I had to prove this:  Let $A$, $B$, and $C$ be sets. Then, $A \times (B \Delta C) = (A \times B) \Delta (A \times C)$.  By the definition of symmetric difference, we know that $$B \Delta C= (B \Delta C) \cup (C \Delta B).$$ So, we can expand the LHS of this theorem to get  $$A \times ( (B \Delta C) \cup (C \Delta B) ).$$ Therefore, \begin{align} A \times (B \Delta C) &amp;= A \times ( (B \Delta C) \cup (C \Delta B) ) \\ &amp;= (A \times (B \Delta C)) \cup (A \times (C \Delta B)) \\ &amp;= ((A \times B) \Delta (A \times C)) \cup ((A \times C) \Delta (A \times B)) \\ &amp;= (A \times B) \Delta (A \times C). \end{align} So, the LHS = RHS.  But I'm not sure how to show that cartesian products are actually distributive over unions and intersections?
14425
I've figured out the following statement is true but I was wondering how you actually go about proving something like this?   $A \times (B \triangle C) = (A \times B) \triangle (A \times C)$
58472
$$ [(\vec{a}+\vec{b})\times(\vec{b}+\vec{c})]\cdot(\vec{c}+\vec{a})=2\vec{c}\cdot(\vec{b}\times\vec{a})$$  I'm supposed to prove that this is true for all vectors $a,b,c$, but I keep getting $4(\vec{a}\times \vec{b})\cdot \vec{c}=0$, which is obviously not true.   Seems to me like there is a mistake in the problem, that it should be $\vec{a}\times\vec{b}$ instead of $\vec{b}\times\vec{a}$. Or I'm doing it wrong?  I'm using facts that $\vec{a}\times\vec{a}=0$ and $[\vec{a},\vec{b},\vec{c}]=[\vec{c},\vec{b},\vec{a}]=-[\vec{b},\vec{a},\vec{c}]$.
5829
How do we calculate the Cartesian product between $n$ polygons? Is it always a polytope? If so, can we say anything about its faces?  For example, $n$ squares: $\{S_1:[a_1, a_2] \times [b_1, b_2], \ldots, S_n:[a_k, a_{k+1}] \times [b_{k}, b_{k+1}]\}$ so, $S_1 \times \ldots \times S_n \in \mathcal{R}^n$ forms a $n-$dimensional box.
6998
I am interested in the following question:  Does $G \times H \cong G$ necessarily imply that $H$ is trivial?  In order to demonstrate it is not, I attempted to construct an example, but I am not entirely sure it is valid:  Let $G = \langle \mathbb{Z}; [a,b] \ (\forall a,b \in \mathbb{Z})\rangle$.  Then $G \times G = \langle \mathbb{Z} \cup \mathbb{Z}; \ \{[a_1, b_1]\} \cup \{ [a_2, b_2]\} \cup \{[a_1,b_2]\}\rangle$, where my notation is intent to convey that the direct product presentation has the abelian relations from the "first" copy of G, the abelian relations from the "second" copy of G, and the required abelian relations which commute generators of the "first" copy with those of the "second" copy.    In this example, since our cardinalities are equal, $|\mathbb{Z} \times \mathbb{Z}| = |\mathbb{Z}|$, and since this implies that commuting all elements of $G$ and commuting all elements of the same cardinality generating set $G \times G$ requires a bijection between the relator sets, then we should have a bijective group homomorphism $\phi : G \rightarrow G \times G$ so that we have isomorphism, but obviously $G$ is not trivial.  Therefore this provides the necessary counterexample to the original statement.  My first question is: is there anything wrong with this proof?  My second question is: if there is nothing wrong with the proof, is there a more standard way of constructing this type proof?  In general, it would seem that whenever you have isomorphic generating sets and isomorphic relator sets, then this should be enough to describe any number of counterexamples to the claim that $H$ must be trivial.  Is this true?  Finally, since direct products can be viewed as the products in the category Ab, and since all final objects must be isomorphic, would there be an easier way (or maybe just an insightful way) of construct this argument entirely in a category theoretic context?  Thanks!
118934
Proving $(A\cup B)\cap(B\cup C)\cap(C\cup A)=(A\cap B)\cup (A\cap C)\cup (B\cap C)$?   I managed to prove one side by saying:  Let E=LHS and F=RHS     Let $ x \in E$. Then $x \in (A\cup B) \lor \ x \in (B\cup C) \lor \ x \in (C\cup A).$ If $x\notin A$, then $x$ must be in $B$ and $C$. If $x\notin B$, then $x$ must be in $A$ and $C$. If $x \notin C$, then $x$ is in $B$ and $A$. Therefore $x \in F$ and $ E \subset F $.   I'm having trouble proving the other direction. Can I prove it in a similar way saying that let $x \in F$, if $x \notin A$, then $x$ would be in $B$ and $C$? That doesn't make sense to me. Another way I thought of solving it would be:  Let $x \in F. $ Then   if $x \in A$, then  $x \in B \cup  C$, If $x \in B$, then $x \in A \cup C$.  if $x \in C$, then $x \in B \cup A$   And then is this sufficient to show that $ x \in F, thus \ F \subset E \ and \ E=F $?
