In the answers to the question, $\alpha$ is an ordinal smaller than $\kappa$, not a cardinal. Regardless to the continuum hypothesis you are correct that the only cardinals smaller than $\omega_1$ are finite and $\omega$. But as an ordinal, $\omega_1$ is the increasing union of countable ordinals.


2

You prove the first statement is false with a counter-example. Consider the diagonal line in $\Bbb R\times\Bbb R$. It is not of the form $A'\times B'$ because every point is of the form $(x,x)$. The second is also false. Consider $A=[0,1]$, $B=[1,2]$ and $C=B$ and $D=A$. Then the LHS is a union of two squares of side length $1$ while the RHS is a square ...


1

For the first, let $A=\{1,2\}, B=\{a,b\}, X=\{(1,a),(2,b)\}$ For the second, note that the pairs on the left have a first element that might come from $B$ while the pairs on the right have a first element that might come from $C$. Maybe, as in my first example, they are different things.


0

I don't think such a function exists. You can construct an example where all the inputs are non-zero but the function gives back 0: $$A=\{1,2,3,4,5\}, B=\{1,6\},C=\{5,6\} \implies f(_\cdots)=0$$ You can easily construct an example where you have the same cardinalities but the intersection is non trivial and hence $f$ doesn't define a function.


3

No such function exists. To see that, suppose you had such an $f$. Let's compute a few examples. Case I: $$A=\{1,2,3\},\,B=\{3,4,5\},C=\{1,4,6\}$$ Then it is easy to compute everything in your expression. We get $$0=f(3,3,3,1,1,1)$$ Case II: $$A=\{1,2,3\},\,B=\{1,4,5\},C=\{1,6,7\}$$ Then we get $$1=f(3,3,3,1,1,1)$$



Top 50 recent answers are included