63269
I have some trouble trying to understand Halmos' explanations about generalized Cartesian products. I've read these entries already:    This also has the page I'm stuck on.      Is his generalization the same as a Cartesian product of several sets? As here:   What puzzles me is $\{a,b\}$. Where does it come from, what does it do? I know that it is supposed to be arbitrary. How does it identify distinct elements of $X \times Y$? Does some function map each $a \in A$ to an element in $X$? Or is it always the same $a$?  What does he mean by      the set $Z$ of all families $z$, indexed by $\{a,b\}$, such that $z_a \in X$  and $z_b \in Y$   ? What does does an element of $Z$, namely one family $z$, consist of? By family, does he mean an indexing function or the indexed elements? This comes to a very important question for me: Does family mean a function or the set of all indexed elements?   I will try to explain how I understand it thus far: $Z$ contains elements of the form $\{(a,x),(b,y)\}$. The choice of $x$ depends on $a$, so an $x$ in a $z \in Z$ equals the $z_a$ of $X$. Now, we want to map these pairs to $X \times Y$, so we have a function $f$ that takes $z \in Z$ and connects it to the corresponding ordered pair $(z_a, z_b)$, where $z_a \in X$. $X \times Y$ is not the same as $Z$; they only, effectively, contain equal elements.  He goes on:     If $\{X_i\}$ is a family of sets ($i \in I$), the Cartesian product of the family is, by definition, the set of all families $\{x_i\}$ with $x_i \in X_i$ for each $i$ in $I$.   How I understand it: $\{X_i\}$ contains, as elements, several sets, each single one a set $X$ identified by an index/element of $I$. Then, we go over all indeces $i$ in $I$, take from each set $X$, namely $X_i$, an element $x$ and add it to current set of the newly created sets/families. So, the lower index $i$ does not refer to an element of $X_i$ indexed by $i$, but rather calls all elements of $X_i$ $x_i$. We end up with a set, defined as Cartesian product, of sets, in which each element (set) contains ordered elements, each from a different $X_i$. How does the ordering occur?  $\prod_i X_i$, the Cartesian product, becomes equal to $X^I$, if all $X_i$ are the same set $X$. $X^I$ means all possible combinations of mapping from $I$ to $X$. We randomly take an elements $x$ from some $X_i$, put it in a set/family with other randomly obtained $x$, put the entire set into a set denoted $X^I$, and continue doing so, till we have no more options. Then, each set in the new $X^I$ contains a mapping of some $i$ to some $x$. Since we exhausted the options, they are indeed all combinations.  Where did I go wrong? Someone needs to spoonfeed my the answer; I am completely confused by now.
95642
Given a set $\mathcal{S}$ of size $n$, let a $k$-partition of $\mathcal{S}$ be a grouping into $k$ disjoint classes $$(S_1 ,S_2,...,S_k)$$   Where the $S_i$ do not necessarily contain the same number of elements.  Let $C(S_1, S_2, ... S_k)$ be the set of combinations which can be made by taking exactly one element of each class, (i.e. $C(S_1, S_2, ... S_k)$ is the Cartesian Product $S_1 \times S_2 \times ...\times S_k$). For a given $(n,k)$, what is the minimum number of k-partitions $N$ such that  $\bigcup\limits_{i=1}^{N} C(S_1^{(i)}, S_2^{(i)}, ...,S_3^{(i)}) = $ All Possible Combinations  Example:  for $n=4$ and $k=2$, $\mathcal{S} = \{1,2,3,4\}$ we can choose:  $$ S^{(1)} = \left\{\left\{ 1, 2\right\}, \left\{ 3, 4\right\}\right\}$$ $$ S^{(2)} = \left\{\left\{ 1, 3\right\}, \left\{ 2, 4\right\}\right\}$$  Then $$C_{1} = \left\{\left\{ 1, 3\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 2, 4\right\}\right\}$$ $$C_{2} = \left\{\left\{ 1, 2\right\} ,\left\{ 1, 4\right\},\left\{ 2, 3\right\},\left\{ 3, 4\right\}\right\}$$  and $C_1 \cup C_2 =$ All possible combinations and $N=2$.
24150
Question:  Show that if $ \ A \cup B = A$ and $ \ A\cap  B = A$ then $ \ A = B$  My attempt:  Proof by contradiction:  Assume $ \ A \cup B = A$ and $ \ A\cap  B = A$ and $ \ A \neq B$   Case 1: $ \exists \ x \in A, x\notin B$  If $ x \in A \implies x\in A \cap B \implies x \in A \ and \  x\in B  \implies x\in B$, a contradiction.   Case 2: $ \exists \ x \in B, x\notin A$  If $ \ x \in B \implies x\in A \cup B \implies x\in A$, since $ \ A \cup B = A$. Contradiction.   Is this approach correct? Could someone please show me how to do a direct proof?
108420
$A, B, C, D$ are sets, where $A \not = \emptyset$ and $B \not = \emptyset$. Equation $(A \times B) \cup (B \times A) = C \times D$ is true.  Prove that $A = B = C = D$.  I tried to find $A, B, C, D$ separately, but that didn't get me anywhere. Any ideas how to solve it?
37976
(1) If $X$ and $Y$ are two sets, we define the Cartesian product $X \times Y$ as the set of ordered pairs $(x,y)$, such that $x \in X$ and $y \in Y$.  (2) On the other hand [Folland, Real Analysis, page 4], if $\{X_\alpha\}_{\alpha \in A}$ is infinite indexed family of sets, their Cartesian product $$ \prod_{\alpha \in A}X_\alpha $$ is defined as the set of maps $f: A \to \bigcup\limits_{\alpha \in A} X_\alpha$ such that $f(\alpha) \in X_\alpha$ for every $\alpha \in A$.  After saying this, Folland remarks:     it should be noted, and promptly forgotten, that when $A = \{1,2\}$, the previous definition  of $X_1 \times X_2$ [that's (1) above] is set-theoretically different from the present definition of $\prod_1^2 X_j$ [that's (2) above]. Indeed, the latter concept depends on the mappings, which are defined in terms of the former one.   I am not grasping this remark. Specifically, here are my questions.  Question 1: How is (2) set-theoretically different from (1)? A simple illustrative example?  Question 2: If (1) is extended to infinite families, which definition would be stronger? A simple illustrative example?  Question 3: Why should this be "promptly forgotten"?  I'll probably have more questions depending on the type of answers I'll get to these.  Thanks!
123016
Is it always true that  $(A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)$?  I don't believe this is true. I have tried to draw pictures to help me get on the right path, but I think that the union makes this untrue. for example, if $a \in A_1$ and $b \in B_2$, then $(a,b)$ would not be in $(A_1\times B_1) \cup (A_2 \times B_2)$. Is this a correct assumption?
26847
Given two sets $A$ and $\varnothing$, where $\varnothing$ is the empty set. The Cartesian product is defined by  $$A \times \varnothing = \{ (x,y) \mid x \in A, y \in \varnothing \}$$  Why is $ A \times \varnothing = \varnothing$?
45250
How can I proof the following statement?     Any $n \times n$ matrix $A$ can be written as a sum   $$   A = B + C $$   where $B$ is symmetric and $C$ is skew-symmetric.   I tried to work out the properties of a matrix to be symmetric or skew-symmetric, but I could not prove this. Does someone know a way to prove it?  Thank you.  PS: The question  may be similiar, in fact gives a hint to a solution, but if someone does not mind in expose another way, our a track to reach to what is mentioned in the question of the aforementioned link.
100292
Let $a,b$ and $c$ be vectors in $\mathbb{R}^3$. How do I show that    $$\|a-b\| \le \|a-c\|+\|c-b\|$$  and   $$\|a \times b\|^2=\|a\|^2\|b\|^2-(a\cdot b)^2$$ ?
63087
Suppose $A,B,C$ are sets.  Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$.  I proved that for $x\in A\setminus B$, $x\in A\cup C$ and for $x\in B\setminus A$, $x\in B\cup C$. How do I show that $A\cup C=B\cup C$, and that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$?
87286
Just reading about the cross product. It says here: $$A\times B= \{(x,y)|x\in A, y\in B\}$$ Does this mean that the result of $A\times B$ is a new set of ordered pairs? If so, what happens if both A and B share a common member? What would the result be, since an ordered pair cannot include duplicates?    Edit: Quoting the free textbook I'm looking at online,     If we are given that A is a set and no other information about A, then there is no   ordering to the elements of A. Thus, we cannot speak of “the second element of the set A”   unless we have specified an ordering of the elements of A. If we wish to regard A as ordered in some way, then we specify this fact explicitly: “The elements of A are ordered a, b, c,” or “A = (a, b, c).” The latter notation replaces the braces with parentheses and designates that A is ordered, left to right, as indicated. We call this an ordered set. An ordered set is also called a linear order. Various other names are also used: list, vector, string, word — all with no repeated elements. Of course, you’ve seen repeated elements in vectors, for example the point in the plane at the coordinates (1,1). That’s fine, it’s just not an ordered set.       Why is it okay to specify a set S = {a, b, c, a} where the element a has been repeated,   but it is not okay to have repeated elements in an ordering of S?   When we say S  =   {a, b, c, a}, we know that S contains just the three elements a, b and c.  If we were to talk   about the ordered set (a, b, c, a) it would not make sense because it would say that the   element a is in two places at once: the first position and the last position.
8158
How to prove $( A \times B ) \cap ( C \times D ) = ( A \cap C ) \times ( B \cap D )$ ?